# Factor theorem and zeroes of a polynomial

What should be added to $x^3-2x^2-12x-9$ such that it is completely divisible by $x^2+x-6$? I factorized $x^2+x-6$ as $(x+3)(x-2)$. I am unable to understand how to make use of factor theorem to arrive at the solution of this problem. But by actual division we get that $3x+27$ must be added such that $f(-3)=0$ and $f(2)=0$.

Please give me the solution through factor theorem only.

• Try long division May 23, 2014 at 6:32
• Try remainder theorem(almost same as factor theorem) May 23, 2014 at 8:09

Hint: why dont you find the remainder. Subtracting the remainder from your polynomial will give you your answer..

• That is true. But by using how can I find the solution using factor theorem May 23, 2014 at 6:40

If a number is multiple of $6$, it must be multiple of $2$ and $3$.
Similarly, if you want a polynomial to be divisible by $(x-2)(x+3)$, you can make it to be divisible by $x-2$ and $x+3$.

So you can consider $x^3-12x^2-2x-9+(ax+b)$ or $x^3-12x^2+(a-2)x+(b-9)$. Now, substitute $x$ by $2$ and this must be $0$. Then substitute by $-3$ and equalize to $0$. (This is the part where you use the factor theorem).

You get so a system of two equations with two unknowns (namely, $a$ and $b$). Solve it and you are done.

Add $ax+b$ to the given polynomial to get $$f(x)=(x^3-2x^2-12x-9)+(ax+b)\ .$$ If $x-2$ is a factor of this then $f(2)=0$, that is, $$-33+2a+b=0\ .$$ Can you use the factor theorem again to find another equation connecting $a$ and $b$, and hence to solve the problem?

• This is convincing. But how does one decide that one should add a*x+b. Please elaborate May 23, 2014 at 7:06
• this is convincing. How do you decide a*x+b May 23, 2014 at 7:11
• You have two independent conditions (divisibility by $x-2$ and by $x+3$), so you need two unknowns to be sure that you can find a solution, hence $ax+b$. In fact there are other answers too - for example, if you have found $f(x)$ which is divisible by $x^2+x-6$, and if you then add $x^2+x-6$ again, you get $f(x)+(x^2+x-6)$ which is also divisible by $x^2+x-6$. You can find many other examples too. But adding $ax+b$ is the simplest way. Of course it might turn out that $a=0$, so you could have just added $b$. But you can't be sure of this in advance, so you need $ax+b$ just in case. May 23, 2014 at 7:31
• thank you Sir for helping me resolve May 23, 2014 at 12:41