Question:
Let $(x-1)^3$ divides $p(x)+1$ and $(x+1)^3$ divides $p(x)-1$. Find the polynomial $p(x)$ of degree $5$ .
Please help me with this question, I do not know how to proceed.
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1Can you please show your attempt (your work) to solve the problem and where are you stuck at? – The 2nd Feb 20 '20 at 11:40
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Sorry ,can't have any progress as do not know how to solve such questions – Aayush Feb 20 '20 at 11:41
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I do not know if it's correct or not but I arrived at this : p[(x+1)^3]-p[(x-1)^3]=2 – Aayush Feb 20 '20 at 11:44
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The solution you propose in a comment below is not correct. Indeed for your solution we have $p(x)-1=x^5 -5x^3 +3x -1=(x + 1) (x^4 - x^3 - 4 x^2 + 4 x - 1)$ so it is not divisible by $(x+1)^3$. – lulu Feb 20 '20 at 12:14
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Oh yes you are correct – Aayush Feb 20 '20 at 12:21
3 Answers
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From the problem, we can know that: $\text{deg } P' = 4$ (highest degree of derivative of $P(x)$ is equal to $4$); $(x-1)^2$ and $(x+1)^2$ divides $P'(x)$
Therefore:
$$P'(x)=a(x-1)^{2}(x+1)^{2}$$ $$=a(x^{4}-2x^{2}+1)$$ $$ \implies P(x)=a \left(\frac{x^{5}}{5} - \frac{2}{3} x^{2} + x + b\right) \text{ (1)}$$
Since $(x-1)^3$ and $(x+1)^3$ divides $P(x)$, we get $P(1)=-1$ and $P(-1)=1$. Subsitute it into $(1)$ and we get $a=-\dfrac{15}{8}, b=0$
Therefore, the answer is:
$$P(x)=-\dfrac{1}{8}(3x^{5} - 10 x^{2} +15x)$$
The 2nd
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Sorry , that was wrong. But I have a doubt there. Can you explain me how the deg. Of p=4. And how you chose (x-1)^2 And (x+1)^2 as factors of p'(x). Please explain this to me , rest I understood.... – Aayush Feb 20 '20 at 12:25
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As P(x) has degree of 5, if you take the derivative of P(x), its degree will be down by 1 (by the power rule). – The 2nd Feb 20 '20 at 12:29
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$P'(x) = q(x)(x-1)^2$ ($(x-1)^2$ divides P'(x)), and $P'(x) = q(x)(x+1)^2$ ($(x+1)^2$ divides P'(x)). But P'(x) has a degree of 4, so $P'(x) =a (x+1)^2 (x-1)^2$ – The 2nd Feb 20 '20 at 12:35
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COMMENT.-The unknown factors of the polynomial $p(x)$ must be quadratics and we choose all polynomial unitary. We have
$$\begin{cases}p(x)+1=(x-1)^3(x^2+ax+b)\\p(x)=x^5+(a-3)x^4+(b-3a+3)x^3+(-3b+3a-1)x^2+(3b-a)x-b-1\\p(x)-1=(x+1)^3(x^2+cx+d)\\p(x)=x^5+(c+3)x^4+(d+3c+3)x^3+(3d+3c-1)x^2+(3d+c)x+d+1\end{cases}$$ You now have to match the coefficients what gives you $a = c + 6$ and $b = - (d + 2)$ reducing the equations to two unknowns $c$ and $d$ so you can easily find the values $a,b,c$ and $d$.
Piquito
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When $(x-a)^3$ divides a polynomial $f(x)$, then $f(a)=0, f'(a)=0, f''(a)=0$ For the case here we get $p(1)=-1, p'(1)=0, p''(1)=0$ and $p(-1)=-1, p'(-1)=0, p''(-1)=0$. This means we can determine $p(x)$ which is rth degree polynomial. Get $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$, we get the following equationbs: $$a+b+c+d+e+f=-1~~~~(1)$$ $$-a+b-c+d-e+f=1~~~(2)$$ $$5a+4b+3c+2d+e=0~~~(3)$$ $$5a-4b+3c-2d+e=0~~~~(4)$$ $$10a+6b+3c+d=0~~~~~(5)$$ $$-10a+6b-3c+d=0~~~~(6)$$ Solving these equations, we get $a=-\frac{3}{8}, b=0, c=\frac{5}{4}, d=0, e=-\frac{15}{8},f=0.$ So the polynomial $$P(x)=-\frac{3}{8}x^5+\frac{5}{4}x^3-\frac{15}{8}x$$
