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]}