How do you factor # 20x^4+16x^3-5x-4#?

1 Answer
George C.
Oct 2, 2016

#20x^4+16x^3-5x-4 = (4x^3-1)(5x+4)#

#color(white)(20x^4+16x^3-5x-4) = (root(3)(4)x-1)(2root(3)(2)x^2+root(3)(4)x+1)(5x+4)#

Explanation:

Notice that the ratio of the first and second terms is the same as that of the third and fourth terms. So this quadrinomial will factor by grouping:

#20x^4+16x^3-5x-4 = (20x^4+16x^3)-(5x+4)#

#color(white)(20x^4+16x^3-5x-4) = 4x^3(5x+4)-1(5x+4)#

#color(white)(20x^4+16x^3-5x-4) = (4x^3-1)(5x+4)#

We can factor #4x^3-1# as a difference of cubes using irrational coefficients:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

So:

#4x^3-1 = (root(3)(4)x)^3-1^3#

#color(white)(4x^3-1) = (root(3)(4)x-1)((root(3)(4)x)^2+root(3)(4)x+1)#

#color(white)(4x^3-1) = (root(3)(4)x-1)(root(3)(16)x^2+root(3)(4)x+1)#

#color(white)(4x^3-1) = (root(3)(4)x-1)(2root(3)(2)x^2+root(3)(4)x+1)#

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