Finding the last root of $p(x) = x^5 + a_3 x^3 + a_2 x^2 + a_1x + a_0$ given that...

Question

The polynomial $p(x) = x^5 + a_3 x^3 + a_2 x^2 + a_1x + a_0$ has real coefficients and has 2 roots of $x = -3$, and two roots of $x=4$. What is the last root, and how many times does it occur?

At first I expanded $(x+3)^2(x-4)^2$ to get a divisor to divide the polynomial with, but I could not get around the fact that the coefficients are not given, so I can't get any concrete number. Here's the expansion: $$x^4 - 2x^3 - 23x^2 + 24x + 144$$ I don't quite get how to use it to help me find the last root.

Answer 1

Using Vieta formula $$x_1+x_2+x_3+x_4+x_5=-{a_4\over a_5} =-{0\over 1}=0$$ you can get $x_5$. And then you can also calculate all other coefficients $a_0,a_1...$.

Answer 2

Hints:

1. Coefficient of $x^4$ in $p(x)$ is zero.
2. The polynomial $p(x)$ can be written as $(x-a)(x-4)^2(x+3)^2$,where $a$ is the last root.

Answer 3

The leading coefficient is $1$. So the sum of the roots is $-a_4=0$. That’s all you need!

Answer 4

If $(x-r)$ is a repeated root of $p(x)$ then $p(r) =p'(r) =0$.

So $p(-3)=p(4)=p'(-3)=p'(4)=0$, allowing you to form a system of four linear equations in terms of four unknowns. Solve for the coefficients and find the last root (which has to be a single root as the polynomial is quintic).

Answer 5

You know that $q(x)=x^4-2x^3-23x^2+24x+144$ is a factor of $p(x)=q(x)r(x)$

Consider $s(x)=p(x)-xq(x)=q(x)(r(x)-x)$ which also has a factor $q(x)$

$$p(x)-xq(x)=x^5+a_3x^3+a_2x^2+a_1x+a_0-x^5+2x^4+23x^3-24x^2-144x=$$$$=2x^4+(a_3+23)x^3+(a_2-24)x^2+(a_1-144)x+a_0$$

Now this is divisible by $q(x)$, and comparing the coefficients of $x^4$ it is clearly equal to $2q(x)$

So $p(x)=(x+2)q(x)$ and you can get the unknown coefficients by comparing coefficients in $s(x)$.

This is simply a way of writing out the polynomial division involved here.