Given that $ P(x) = x^5 + ax^4 + bx^3 + c^2 + dx + e$ is divisible by both $x - 2$ and

Question

Given that $ P(x) = x^5 + ax^4 + bx^3 + c^2 + dx + e$ is divisible by both $x - 2$ and $ (x + 1)^2 $ Also, $ P(0) = 6$, $P'(-1) = 0$. Find the polynomial. Here $c^2$ is a constant term; there is no $x^2$ term.

Accepted Answer

$P(x)=x^5-\dfrac23x^4-6x^3+\dfrac{31}{3}x+6$

Polynomials, Cubic Equations and Partial Fractions - Forming Polynomials from Roots and Factors

Polynomials, Cubic Equations and Partial Fractions - Forming Polynomials from Roots and Factors

Solution:

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