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A polynomial function P(x)P(x) of degree 4 with real coefficients is symmetric about the line x=2x = 2 in the xyxy-plane. If P(x)P(x) is divisible by x2−4x+3x^2 - 4x + 3, the remainder when P(x)P(x) is divided by x−4x - 4 is 3636, and P(2)=−8P(2) = -8, what is the value of P(5)P(5)?

Cevap: 136

Cevap

The value of P(5)P(5) is 136.
The correct answer of 136 is found by representing the symmetric fourth-degree polynomial as P(x)=a(x−2)4+b(x−2)2+cP(x) = a(x - 2)^4 + b(x - 2)^2 + c, solving for the coefficients using the roots at x=3x=3, the point (2,−8)(2, -8), and the remainder point (4,36)(4, 36), and then evaluating the resulting expression (x−2)4+7(x−2)2−8(x-2)^4 + 7(x-2)^2 - 8 at x=5x=5.

Adım Adım Çözüm

1
Express the fourth-degree polynomial in a form that reflects its symmetry.
P(x)=a(x−2)4+b(x−2)2+cP(x) = a(x - 2)^4 + b(x - 2)^2 + c
Since the graph of P(x)P(x) is symmetric about the line x=2x = 2, the polynomial expression in terms of (x−2)(x - 2) must contain only even powers.
2
Use the divisibility condition to establish an equation for the coefficients.
P(3)=a(3−2)4+b(3−2)2+c=a+b+c=0P(3) = a(3-2)^4 + b(3-2)^2 + c = a + b + c = 0
The divisor x2−4x+3x^2 - 4x + 3 factors into (x−1)(x−3)(x-1)(x-3). By the Factor Theorem, P(3)=0P(3) = 0 and P(1)=0P(1) = 0.
3
Use the given value P(2)=−8P(2) = -8 to find the constant term cc.
c=−8c = -8
Substituting x=2x = 2 into P(x)=a(x−2)4+b(x−2)2+cP(x) = a(x - 2)^4 + b(x - 2)^2 + c makes the terms with (x−2)(x-2) equal to zero, leaving P(2)=cP(2) = c.
4
Formulate a system of linear equations for aa and bb.
a+b=8a + b = 8 and 4a+b=114a + b = 11
Substituting c=−8c = -8 into a+b+c=0a + b + c = 0 gives a+b=8a + b = 8. By the Remainder Theorem, P(4)=36P(4) = 36, which gives a(4−2)4+b(4−2)2−8=36  ⟹  16a+4b=44  ⟹  4a+b=11a(4-2)^4 + b(4-2)^2 - 8 = 36 \implies 16a + 4b = 44 \implies 4a + b = 11.
5
Solve the system of equations for aa and bb.
a=1a = 1 and b=7b = 7
Subtracting a+b=8a + b = 8 from 4a+b=114a + b = 11 yields 3a=3  ⟹  a=13a = 3 \implies a = 1, which then gives b=7b = 7.
6
Evaluate the polynomial at x=5x = 5.
P(5)=136P(5) = 136
Substitute a=1a = 1, b=7b = 7, c=−8c = -8, and x=5x = 5 into the symmetric polynomial form: P(5)=(5−2)4+7(5−2)2−8=34+7(32)−8=81+63−8=136P(5) = (5-2)^4 + 7(5-2)^2 - 8 = 3^4 + 7(3^2) - 8 = 81 + 63 - 8 = 136.

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