Question

How do you factor `2x^4-19x^3+57x^2-64x+20`?

Answer 1

Use the rational root theorem to help find the first couple of factors and separate them out to find:

`2x^4-19x^3+57x^2-64x+20 = (2x-1)(x-2)(x-2)(x-5)`

Explanation:

Let `f(x) = 2x^4-19x^3+57x^2-64x+20`

By the rational root theorem any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p` and `q` with `p` a divisor of the constant term `20` and `q` a divisor of the coefficient `2` of the leading term.

So the only possible rational roots are:

`+-1/2`, `+-1`, `+-2`, `+-5/2`, `+-4`, `+-5`, `+-10`, `+-20`

Trying these in turn we find:

`f(1/2) = 1/8-19/8+57/4-32+20 = (1-19+114-256+160)/8 = 0`

So `x=1/2` is a zero and `(2x-1)` is a factor:

`2x^4-19x^3+57x^2-64x+20 = (2x-1)(x^3-9x^2+24x-20)`

Having 'used up' the factor of `2` on the leading coefficient, the remaining possible rational roots are:

`+-1`, `+-2`, `+-4`, `+-5`, `+-10`, `+-20`

Let `g(x) = x^3-9x^2+24x-20`

We find:

`g(1) = 1-9+24-20 = -4`

`g(-1) = -1-9-24-20 = -54`

`g(2) = 8-36+48-20 = 0`

So `x=2` is a zero and `(x-2)` is a factor:

`(2x-1)(x^3-9x^2+24x-20) = (2x-1)(x-2)(x^2-7x+10)`

The remaining quadratic can be factored by noticing that `2+5 = 7` and `2xx5 = 10`, hence:

`(2x-1)(x-2)(x^2-7x+10) = (2x-1)(x-2)(x-2)(x-5)`

graph{2x^4-19x^3+57x^2-64x+20 [-7.42, 12.58, -5.8, 4.2]}