# If $x^2-1$ is a factor of $p(x)=ax^4+bx^3+cx^2+dx+e$, prove that: $a+b+c=b+d=0$

If $x^2-1$ is a factor of $p(x)=ax^4+bx^3+cx^2+dx+e$, prove that: $a+c+e=b+d=0$.

My Working. Here, $x^2-1$ is a factor of $p(x)$ $$x=+/- 1$$. When $x=1$, $$p(1)=a+b+c+d+e$$ $$0=a+b+c+d+e$$. When $x=-1$, $$p(-1)=a-b+c-d+e$$ $$0=a-b+c-d+e$$.

What should I do next?

• Have you tried setting $0=0$? It would be the obvious next step.
– Nij
Sep 6, 2016 at 0:11
• @Nij, Doing that, I get $b+d=-(b+d)$.
– pi-π
Sep 6, 2016 at 0:16
• You might have a mistake in you question :It should be $a+c+e=b+d=0$ right?
– 王李远
Sep 6, 2016 at 0:24

Now $$a+b+c+d+e=a-b+c-d+e$$ so $$2b+2d=0 \Rightarrow b+d=0$$ and immediately $$a+b+c+d+e=a+c+e+(b+d)=0 \Rightarrow a+c+e=0$$ Similarly $$a+b+c+d+e+a-b+c-d+e=0+0=0$$ so $$2(a+c+e)=0 \Rightarrow a+c+e=0$$

• Are we done with this?
– pi-π
Sep 6, 2016 at 0:19
• @user354073 Not quite. You ask for $a+b+c=0$, while so far only the sums of all five, the odd-indexed and the even-indexed, elements is zero.
– Nij
Sep 6, 2016 at 0:21
• Then what should be done further to get the proof?
– pi-π
Sep 6, 2016 at 0:22

What you did so far is good. Now compute

$$p(1) - p(-1)$$

and

$$p(1) + p(-1)$$

using the expressions you have for each one. What do you get and what can you conclude?

$p(x)$ has the form $$k\cdot (x-x_0)\cdot(x-x_1)\cdot(x^2-1)=\\kx^4-k(x_0+x_1)x^3+k(x_0x_1-1)x^2+k(x_0+x_1)x-kx_0x_1$$

for some $k,x_0,x_1$. Hence, we have $$a=k\\ b=-k(x_0+x_1)\\ c=k(x_0x_1-1)\\ d=k(x_0+x_1)\\ e=-kx_0x_1$$

So unless I'm mistaken something's wrong here. We do get $b+d=0$, but we get $a+b+c=k(x_0x_1-x_0-x_1)$.

• @Nij Did you downvote? Because the conclusion I drew is the same as yours. The question asked to show $a+b+c=0$, which is not generally true, while my answer also finds that $a+c+e=0$. Sep 6, 2016 at 0:32
• No, I didn't downvote. I reread question to check my own as well; I believe OP wrote the question wrong and your answer supports that thought.
– Nij
Sep 6, 2016 at 0:51