How do you factor: $2x^4+x^3-7x^2-x+5$

Question

I don’t know how factor this, I’ve never factored something with a degree higher than 3.

Answer 1

In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.

You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.

The possible root combinations in this case are $-1,1,-\frac{1}{2},\frac{1}{2},-5,5,-\frac{5}{2}$, and $\frac{5}{2}$.

A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.

Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.

Answer 2

Hint $\ $ It is $\ (2x^4-7x^2+5) + (x^3-x)\, $ and both clearly have roots $\,x = \pm 1$

Answer 3

Hint:

This polynoial is divisible by $x^2-1$: \begin{align} 2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\\ &=2x^2(x^2-1)+ x^2(x-5)-x+5 \\ &= (x^2-1)(2x^2+x-5). \end{align}+

Answer 4

Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:

Divide out:

$${2 x^4+x^3-7 x^2-x+5 \over x-1} = 2 x^3+3 x^2-4 x-5$$

Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:

$${2 x^3+3 x^2-4 x-5 \over x+1} = 2 x^2+x-5.$$

So:

$$(x-1)(x+1)(2 x^2+x-5)$$

Use the quadratic equation:

$$(x-1)(x+1)(x - 1/4 (-1 - \sqrt{41}))(x+1/4 (-1 - \sqrt{41}))$$