# Find the value of a+b+c

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.

All you need here is the following fact:

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

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

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.

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 $0$.

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 $a+b+c=0$