Factoring Polynomial with Four Terms

Question

To begin, I have the following equation: $f(x)=2x^2-5x$. Now I need to find the roots of $f(f(x))$. I also know that $f(x(x))$ simplifies to $8x^4-40x^3+40x^2+25x$. It's very obvious to me that a $x$ can be factored out, but how can I further factor $8x^3-40x^2+40x+25$?

Answer 1

let's call $y=2x^2-5x$. So you are looking for the roots of $f(y)=0$. We know that $f(y)=2y^2-5y$, so the solutions for $f(y)=0$ are $y=0$ and $y=5/2$. Now, we can find 4 roots of $f(f(x))$ by solving these two quadratic equations:

• $y=2x^2-5x=0$
• $y=2x^2-5x=5/2$

Which give us:

• $x=0$
• $x=5/2$
• $x=2.927050983$
• $x=-0.4270509831$

Answer 2

It's $+25$ not $-25$. Now employ the rational roots theorem to get one of the rational roots. Then use long division of polynomials to get it down to a quadratic, then invoke the quadratic formula.

If you use hardmath's hint you can get the rational root fairly expediently, and with mm-aops' hint you can remove the need for polynomial long division.