8
$\begingroup$

If $x+1$ is a factor of $ax^4 + bx^2 + c$, find the value of $a + b + c$? I know that it is equal to zero, but I have to know How to do it.

$\endgroup$
37
$\begingroup$

All you need here is the following fact:

If $(x-r)$ is a factor of the polynomial $P(x)$, then $P(r) = 0$.

$\endgroup$
14
$\begingroup$

Hint: Try dividing $ax^4+bx^2+c$ by $x+1$ through long division. What is the remainder? What should it be?

$\endgroup$
6
$\begingroup$

If you are familiar with synthetic division, you can use that technique to divide $ax^4 + bx^2 + c$ by $x+1$ and find the remainder. Otherwise, you could use long division.

Regardless of what technique you use, for $x+1$ to cleanly divide $ax^4 + bx^2 + c$, then the remainder must be zero. Once you find the remainder, assign values to $a, b, c$ in order to make that remainder zero.

EDIT: Leaving this here as an alternate solution. TonyK's approach is far superior.

$\endgroup$
1
$\begingroup$

As a shortcut for the long division, note that if $x+1$ is a factor of $ax^4+bx^2+c=0$, then $x-1$ is also a factor. Divide $ax^4+bx^2+c$ by $(x+1)(x-1)=x^2-1$ to get the remainder $a+b+c$, which must be equal to $0$.

$\endgroup$
0
$\begingroup$

Let $p(x)=ax^4+bx^2+c$. We are given $p(x)=(x+1)q(x)$ for $q(x)$ some cubic polynomial. Now evaluate it at $x=-1$ and we get $p(-1)=a(-1)^4+b(-1)^2+c=a+b+c=0\cdot q(x)=0$ so $a+b+c=0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.