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

##### 1 Answer

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

By the rational root theorem any rational zeros of

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

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

Having 'used up' the factor of

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

Let

We find:

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

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

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

So

#(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

#(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]}