answer：First, I need to determine the total weight of the 27 carrots that remain on the scale. Given that their average weight is 200 grams, I can calculate the total weight by multiplying the number of carrots by the average weight.Next, I should convert the initial total weight of the 30 carrots from kilograms to grams to ensure consistent units for accurate calculations.After obtaining the total weight of the 27 carrots, I can find the total weight of the 3 removed carrots by subtracting the weight of the remaining carrots from the initial total weight.Finally, to find the average weight of the 3 removed carrots, I will divide the total weight of the removed carrots by the number of carrots removed.

answer：Alright, so I have this problem where I need to find all real polynomials ( p(x) ) such that ( p(x^2) = p(x)p(x-1) ) for all ( x ). Hmm, okay, let me try to break this down step by step.First, I know that polynomials are expressions of the form ( p(x) = a_nx^n + a_{n-1}x^{n-1} + dots + a_1x + a_0 ), where the coefficients ( a_i ) are real numbers. The equation given is a functional equation, meaning it relates the polynomial evaluated at different points.So, the equation is ( p(x^2) = p(x)p(x-1) ). That means if I plug ( x^2 ) into the polynomial ( p ), it should be equal to the product of ( p(x) ) and ( p(x-1) ). Interesting. I need to find all such polynomials.Maybe I should start by considering the degree of the polynomial. Let's say ( p(x) ) is a polynomial of degree ( n ). Then, ( p(x^2) ) would be a polynomial of degree ( 2n ) because substituting ( x^2 ) into a degree ( n ) polynomial gives degree ( 2n ).On the other hand, ( p(x)p(x-1) ) is the product of two polynomials, each of degree ( n ). So, the product would be a polynomial of degree ( 2n ). So, at least the degrees on both sides match, which is a good consistency check.But that doesn't tell me much about the specific form of ( p(x) ). Maybe I should consider specific cases, like low-degree polynomials, to see if I can spot a pattern or find some constraints.Let's start with constant polynomials. If ( p(x) = c ) where ( c ) is a constant, then the equation becomes ( c = c cdot c ), which simplifies to ( c = c^2 ). The solutions to this are ( c = 0 ) or ( c = 1 ). So, constant polynomials ( p(x) = 0 ) and ( p(x) = 1 ) satisfy the equation. Okay, that's straightforward.Next, let's consider linear polynomials. Suppose ( p(x) = ax + b ), where ( a ) and ( b ) are real numbers, and ( a neq 0 ) since it's linear. Then, let's compute both sides of the equation.First, ( p(x^2) = a(x^2) + b = ax^2 + b ).On the other hand, ( p(x)p(x-1) = (ax + b)(a(x-1) + b) ). Let's expand this:( (ax + b)(a(x - 1) + b) = (ax + b)(ax - a + b) ).Multiplying these out:( ax cdot ax = a^2x^2 ),( ax cdot (-a + b) = a(-a + b)x ),( b cdot ax = abx ),( b cdot (-a + b) = -ab + b^2 ).So, combining like terms:( a^2x^2 + [a(-a + b) + ab]x + (-ab + b^2) ).Simplify the coefficients:For the ( x ) term: ( a(-a + b) + ab = -a^2 + ab + ab = -a^2 + 2ab ).For the constant term: ( -ab + b^2 ).So, ( p(x)p(x-1) = a^2x^2 + (-a^2 + 2ab)x + (-ab + b^2) ).Now, set this equal to ( p(x^2) = ax^2 + b ):( a^2x^2 + (-a^2 + 2ab)x + (-ab + b^2) = ax^2 + b ).Now, equate the coefficients of corresponding powers of ( x ):1. For ( x^2 ): ( a^2 = a ). So, ( a^2 - a = 0 ) which implies ( a(a - 1) = 0 ). Thus, ( a = 0 ) or ( a = 1 ).But since we're considering linear polynomials, ( a neq 0 ). So, ( a = 1 ).2. For ( x ): ( -a^2 + 2ab = 0 ). Since ( a = 1 ), this becomes ( -1 + 2b = 0 ), so ( 2b = 1 ), hence ( b = frac{1}{2} ).3. For the constant term: ( -ab + b^2 = b ). Substituting ( a = 1 ) and ( b = frac{1}{2} ):( -1 cdot frac{1}{2} + left(frac{1}{2}right)^2 = -frac{1}{2} + frac{1}{4} = -frac{1}{4} ).But the right-hand side is ( b = frac{1}{2} ). So, ( -frac{1}{4} = frac{1}{2} ), which is not true.Hmm, that's a contradiction. So, even though ( a = 1 ) and ( b = frac{1}{2} ) satisfy the coefficients for ( x^2 ) and ( x ), they don't satisfy the constant term. Therefore, there are no linear polynomials that satisfy the given equation.Okay, so constant polynomials ( p(x) = 0 ) and ( p(x) = 1 ) work, but no linear polynomials. Let's move on to quadratic polynomials.Let ( p(x) = ax^2 + bx + c ), where ( a neq 0 ). Then, compute both sides.First, ( p(x^2) = a(x^2)^2 + b(x^2) + c = ax^4 + bx^2 + c ).Next, ( p(x)p(x-1) ). Let's compute ( p(x-1) ):( p(x-1) = a(x-1)^2 + b(x-1) + c = a(x^2 - 2x + 1) + b(x - 1) + c = ax^2 - 2ax + a + bx - b + c ).Simplify:( ax^2 + (-2a + b)x + (a - b + c) ).Now, multiply ( p(x) ) and ( p(x-1) ):( (ax^2 + bx + c)(ax^2 + (-2a + b)x + (a - b + c)) ).This will be a bit tedious, but let's go through it term by term.First, multiply ( ax^2 ) with each term in ( p(x-1) ):( ax^2 cdot ax^2 = a^2x^4 ),( ax^2 cdot (-2a + b)x = a(-2a + b)x^3 ),( ax^2 cdot (a - b + c) = a(a - b + c)x^2 ).Next, multiply ( bx ) with each term in ( p(x-1) ):( bx cdot ax^2 = abx^3 ),( bx cdot (-2a + b)x = b(-2a + b)x^2 ),( bx cdot (a - b + c) = b(a - b + c)x ).Finally, multiply ( c ) with each term in ( p(x-1) ):( c cdot ax^2 = acx^2 ),( c cdot (-2a + b)x = c(-2a + b)x ),( c cdot (a - b + c) = c(a - b + c) ).Now, let's collect like terms:- ( x^4 ): ( a^2x^4 ).- ( x^3 ): ( a(-2a + b)x^3 + abx^3 = [ -2a^2 + ab + ab ]x^3 = (-2a^2 + 2ab)x^3 ).- ( x^2 ): ( a(a - b + c)x^2 + b(-2a + b)x^2 + acx^2 ). Let's compute each coefficient: ( a(a - b + c) = a^2 - ab + ac ), ( b(-2a + b) = -2ab + b^2 ), ( ac ). So, total coefficient for ( x^2 ): ( a^2 - ab + ac - 2ab + b^2 + ac = a^2 - 3ab + 2ac + b^2 ).- ( x ): ( b(a - b + c)x + c(-2a + b)x ). Compute coefficients: ( b(a - b + c) = ab - b^2 + bc ), ( c(-2a + b) = -2ac + bc ). So, total coefficient for ( x ): ( ab - b^2 + bc - 2ac + bc = ab - b^2 - 2ac + 2bc ).- Constant term: ( c(a - b + c) = ac - bc + c^2 ).So, putting it all together, ( p(x)p(x-1) ) is:( a^2x^4 + (-2a^2 + 2ab)x^3 + (a^2 - 3ab + 2ac + b^2)x^2 + (ab - b^2 - 2ac + 2bc)x + (ac - bc + c^2) ).Now, set this equal to ( p(x^2) = ax^4 + bx^2 + c ).So, equate coefficients:1. For ( x^4 ): ( a^2 = a ). So, ( a^2 - a = 0 ), which gives ( a(a - 1) = 0 ). Thus, ( a = 0 ) or ( a = 1 ). But since it's a quadratic polynomial, ( a neq 0 ). So, ( a = 1 ).2. For ( x^3 ): ( -2a^2 + 2ab = 0 ). Substituting ( a = 1 ), we get ( -2 + 2b = 0 ), so ( 2b = 2 ), hence ( b = 1 ).3. For ( x^2 ): ( a^2 - 3ab + 2ac + b^2 = b ). Substituting ( a = 1 ) and ( b = 1 ):( 1 - 3(1)(1) + 2(1)c + (1)^2 = 1 - 3 + 2c + 1 = (-1) + 2c = b ).But ( b = 1 ), so:( -1 + 2c = 1 ) => ( 2c = 2 ) => ( c = 1 ).4. For ( x ): ( ab - b^2 - 2ac + 2bc = 0 ). Substituting ( a = 1 ), ( b = 1 ), ( c = 1 ):( (1)(1) - (1)^2 - 2(1)(1) + 2(1)(1) = 1 - 1 - 2 + 2 = 0 ). Okay, that works.5. For the constant term: ( ac - bc + c^2 = c ). Substituting ( a = 1 ), ( b = 1 ), ( c = 1 ):( (1)(1) - (1)(1) + (1)^2 = 1 - 1 + 1 = 1 ). And ( c = 1 ), so that's consistent.So, all coefficients match when ( a = 1 ), ( b = 1 ), and ( c = 1 ). Therefore, the quadratic polynomial ( p(x) = x^2 + x + 1 ) satisfies the equation.Alright, so we have constant polynomials ( p(x) = 0 ) and ( p(x) = 1 ), and the quadratic polynomial ( p(x) = x^2 + x + 1 ). Maybe higher-degree polynomials can be built from these?Wait, let's think about the functional equation again: ( p(x^2) = p(x)p(x-1) ). If ( p(x) ) is a product of such polynomials, does the equation hold? For example, if ( p(x) = (x^2 + x + 1)^n ), would that satisfy the equation?Let me test this. Suppose ( p(x) = (x^2 + x + 1)^n ). Then,( p(x^2) = (x^4 + x^2 + 1)^n ).On the other hand,( p(x)p(x-1) = (x^2 + x + 1)^n cdot ((x-1)^2 + (x-1) + 1)^n ).Simplify ( (x-1)^2 + (x-1) + 1 ):( (x^2 - 2x + 1) + (x - 1) + 1 = x^2 - 2x + 1 + x - 1 + 1 = x^2 - x + 1 ).So, ( p(x)p(x-1) = (x^2 + x + 1)^n (x^2 - x + 1)^n = [(x^2 + x + 1)(x^2 - x + 1)]^n ).Multiply ( (x^2 + x + 1)(x^2 - x + 1) ):( x^4 - x^3 + x^2 + x^3 - x^2 + x + x^2 - x + 1 ).Simplify:- ( x^4 ),- ( -x^3 + x^3 = 0 ),- ( x^2 - x^2 + x^2 = x^2 ),- ( x - x = 0 ),- ( +1 ).So, the product is ( x^4 + x^2 + 1 ).Therefore, ( p(x)p(x-1) = (x^4 + x^2 + 1)^n = p(x^2) ).So, yes, ( p(x) = (x^2 + x + 1)^n ) satisfies the equation for any non-negative integer ( n ).What about the zero polynomial? If ( p(x) = 0 ), then ( p(x^2) = 0 ), and ( p(x)p(x-1) = 0 cdot 0 = 0 ). So, that works too.Similarly, the constant polynomial ( p(x) = 1 ) satisfies ( p(x^2) = 1 = 1 cdot 1 = p(x)p(x-1) ).Are there any other polynomials? Suppose ( p(x) ) is a product of ( (x^2 + x + 1)^n ) and some other polynomial. Would that still satisfy the equation?Let me think. Suppose ( p(x) = q(x)(x^2 + x + 1)^n ), where ( q(x) ) is another polynomial. Then,( p(x^2) = q(x^2)(x^4 + x^2 + 1)^n ).On the other hand,( p(x)p(x-1) = q(x)(x^2 + x + 1)^n cdot q(x-1)(x^2 - x + 1)^n = q(x)q(x-1)(x^4 + x^2 + 1)^n ).So, for ( p(x^2) = p(x)p(x-1) ), we need:( q(x^2)(x^4 + x^2 + 1)^n = q(x)q(x-1)(x^4 + x^2 + 1)^n ).Dividing both sides by ( (x^4 + x^2 + 1)^n ) (assuming it's non-zero, which it is except at roots), we get:( q(x^2) = q(x)q(x-1) ).So, ( q(x) ) must also satisfy the same functional equation. Therefore, if ( q(x) ) is not a constant polynomial, it must also be of the form ( (x^2 + x + 1)^m ) for some ( m ). But then, ( p(x) ) would just be ( (x^2 + x + 1)^{n + m} ), which is still a power of ( x^2 + x + 1 ).Therefore, the only solutions are the zero polynomial and polynomials of the form ( (x^2 + x + 1)^n ) for non-negative integers ( n ).Wait, but what about the constant polynomial ( p(x) = 1 )? It can be considered as ( (x^2 + x + 1)^0 ), so it's included in the general form.Also, the zero polynomial is a separate case because it can't be expressed as a power of ( x^2 + x + 1 ), but it trivially satisfies the equation.So, putting it all together, the real polynomials satisfying ( p(x^2) = p(x)p(x-1) ) for all ( x ) are:- The zero polynomial ( p(x) = 0 ).- The polynomials ( p(x) = (x^2 + x + 1)^n ) for some non-negative integer ( n ).I think that's all. I don't see any other polynomials that could satisfy the equation without leading to contradictions in the coefficients or degrees.

answer：Alright, so I've got this geometry problem here, and I need to show that AR is perpendicular to PQ. Let me try to unpack the problem step by step.First, let's parse the given information:We have triangle ABC with points P and Q inside it. The conditions given are:1. ∠APC = ∠AQB = 90°. So, both P and Q lie on the circles with diameters AC and AB, respectively.2. ∠ACP = ∠PBC. That means the angle at C between AC and CP is equal to the angle at B between BC and BP.3. ∠ABQ = ∠QCB. Similarly, the angle at B between AB and BQ is equal to the angle at C between BC and CQ.Lines BP and CQ intersect at point R. We need to show that AR is perpendicular to PQ.Okay, so let me try to visualize this. Triangle ABC, with points P and Q inside. P is such that APC is a right angle, so P lies on the circle with diameter AC. Similarly, Q lies on the circle with diameter AB since AQB is a right angle.Also, the angles ∠ACP and ∠PBC are equal, and ∠ABQ and ∠QCB are equal. These angle conditions might imply some similarity or congruence between triangles.Maybe I can draw the circles first. The circle with diameter AC will pass through P, and the circle with diameter AB will pass through Q. Then, BP and CQ intersect at R.I think I need to find some relationships between these points. Maybe using cyclic quadrilaterals or similar triangles.Let me consider the first angle condition: ∠ACP = ∠PBC. Let's denote this angle as θ. So, ∠ACP = θ and ∠PBC = θ.Similarly, the second angle condition: ∠ABQ = ∠QCB. Let's denote this angle as φ. So, ∠ABQ = φ and ∠QCB = φ.Now, since P lies on the circle with diameter AC, triangle APC is right-angled at P. Similarly, triangle AQB is right-angled at Q.Maybe I can use these right angles to find some perpendicular lines or to find some cyclic quadrilaterals.Let me consider triangle APC. Since ∠APC = 90°, then AP is perpendicular to PC. Similarly, in triangle AQB, AQ is perpendicular to QB.Wait, but how does that help me? Maybe I need to look for similar triangles.Looking at the angles, since ∠ACP = ∠PBC = θ, perhaps triangles ACP and PBC are similar? Let me check.In triangle ACP and triangle PBC:- ∠ACP = ∠PBC = θ- Both have a right angle? Wait, triangle ACP has a right angle at P, but triangle PBC doesn't necessarily have a right angle.Hmm, maybe that's not the way. Alternatively, maybe triangle ACP is similar to triangle PBC in some way.Wait, let me think about the sides. If ∠ACP = ∠PBC, and both share angle at C and B respectively, maybe there's some proportionality.Alternatively, maybe I can use trigonometric ratios. Let me denote some lengths.Let me denote AC = b, AB = c, BC = a. Let me denote AP = x, PC = y, AQ = m, QB = n.But maybe that's getting too ahead of myself. Let me try to find some cyclic quadrilaterals.Since ∠APC = 90°, point P lies on the circle with diameter AC. Similarly, Q lies on the circle with diameter AB.Now, lines BP and CQ intersect at R. So, R is the intersection point inside the triangle.I need to show that AR is perpendicular to PQ. So, maybe I can show that the product of their slopes is -1, but since this is a synthetic geometry problem, I should look for a synthetic proof.Perhaps I can construct some perpendicular lines or use properties of cyclic quadrilaterals.Wait, another thought: since both P and Q lie on circles with diameters AC and AB, maybe lines AP and AQ have some special properties.Alternatively, maybe I can use the fact that angles at P and Q are right angles to find some orthocenters or something.Wait, let me think about the orthocenter. In triangle ABC, the orthocenter is the intersection of the altitudes. But I don't know if P or Q are orthocenters.Alternatively, maybe I can use the fact that AP and AQ are related through the given angle conditions.Wait, another approach: maybe use Ceva's theorem. Since lines BP and CQ intersect at R, maybe I can apply Ceva's theorem to find some concurrency or ratio.But Ceva's theorem usually involves three cevians, so maybe I need another cevian to apply it.Alternatively, maybe use Menelaus' theorem.Wait, perhaps I can use the given angle conditions to show that certain triangles are similar, leading to some proportional sides, which can then be used to show that AR is perpendicular to PQ.Let me try to explore the angle conditions more.Given ∠ACP = ∠PBC = θ.So, in triangle ACP, angle at C is θ, and in triangle PBC, angle at B is θ.Similarly, in triangle ABQ, angle at B is φ, and in triangle QCB, angle at C is φ.Maybe I can consider the triangles ACP and PBC.In triangle ACP:- ∠ACP = θ- ∠APC = 90°- So, ∠CAP = 90° - θIn triangle PBC:- ∠PBC = θ- ∠BPC = 180° - θ - ∠BCPWait, but I don't know ∠BCP.Alternatively, maybe I can use the Law of Sines in these triangles.In triangle ACP:- AP / sin θ = AC / sin ∠APC = AC / 1 (since ∠APC = 90°)- So, AP = AC sin θSimilarly, in triangle PBC:- BP / sin ∠BCP = BC / sin ∠BPC- But I don't know ∠BCP or ∠BPC.Hmm, maybe that's not helpful.Wait, another idea: since ∠ACP = ∠PBC, maybe lines AP and BP are isogonal conjugates with respect to angle C.Similarly, since ∠ABQ = ∠QCB, lines AQ and CQ are isogonal conjugates with respect to angle B.Isogonal conjugates might have some properties that can help here.Alternatively, maybe I can use the fact that P and Q lie on the circles with diameters AC and AB, respectively, and use some properties of these circles.Wait, since P is on the circle with diameter AC, then AP is perpendicular to PC. Similarly, AQ is perpendicular to QB.So, AP ⊥ PC and AQ ⊥ QB.Wait, so AP and PC are perpendicular, and AQ and QB are perpendicular.Maybe I can use these perpendicularity conditions to find some relationships between the lines.Alternatively, maybe I can construct some rectangles or other quadrilaterals.Wait, another thought: since AP ⊥ PC and AQ ⊥ QB, maybe points P and Q are the feet of some altitudes, but I don't think that's necessarily the case here.Wait, but in triangle ABC, if P were the foot of the altitude from A, then AP would be perpendicular to BC, but here AP is perpendicular to PC, which is different.Similarly, AQ is perpendicular to QB, which is also different from the altitude.Hmm, maybe I need to consider the orthocenter of some other triangle.Alternatively, maybe I can use coordinate geometry. Let me try assigning coordinates to the triangle and see if I can compute the necessary slopes.Let me place triangle ABC in the coordinate plane. Let me set point A at (0,0), point B at (c,0), and point C at (d,e). Then, points P and Q can be expressed in terms of coordinates.But this might get messy, but let me try.Let me denote:- A = (0,0)- B = (c,0)- C = (d,e)Then, the circle with diameter AC has center at (d/2, e/2) and radius sqrt((d/2)^2 + (e/2)^2).Similarly, the circle with diameter AB has center at (c/2, 0) and radius c/2.Point P lies on the circle with diameter AC, so its coordinates satisfy (x - d/2)^2 + (y - e/2)^2 = (d/2)^2 + (e/2)^2.Simplifying, x^2 - d x + y^2 - e y = 0.Similarly, point Q lies on the circle with diameter AB, so its coordinates satisfy (x - c/2)^2 + y^2 = (c/2)^2.Simplifying, x^2 - c x + y^2 = 0.Now, the angle conditions: ∠ACP = ∠PBC and ∠ABQ = ∠QCB.Let me try to express these angle conditions in terms of coordinates.First, ∠ACP = ∠PBC.Let me compute the slopes of lines CP and BP.Point C is (d,e), point P is (x_p, y_p), point B is (c,0), point Q is (x_q, y_q).The slope of CP is (y_p - e)/(x_p - d).The slope of BP is (y_p - 0)/(x_p - c) = y_p/(x_p - c).Similarly, the slope of BC is (0 - e)/(c - d) = -e/(c - d).Wait, but ∠ACP is the angle between AC and CP.Vector AC is from A(0,0) to C(d,e), so direction vector (d,e).Vector CP is from C(d,e) to P(x_p, y_p), direction vector (x_p - d, y_p - e).The angle between AC and CP is θ, so the tangent of θ is the magnitude of the cross product divided by the dot product.Similarly, ∠PBC is the angle at B between BC and BP.Vector BC is from B(c,0) to C(d,e), direction vector (d - c, e).Vector BP is from B(c,0) to P(x_p, y_p), direction vector (x_p - c, y_p).The tangent of θ is the magnitude of the cross product divided by the dot product.So, setting these equal:| (d)(y_p - e) - (e)(x_p - d) | / [d(x_p - d) + e(y_p - e)] = | (d - c)(y_p) - e(x_p - c) | / [(d - c)(x_p - c) + e y_p]This seems complicated, but maybe I can find a relationship between x_p and y_p.Alternatively, maybe I can use the fact that AP is perpendicular to PC, so the product of their slopes is -1.Slope of AP: (y_p - 0)/(x_p - 0) = y_p/x_p.Slope of PC: (y_p - e)/(x_p - d).Their product should be -1:(y_p/x_p) * [(y_p - e)/(x_p - d)] = -1So,(y_p (y_p - e)) / (x_p (x_p - d)) = -1Which simplifies to:y_p^2 - e y_p = -x_p^2 + d x_pOr,x_p^2 + y_p^2 - d x_p - e y_p = 0But from earlier, since P is on the circle with diameter AC, we have x_p^2 - d x_p + y_p^2 - e y_p = 0. Wait, that's the same equation. So, that doesn't give us new information.Hmm, maybe I need to use the angle condition ∠ACP = ∠PBC.Let me denote the slopes as follows:Slope of AC: e/d.Slope of CP: (y_p - e)/(x_p - d).The angle between AC and CP is θ, so tan θ = |(m2 - m1)/(1 + m1 m2)|, where m1 is slope of AC, m2 is slope of CP.Similarly, slope of BC: (e - 0)/(d - c) = e/(d - c).Slope of BP: y_p/(x_p - c).The angle between BC and BP is also θ, so tan θ = |(m4 - m3)/(1 + m3 m4)|, where m3 is slope of BC, m4 is slope of BP.Setting these equal:|( ( (y_p - e)/(x_p - d) ) - (e/d) ) / (1 + (e/d)( (y_p - e)/(x_p - d) ))| = |( (y_p/(x_p - c) ) - (e/(d - c)) ) / (1 + (e/(d - c))(y_p/(x_p - c)) )|This is getting really messy. Maybe coordinate geometry isn't the best approach here.Let me try a different approach. Maybe using vectors or complex numbers.Alternatively, maybe I can use properties of cyclic quadrilaterals.Wait, since P is on the circle with diameter AC, then AP ⊥ PC. Similarly, Q is on the circle with diameter AB, so AQ ⊥ QB.So, AP ⊥ PC and AQ ⊥ QB.Now, let me consider the orthocenter of triangle ABC. The orthocenter is the intersection of the altitudes. But I don't know if P or Q are related to the orthocenter.Wait, another idea: maybe P and Q are the feet of some symmedians or something like that.Alternatively, maybe I can use the fact that BP and CQ intersect at R, and then use some properties of R.Wait, perhaps I can use Ceva's theorem. Ceva's theorem states that for concurrent cevians, the product of certain ratios equals 1.But in this case, we have two cevians BP and CQ intersecting at R. To apply Ceva's theorem, we need a third cevian. Maybe AR is the third cevian, but I don't know if it's concurrent.Alternatively, maybe I can use Menelaus' theorem for the transversal R.Wait, perhaps I can consider triangle ABC with transversal R-P-Q or something.Alternatively, maybe I can use the properties of the orthocenter or centroid, but I'm not sure.Wait, another thought: since AP ⊥ PC and AQ ⊥ QB, maybe PQ is related to the orthocenter in some way.Alternatively, maybe I can construct the orthocenter of triangle APQ and see if it relates to R.Wait, perhaps I can consider the nine-point circle or something like that.Alternatively, maybe I can use the fact that AR is the Euler line or something.Wait, maybe I'm overcomplicating this. Let me try to think of some simpler properties.Since AP ⊥ PC and AQ ⊥ QB, maybe I can consider the reflections of A over BP and CQ.Wait, another idea: maybe use the fact that in triangle ABC, the orthocenter H has the property that AH ⊥ BC, BH ⊥ AC, etc. But I don't know if P or Q are related to H.Wait, maybe I can consider the pedal triangle of point R. Since R is the intersection of BP and CQ, the pedal triangle might have some properties.Alternatively, maybe I can use the fact that AR is the symmedian or something.Wait, perhaps I can use the fact that since AP ⊥ PC and AQ ⊥ QB, then P and Q are the feet of the altitudes from A in some other triangles.Wait, another approach: since AP ⊥ PC, then P lies on the circle with diameter AC, and similarly Q lies on the circle with diameter AB. Maybe the radical axis of these two circles is relevant here.The radical axis of two circles is the locus of points with equal power with respect to both circles. The radical axis is perpendicular to the line joining the centers.But I'm not sure how that helps here.Wait, another thought: maybe the midpoint of PQ lies on AR, and since AR is perpendicular to PQ, then AR is the perpendicular bisector of PQ.But I need to show that AR is perpendicular to PQ, not necessarily that it bisects it.Alternatively, maybe I can show that the midpoint of PQ lies on AR, and then use some properties.Wait, perhaps I can use the fact that in triangle ABC, the cevians BP and CQ intersect at R, and then use some properties of R.Wait, another idea: maybe use the fact that since ∠ACP = ∠PBC and ∠ABQ = ∠QCB, then triangles PBC and QCB are similar in some way.Wait, let me consider triangles PBC and QCB.In triangle PBC:- ∠PBC = θ- ∠BPC = 180° - θ - ∠BCPIn triangle QCB:- ∠QCB = φ- ∠CQB = 180° - φ - ∠CBQBut I don't see a direct similarity here.Wait, maybe I can use the Law of Sines in these triangles.In triangle PBC:- PB / sin ∠BCP = BC / sin ∠BPCIn triangle QCB:- QC / sin ∠CBQ = BC / sin ∠CQBBut I don't know the angles ∠BCP or ∠CBQ.Wait, but from the given angle conditions, ∠ACP = ∠PBC = θ and ∠ABQ = ∠QCB = φ.So, in triangle PBC, ∠PBC = θ, and in triangle QCB, ∠QCB = φ.Maybe I can relate these angles to other parts of the triangle.Wait, another thought: since AP ⊥ PC and AQ ⊥ QB, maybe lines AP and AQ are related to the altitudes of some other triangles.Alternatively, maybe I can use the fact that AP and AQ are the altitudes of triangles APC and AQB, respectively.Wait, perhaps I can consider the orthocenters of triangles APC and AQB.In triangle APC, the orthocenter is A, since AP is perpendicular to PC.Similarly, in triangle AQB, the orthocenter is A, since AQ is perpendicular to QB.So, A is the orthocenter of both triangles APC and AQB.Hmm, interesting. So, A is the orthocenter for both these triangles.Now, maybe I can use some properties of orthocenters.Wait, another idea: since A is the orthocenter of triangle APC, then the reflection of A over PC lies on the circumcircle of triangle APC.Similarly, the reflection of A over QB lies on the circumcircle of triangle AQB.But I'm not sure how that helps.Wait, perhaps I can use the fact that the reflection of the orthocenter over a side lies on the circumcircle.So, reflecting A over PC would give a point on the circumcircle of APC, which is the circle with diameter AC.Similarly, reflecting A over QB would give a point on the circle with diameter AB.But I'm not sure how that helps with AR and PQ.Wait, maybe I can consider the nine-point circle, which passes through the midpoints of the sides, the feet of the altitudes, and the midpoints between the orthocenter and each vertex.But since A is the orthocenter for both triangles APC and AQB, maybe the nine-point circles for these triangles pass through some relevant points.Alternatively, maybe I can consider the Euler lines of these triangles, but I'm not sure.Wait, perhaps I can use the fact that since A is the orthocenter, then the Euler line of triangle APC passes through A, the centroid, and the circumcenter.But I'm not sure.Wait, maybe I'm overcomplicating this. Let me try to think of some simpler properties.Since AP ⊥ PC and AQ ⊥ QB, maybe I can consider the cyclic quadrilaterals formed by these perpendiculars.Wait, in triangle APC, since AP ⊥ PC, then APC is a right triangle, so the circumcircle of APC has AC as its diameter.Similarly, the circumcircle of AQB has AB as its diameter.Now, since P is on the circumcircle of APC, and Q is on the circumcircle of AQB, maybe I can use some properties of these circles.Wait, another idea: maybe the points P and Q are the midpoints of some arcs or something.Alternatively, maybe I can use the fact that the angles at P and Q are right angles to find some relationships.Wait, perhaps I can use the fact that the power of point R with respect to both circles is equal.Since R lies on BP and CQ, which are cevians intersecting at R, maybe the power of R with respect to both circles can be expressed in terms of the lengths of segments.Wait, the power of R with respect to the circle with diameter AC is equal to RP * RB = RC * RQ.Wait, no, the power of R with respect to the circle with diameter AC would be RA * RC (if R lies on the radical axis), but I'm not sure.Wait, actually, the power of R with respect to the circle with diameter AC is equal to RP * RB, since BP is a secant line passing through P and B.Similarly, the power of R with respect to the circle with diameter AB is equal to RQ * RC, since CQ is a secant line passing through Q and C.But since both powers are equal to the power of R with respect to both circles, maybe I can set them equal.So, RP * RB = RQ * RC.Hmm, that's an interesting relation. Maybe I can use this to find some ratio.But I'm not sure how to proceed from here.Wait, another thought: since AR is supposed to be perpendicular to PQ, maybe I can show that the product of their slopes is -1, but in a synthetic way.Alternatively, maybe I can show that AR is the altitude of some triangle, making it perpendicular.Wait, perhaps I can consider triangle APQ and show that AR is its altitude.But I need to show that AR is perpendicular to PQ, not necessarily that it's an altitude.Wait, maybe I can use the fact that in triangle APQ, AR is the symmedian or something.Alternatively, maybe I can use the fact that AR is the radical axis of some circles.Wait, perhaps I can consider the circles with diameters AP and AQ, but I'm not sure.Wait, another idea: since AP ⊥ PC and AQ ⊥ QB, maybe lines PC and QB are related in some way.Wait, perhaps I can consider the orthocenter of triangle APQ, but I'm not sure.Wait, maybe I can use the fact that since AP ⊥ PC and AQ ⊥ QB, then PC and QB are the altitudes of some triangles.Wait, another approach: maybe use trigonometric Ceva's theorem.Trigonometric Ceva's theorem states that for concurrent cevians, the product of the sines of the angles is equal.But I'm not sure if that applies here.Wait, perhaps I can use the fact that ∠ACP = ∠PBC and ∠ABQ = ∠QCB to set up some ratios.Let me denote:In triangle ABC, let me denote the angles at A, B, C as α, β, γ respectively.Then, from the given angle conditions:∠ACP = ∠PBC = θ∠ABQ = ∠QCB = φSo, in triangle ABC, we have:At point C: ∠ACB = γ = ∠ACP + ∠PCB = θ + ∠PCBAt point B: ∠ABC = β = ∠PBC + ∠ABQ = θ + φSimilarly, at point A: ∠BAC = α = ∠BAQ + ∠QAC = something, but I'm not sure.Wait, maybe I can express the angles in terms of θ and φ.From point B: ∠ABC = β = θ + φFrom point C: ∠ACB = γ = θ + ∠PCBBut I don't know ∠PCB.Wait, maybe I can relate ∠PCB to φ.Wait, from the angle condition at Q: ∠ABQ = ∠QCB = φSo, ∠QCB = φ, which is part of ∠ACB.So, ∠ACB = γ = ∠QCB + ∠QCP = φ + ∠QCPBut I don't know ∠QCP.Wait, maybe I can find a relationship between θ and φ.Wait, in triangle ABC, the sum of angles is α + β + γ = 180°We have β = θ + φAnd γ = θ + something, but I don't know.Wait, maybe I can use the Law of Sines in triangles PBC and QCB.In triangle PBC:PB / sin ∠BCP = BC / sin ∠BPCIn triangle QCB:QC / sin ∠CBQ = BC / sin ∠CQBBut I don't know the angles ∠BCP or ∠CBQ.Wait, but from the given angle conditions:∠ACP = θ, so ∠BCP = γ - θSimilarly, ∠ABQ = φ, so ∠CBQ = β - φ = (θ + φ) - φ = θWait, that's interesting.So, in triangle PBC:∠BCP = γ - θ∠PBC = θSo, ∠BPC = 180° - θ - (γ - θ) = 180° - γSimilarly, in triangle QCB:∠CBQ = θ∠QCB = φSo, ∠CQB = 180° - θ - φWait, but from triangle ABC, β = θ + φ, so ∠CQB = 180° - βHmm, interesting.So, in triangle PBC, ∠BPC = 180° - γIn triangle QCB, ∠CQB = 180° - βNow, let me consider the Law of Sines in these triangles.In triangle PBC:PB / sin(γ - θ) = BC / sin(180° - γ) = BC / sin γSo, PB = BC * sin(γ - θ) / sin γSimilarly, in triangle QCB:QC / sin θ = BC / sin(180° - β) = BC / sin βSo, QC = BC * sin θ / sin βNow, from triangle ABC, using the Law of Sines:BC / sin α = AB / sin γ = AC / sin βSo, BC = AB sin α / sin γWait, maybe I can express PB and QC in terms of AB, AC, etc.But this might not be directly helpful.Wait, another idea: since we have expressions for PB and QC, maybe we can find the ratio PB/QC.From above:PB = BC * sin(γ - θ) / sin γQC = BC * sin θ / sin βSo, PB / QC = [sin(γ - θ) / sin γ] / [sin θ / sin β] = [sin(γ - θ) sin β] / [sin γ sin θ]Hmm, not sure.Wait, but from triangle ABC, β = θ + φ, and γ = θ + something.Wait, earlier, I had γ = θ + ∠PCB, but ∠PCB is part of ∠ACB.Wait, maybe I can express γ in terms of θ and φ.Wait, from point C: ∠ACB = γ = ∠ACP + ∠PCB = θ + ∠PCBBut ∠PCB is part of ∠BCP, which in triangle PBC is ∠BCP = γ - θWait, that's the same as before.Hmm, maybe I'm going in circles.Wait, another idea: since we have PB and QC expressed in terms of BC, θ, γ, β, maybe we can relate them to the coordinates or something.Alternatively, maybe I can use Ceva's theorem in triangle ABC with cevians BP, CQ, and some third cevian.But I don't know the third cevian.Wait, Ceva's theorem states that for cevians AP, BQ, CR to be concurrent, (AF/FB) * (BD/DC) * (CE/EA) = 1But in our case, cevians are BP and CQ intersecting at R. So, if we consider the third cevian as AR, then Ceva's condition would be:(AF/FB) * (BD/DC) * (CE/EA) = 1But I don't know the ratios AF/FB, BD/DC, etc.Wait, but maybe I can express these ratios in terms of θ and φ.Wait, from the angle conditions, maybe I can find some ratios.Wait, in triangle ABC, from the Law of Sines:AB / sin γ = BC / sin α = AC / sin βSo, AB = BC sin γ / sin αAC = BC sin β / sin αSimilarly, from triangle PBC:PB = BC sin(γ - θ) / sin γFrom triangle QCB:QC = BC sin θ / sin βSo, maybe I can express PB and QC in terms of AB and AC.Wait, since AC = BC sin β / sin α, then BC = AC sin α / sin βSo, PB = (AC sin α / sin β) * sin(γ - θ) / sin γSimilarly, QC = (AC sin α / sin β) * sin θ / sin βHmm, not sure.Wait, another idea: since we have expressions for PB and QC, maybe we can find the ratio PB/QC.From above:PB = BC sin(γ - θ) / sin γQC = BC sin θ / sin βSo, PB / QC = [sin(γ - θ) / sin γ] / [sin θ / sin β] = [sin(γ - θ) sin β] / [sin γ sin θ]Now, from triangle ABC, we have:sin α / BC = sin β / AC = sin γ / ABSo, sin β = AC sin α / BCSimilarly, sin γ = AB sin α / BCSo, substituting back:PB / QC = [sin(γ - θ) * (AC sin α / BC)] / [ (AB sin α / BC) * sin θ ] = [sin(γ - θ) AC] / [AB sin θ]Hmm, interesting. So, PB / QC = [AC / AB] * [sin(γ - θ) / sin θ]But from triangle ABC, AC / AB = sin β / sin γSo, PB / QC = (sin β / sin γ) * [sin(γ - θ) / sin θ]Hmm, not sure.Wait, but from earlier, β = θ + φ, and γ = θ + something.Wait, maybe I can express sin(γ - θ) as sin(something).Wait, γ = θ + ∠PCB, so γ - θ = ∠PCBBut I don't know ∠PCB.Wait, but from triangle PBC, ∠BCP = γ - θ, and we had ∠BPC = 180° - γSo, maybe I can relate ∠PCB to other angles.Wait, another idea: maybe use the fact that in triangle PBC, ∠BPC = 180° - γ, and in triangle QCB, ∠CQB = 180° - βSo, points P and Q lie on circles such that angles at P and Q are 180° - γ and 180° - β respectively.Wait, maybe I can use the fact that points P and Q lie on the circumcircles of triangles BPC and CQB, but I'm not sure.Wait, another thought: since ∠BPC = 180° - γ, then point P lies on the circumcircle of triangle BPC, but that's trivial.Wait, maybe I can consider the circumcircle of triangle BPC and see if it passes through some other point.Alternatively, maybe I can consider the circumcircle of triangle CQB.Wait, another idea: since ∠BPC = 180° - γ, and ∠CQB = 180° - β, maybe these angles relate to the angles at A.Wait, from triangle ABC, α + β + γ = 180°, so 180° - γ = α + β, and 180° - β = α + γWait, no, that's not correct. 180° - γ = α + β, and 180° - β = α + γWait, actually, no:In triangle ABC, α + β + γ = 180°, so:180° - γ = α + β180° - β = α + γYes, that's correct.So, ∠BPC = 180° - γ = α + βAnd ∠CQB = 180° - β = α + γHmm, interesting.So, in triangle PBC, ∠BPC = α + βIn triangle QCB, ∠CQB = α + γNow, maybe I can use these angles to find some relationships.Wait, another idea: since ∠BPC = α + β, and ∠CQB = α + γ, maybe these angles can be related to angles at A.Wait, perhaps I can consider the circumcircle of triangle APQ and see if R lies on it or something.Alternatively, maybe I can use the fact that AR is the angle bisector or something.Wait, another thought: since ∠BPC = α + β, and ∠CQB = α + γ, maybe the lines BP and CQ intersect at R such that ∠PRQ = α + something.Wait, perhaps I can consider the angles at R.Wait, at point R, we have lines BP and CQ intersecting.So, ∠PRQ is equal to ∠PBC + ∠QCB = θ + φBut from earlier, β = θ + φ, so ∠PRQ = βHmm, interesting.Similarly, maybe ∠QRP = γ or something.Wait, not sure.Wait, another idea: since ∠PRQ = β, and ∠BAC = α, maybe there's some relationship.Wait, perhaps I can consider triangle PRQ and see if it's similar to triangle ABC or something.Alternatively, maybe I can use the fact that ∠PRQ = β and ∠BAC = α to find some relationship.Wait, another thought: since ∠PRQ = β, and ∠BAC = α, maybe the sum α + β + γ = 180° can be used somehow.Wait, but I'm not sure.Wait, another idea: since ∠PRQ = β, and ∠BAC = α, maybe the lines AR and PQ form some angle that can be related to α and β.Wait, perhaps I can use the fact that the sum of angles around R is 360°, so ∠PRQ + ∠QRP + ∠RPQ = 360°, but I don't know the other angles.Wait, maybe I can use the fact that AR is the angle bisector or something.Wait, another idea: since AP ⊥ PC and AQ ⊥ QB, maybe the quadrilateral APCQ is cyclic or something.Wait, but APCQ would have two right angles at P and Q, so it's cyclic only if the other angles are supplementary, which I don't know.Wait, another thought: since AP ⊥ PC and AQ ⊥ QB, maybe lines PC and QB are perpendicular to AP and AQ respectively, so maybe PC and QB are related in some way.Wait, perhaps I can consider the orthocenter of triangle APQ, but I'm not sure.Wait, another idea: since AP ⊥ PC and AQ ⊥ QB, maybe lines PC and QB are the altitudes of triangle APQ, making A the orthocenter.But I'm not sure.Wait, if PC and QB are altitudes of triangle APQ, then their intersection would be the orthocenter, but I don't know if that's the case.Wait, but PC is perpendicular to AP, and QB is perpendicular to AQ, so if AP and AQ are sides of triangle APQ, then PC and QB would be altitudes.So, yes, PC and QB are altitudes of triangle APQ, intersecting at the orthocenter.But where is the orthocenter? It would be the intersection point of PC and QB, but in our case, PC and QB are not necessarily intersecting at R.Wait, actually, PC is part of BP, and QB is part of CQ, which intersect at R. So, R is the intersection of BP and CQ, which are not necessarily the altitudes of triangle APQ.Wait, but PC is part of BP, and QB is part of CQ, so maybe R is the orthocenter of triangle APQ.Wait, if R is the orthocenter of triangle APQ, then AR would be the third altitude, making AR perpendicular to PQ.That would solve the problem!So, if I can show that R is the orthocenter of triangle APQ, then AR is perpendicular to PQ.So, how can I show that R is the orthocenter of triangle APQ?Well, the orthocenter is the intersection of the altitudes. So, if I can show that R lies on two altitudes of triangle APQ, then it must be the orthocenter.We already have that PC is perpendicular to AP, and QB is perpendicular to AQ.But PC is part of BP, and QB is part of CQ.So, if R is the intersection of BP and CQ, and BP contains PC (which is perpendicular to AP), and CQ contains QB (which is perpendicular to AQ), then R lies on two altitudes of triangle APQ.Therefore, R is the orthocenter of triangle APQ.Hence, AR is the third altitude, which is perpendicular to PQ.Therefore, AR ⊥ PQ.So, that's the solution.Wait, let me recap to make sure I didn't make a mistake.1. P lies on the circle with diameter AC, so AP ⊥ PC.2. Q lies on the circle with diameter AB, so AQ ⊥ QB.3. BP and CQ intersect at R.4. Since PC ⊥ AP and QB ⊥ AQ, and PC is part of BP, QB is part of CQ, then R lies on two altitudes of triangle APQ.5. Therefore, R is the orthocenter of triangle APQ.6. Hence, AR is the third altitude, which is perpendicular to PQ.Yes, that makes sense. So, AR is perpendicular to PQ.

answer：First, I need to determine how many ways there are to choose which two courses out of the four will have students. This can be calculated using the combination formula:[binom{4}{2} = frac{4!}{2! times (4-2)!} = frac{4 times 3}{2 times 1} = 6]Next, I need to figure out how many ways to assign the four students to these two chosen courses, ensuring that each course has exactly two students. I can use the combination formula again to choose two students out of four for the first course:[binom{4}{2} = frac{4!}{2! times (4-2)!} = frac{4 times 3}{2 times 1} = 6]However, this counts each assignment twice because choosing students A and B for the first course and students C and D for the second course is the same as choosing students C and D for the first course and students A and B for the second course. Therefore, I need to divide by 2 to correct for this duplication:[frac{6}{2} = 3]Finally, I multiply the number of ways to choose the two courses by the number of ways to assign the students to these courses to get the total number of course selection schemes:[6 times 3 = 18]

answer：Okay, so I have this problem here: "A car covers a distance of 390 km in 4 hours. What is its speed?" Alright, let's see. I remember that speed has something to do with distance and time. I think it's like how fast something is going. Maybe it's distance divided by time? Yeah, that sounds familiar from what I've learned before.So, if the car goes 390 kilometers in 4 hours, I guess I need to find out how many kilometers it goes in one hour. That would be the speed, right? So, if I take the total distance and divide it by the total time, I should get the speed.Let me write that down: Speed = Distance / Time. So, plugging in the numbers, it would be Speed = 390 km / 4 hours. Hmm, let me do that division. 390 divided by 4. Okay, 4 times 90 is 360, and 4 times 97 is 388. Wait, that's not right. Maybe I should do it step by step.So, 4 goes into 39 nine times because 4 times 9 is 36. Subtract 36 from 39, and I get 3. Bring down the 0, making it 30. 4 goes into 30 seven times because 4 times 7 is 28. Subtract 28 from 30, and I have 2. Bring down the next 0, making it 20. 4 goes into 20 five times exactly. So, putting it all together, I have 97.5.Wait, that's a decimal. Is that okay? I mean, speed can be a decimal, right? Like, not everything has to be a whole number. So, 97.5 km/h seems reasonable. Let me double-check my calculations to make sure I didn't make a mistake.Starting over: 390 divided by 4. 4 into 39 is 9, remainder 3. 4 into 30 is 7, remainder 2. 4 into 20 is 5, no remainder. So, yes, it's 97.5 km/h. That makes sense. So, the car's speed is 97.5 kilometers per hour.I guess that's the answer. It feels a bit strange having a decimal in the speed, but I think it's correct. Maybe if I convert it to a fraction, it would look nicer? 97.5 is the same as 97 and a half, or 195/2. But I think 97.5 km/h is fine for the answer.Just to make sure, I'll think about it another way. If the car is going 97.5 km/h, then in one hour it covers 97.5 km. In four hours, that would be 97.5 times 4, which is 390 km. Yep, that matches the distance given in the problem. So, my answer must be right.I think I'm confident now that the speed of the car is 97.5 kilometers per hour.

answer：Okay, so I need to construct a third-degree polynomial ( P(x) ) with integer coefficients, leading coefficient 1, and it should be divisible by 3 for all integer values of ( x ). Hmm, let me think about how to approach this.First, let's recall what a third-degree polynomial looks like. It generally has the form ( P(x) = x^3 + ax^2 + bx + c ), where ( a ), ( b ), and ( c ) are integers because the coefficients must be integers. The leading coefficient is 1, which is already satisfied here.Now, the tricky part is ensuring that ( P(x) ) is divisible by 3 for every integer ( x ). That means ( P(x) equiv 0 mod 3 ) for all integer ( x ). So, I need to make sure that no matter what integer I plug into ( P(x) ), the result is a multiple of 3.How can I ensure that? Maybe I can use some properties of modular arithmetic or look for patterns in polynomials that are always divisible by 3.I remember that for a polynomial to be divisible by a number for all integer inputs, it must satisfy certain congruence conditions. Specifically, if a polynomial is divisible by 3 for all integers ( x ), then it must be congruent to 0 modulo 3 for all ( x ). So, ( P(x) equiv 0 mod 3 ) for all ( x in mathbb{Z} ).One way to achieve this is to have the polynomial factor in such a way that one of the factors is always divisible by 3. For example, if the polynomial can be written as ( (x)(x+1)(x+2) ), then among any three consecutive integers, one of them must be divisible by 3. That seems promising.Let me test this idea. Let's consider ( P(x) = x(x+1)(x+2) ). Expanding this, we get:[P(x) = x(x+1)(x+2) = x(x^2 + 3x + 2) = x^3 + 3x^2 + 2x]So, ( P(x) = x^3 + 3x^2 + 2x ). This is a third-degree polynomial with integer coefficients and leading coefficient 1, which fits the requirements.Now, let's check if this polynomial is divisible by 3 for all integer ( x ). Take any integer ( x ). Then ( x ), ( x+1 ), and ( x+2 ) are three consecutive integers. In any set of three consecutive integers, one of them must be divisible by 3. Therefore, the product ( x(x+1)(x+2) ) will always be divisible by 3. Hence, ( P(x) ) is divisible by 3 for all integer ( x ).Wait, but is this the only way? Maybe I can think of another polynomial that also satisfies these conditions. For example, what if I add a multiple of 3 to the polynomial? Let's say ( P(x) = x^3 + 3x^2 + 2x + 3k ), where ( k ) is an integer. Then, for integer ( x ), ( P(x) ) would still be divisible by 3 because ( 3k ) is a multiple of 3. However, the problem specifies that the polynomial should have integer coefficients and leading coefficient 1, but it doesn't specify anything about the constant term. So, technically, adding ( 3k ) would still satisfy the conditions.But wait, does adding ( 3k ) affect the divisibility by 3? Let's see. If ( P(x) = x^3 + 3x^2 + 2x + 3k ), then ( P(x) equiv x^3 + 0x^2 + 2x + 0 mod 3 ). Simplifying, ( P(x) equiv x^3 + 2x mod 3 ). Is ( x^3 + 2x ) always divisible by 3?Let me test this for some small integers:- For ( x = 0 ): ( 0 + 0 = 0 ), which is divisible by 3.- For ( x = 1 ): ( 1 + 2 = 3 ), which is divisible by 3.- For ( x = 2 ): ( 8 + 4 = 12 ), which is divisible by 3.- For ( x = 3 ): ( 27 + 6 = 33 ), which is divisible by 3.- For ( x = -1 ): ( -1 - 2 = -3 ), which is divisible by 3.- For ( x = -2 ): ( -8 - 4 = -12 ), which is divisible by 3.Hmm, it seems that ( x^3 + 2x ) is always divisible by 3. So, adding ( 3k ) doesn't affect the divisibility by 3. Therefore, ( P(x) = x^3 + 3x^2 + 2x + 3k ) would also satisfy the conditions for any integer ( k ).But the problem asks for an example, so I can choose ( k = 0 ) for simplicity, giving ( P(x) = x^3 + 3x^2 + 2x ). Alternatively, I could choose ( k = 1 ), giving ( P(x) = x^3 + 3x^2 + 2x + 3 ), and so on.However, the simplest form is probably ( P(x) = x(x+1)(x+2) ), which expands to ( x^3 + 3x^2 + 2x ). This polynomial is not only third-degree with integer coefficients and leading coefficient 1 but also ensures divisibility by 3 for all integer ( x ) because it's the product of three consecutive integers.Let me double-check this by plugging in a few more values:- ( x = 4 ): ( 4 times 5 times 6 = 120 ), which is divisible by 3.- ( x = 5 ): ( 5 times 6 times 7 = 210 ), which is divisible by 3.- ( x = -3 ): ( -3 times -2 times -1 = -6 ), which is divisible by 3.Yes, it seems consistent. Therefore, ( P(x) = x(x+1)(x+2) ) is a valid example.Alternatively, I could consider other forms. For instance, using the fact that ( x^3 equiv x mod 3 ) due to Fermat's little theorem. So, ( x^3 - x ) is divisible by 3 for all integers ( x ). Therefore, ( P(x) = x^3 - x + 3k ) would also be divisible by 3 for all integer ( x ). Let's see:( P(x) = x^3 - x + 3k ). For ( k = 0 ), ( P(x) = x^3 - x ). Testing:- ( x = 1 ): ( 1 - 1 = 0 ), divisible by 3.- ( x = 2 ): ( 8 - 2 = 6 ), divisible by 3.- ( x = 3 ): ( 27 - 3 = 24 ), divisible by 3.- ( x = -1 ): ( -1 - (-1) = 0 ), divisible by 3.This also works. So, another example could be ( P(x) = x^3 - x ). However, this polynomial doesn't have the same structure as the product of three consecutive integers, but it still satisfies the conditions.Comparing the two, ( x(x+1)(x+2) ) is perhaps more intuitive because it directly uses the property of consecutive integers, ensuring one is divisible by 3. On the other hand, ( x^3 - x ) relies on Fermat's little theorem, which might be less obvious to someone not familiar with number theory.Since the problem doesn't specify any additional constraints, either polynomial would be a correct answer. However, ( x(x+1)(x+2) ) might be preferable for its clarity and direct application of the consecutive integer property.In conclusion, constructing such a polynomial involves ensuring that the product or structure inherently includes a factor that is always divisible by 3 for any integer input. Both approaches I considered achieve this, but the product of three consecutive integers is a straightforward and clear example.