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Let $ P(x) $ be a $ 7 $ degree polynomial with the coefficient of $ x^7 $ equal to $ 1 $. Let $ a \in\mathbb{R} $ such that $ P(x)-a $ divides with $ (x+1)^4 $ and $ P(x)+a $ divides with $ (x-1)^4 $

$ 1 ) $ Find the coefficient of $ x^5 $

The answer should be $ -\frac{21}{5} $

$ 2 ) $ Find $ a $

The answer should be $ \frac{16}{5} $

SADBOYS
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    Divides "with"? Does $(x+1)^4$ divide $P(x)+a$ for every real $x$? What if $x=-1$? –  Mar 10 '18 at 13:21

3 Answers3

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$(x+1)^4 \mid (P(x)-a) \implies P(x) = f(x)(x+1)^4+a \implies (x+1)^3 \mid P'(x)$.

Similarly we have $(x-1)^3 \mid P'(x)$, so putting these together with $P(x)$ being monic of seventh degree, $P'(x) = 7(x^2-1)^3$.

$\implies P(x) = x^7 -\frac{21}5x^5+7x^3-7x + C$,

Now $P(1) = -a, P(-1) = a \implies C-\frac{16}5=-a, C + \frac{16}5=a \implies C=0, a = \frac{16}5$.

Macavity
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Let $$P(x)=Ax^7+Bx^6+Cx^5+Dx^4+Ex^3+Fx^2+Gx+H$$ so using long division we have

$$\begin{align}&\underline{\small Ax^3+(B-4A)x^2+(C-4B+10A)x+(D-4C+10B-20A)}\\\small x^4+4x^3+6x^2+4x+1\,/ &\small Ax^7+Bx^6+Cx^5+Dx^4+Ex^3+Fx^2+Gx+H\\&\underline{\small Ax^7+4Ax^6+6Ax^5+4Ax^4+Ax^3}\\&\small(B-4A)x^6+(C-6A)x^5+(D-4A)x^4+(E-A)x^3+Fx^2+Gx+H\\&\small\underline{(B-4A)x^6+(4B-16A)x^5+(6B-24A)x^4+(4B-16A)x^3+(B-4A)x^2}\\&\small(C-4B+10A)x^5+(D-6B+20A)x^4+(E-4B+15A)x^3+(F-B+4A)x^2+Gx+H\\&\underline{\small(C-4B+10A)x^5+(4C-16B+40A)x^4+(6C-24B+60A)x^3+(4C-16B+40A)x^2+(C-4B+10A)x}\\&\small(D-4C+10B-20A)x^4+(E-6C+20B-45A)x^3+(F-4C+15B-36A)x^2+(G-C+4B-10A)x+H\\&\small(D-4C+10B-20A)x^4+(4D-16C+40B-80A)x^3+(6D-24C+60B-120A)x^2+(4D-16C+40B-80A)x+(D-4C+10B-20A)\\&\overline{\small(E-4D+10C-20B+35A)x^3+(F-6D+20C-45B+84A)x^2+(G-4D+15C-36B+70A)x+(H-D+4C-10B+20A)}\end{align}$$

so we have the equations $$E-4D+10C-20B+35A=0\\F-6D+20C-45B+84A=0\\G-4D+15C-36B+70A=0\\H-D+4C-10B+20A=0$$

Can you do something similar for $(x-1)^4=x^4-6x^3+4x^2-6x+1$ and solve the eight equations in total?

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Note that $$P(x)=x^7+\sum_{k=0}^6 c_k x^k=(x+1)^4Q_1(x)+a=(x-1)^4Q_2(x)-a$$ Hence the coefficients $c_0,c_1,\dots, c_6$ and $a$ ($8$ unknowns) satisfy the following linear system ($8$ equations): $$\begin{cases} P(1)=-a\\ P'(1)=0\\ P''(1)=0\\ P'''(1)=0\\ P(-1)=a\\ P'(-1)=0\\ P''(-1)=0\\ P'''(-1)=0 \end{cases}$$ Then $$0=-P''(1)+P''(-1)+P'''(1)+P'''(-1)=336+80c_5\implies c_5 = -21/5.$$ By solving the system we obtain also the other coefficients: $$ c_0 = 0,\; c_1= -7,\; c_2= 0,\; c_3= 7,\; c_4= 0,\; c_5 = -21/5,\; c_6 = 0,\;a = 16/5.$$

Robert Z
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