Given a root and divisor, how can I find the value of 2 variables in a polynomial?

Question

Let $P\left(x\right)=x^3-2ax^2+bx+18$. Knowing that this polynomial is, in an instance, divisible by $x^2-x-6$, and in another, has a double root of -1, what are the values of a and b? I'm having problems solving this kind of problem. What's the best approach to solve these?

I know we're not really supposed to post textbook problems, but I'm learning math all by my own and I can't find an answer anywhere else. It would be very appreciated if you explain the full logic and implications for the solution and not merely an algorithm.

Answer 1

You actually have two different problems there. Telling you that $x^2-x-6$ is a factor is sufficient to determine $a$ and $b$, as is telling you that $-1$ is a double root. You will get different values for $a$ and $b$ in the two cases.

For the first, just perform the division. You need to have the second stage, the one involving $x^2$, come out even. The constants will tell you what constant you are multiplying by and you need $a$ and $b$ correct to come out.

For the second, you are told that $(x+1)^2$ is a factor. You can do it the same way, by division.

Answer 2

Hint: Write $$x^3-2ax^2+bx+18=(x^2-x-6)(x-c)$$ You will get the System $$18=6c$$, $$c-6=b$$,$$-c-1=-2a$$ Expanding the right side we get

$$x^3-x^2(c+1)+x(c-6)+6c$$ and now compare the coeficients.

Answer 3

When it has a double root it has a common root with its derivative. Take derivative and and put $-1$ in it it gives you one equation.The roots of divisor are -3 and 2 .putting in polynomial gives two more equations which with first equation make a system of equation which gives a and b.I think it must have another constant to be found.