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

1 Answer
Jan 30, 2016

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 #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]}