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

1 Answer
Jan 30, 2016

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]}