Is it possible to factor y=x^4 - 2x^3 - 13x^2 + 14x + 24 y=x4−2x3−13x2+14x+24? If so, what are the factors?

1 Answer
George C.
Jan 19, 2016

Use the rational root theorem to help find the first two factors, then divide and factor the remaining quadratic to find:

y = x^4-2x^3-13x^2+14x+24y=x4−2x3−13x2+14x+24

=(x+1)(x-2)(x-4)(x+3)=(x+1)(x−2)(x−4)(x+3)

Explanation:

Let f(x) = x^4-2x^3-13x^2+14x+24f(x)=x4−2x3−13x2+14x+24

By the rational root theorem, any rational roots of f(x) = 0f(x)=0 must be of the form p/qpq for some integers pp and qq with pp as factor of the constant term 2424 and qq a factor of the coefficient 11 of the leading term.

That means that the only possible rational zeros are:

+-1±1, +-2±2, +-3±3, +-4±4, +-6±6, +-8±8, +-12±12, +-24±24

Try the first few in turn:

f(1) = 1-2-13+14+24 = 24f(1)=1−2−13+14+24=24

f(-1) = 1+2-13-14+24 = 0f(−1)=1+2−13−14+24=0

f(2) = 16-16-52+28+24 = 0f(2)=16−16−52+28+24=0

So x=-1x=−1 and x=2x=2 are zeros and (x+1)(x+1) and (x-2)(x−2) are factors:

x^4-2x^3-13x^2+14x+24x4−2x3−13x2+14x+24

=(x+1)(x^3-3x^2-10x+24)=(x+1)(x3−3x2−10x+24)

=(x+1)(x-2)(x^2-x-12)=(x+1)(x−2)(x2−x−12)

To factor the remaining quadratic, find a pair of factors of 1212 that differ by 11. The pair 4, 34,3 works, hence we find:

=(x+1)(x-2)(x-4)(x+3)=(x+1)(x−2)(x−4)(x+3)

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