How do you factor y=x3−5x2−2x+24 ?

1 Answer
George C.
Jan 2, 2016

Use the rational root theorem to get started, then factor the remaining quadratic to find:

x3−5x2−2x+24=(x+2)(x−4)(x−3)

Explanation:

Let f(x)=x3−5x2−2x+24

By the rational root theorem, any rational zeros of f(x) must be expressible in the for pq for integers p, q with p a divisor of the constant term 24 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational zeros are the factors of 24, namely:

±1,±2,±3,±4,±6,±12,±24

Try each in turn:

f(1)=1−5−2+24=18

f(−1)=−1−5+2+24=20

f(2)=8−20−4+24=8

f(−2)=−8−20+4+24=0

So x=−2 is a zero and (x+2) is a factor.

x3−5x2−2x+24=(x+2)(x2−7x+12)

We can factor x2−7x+12 by noting that 4×3=12 and 4+3=7, so:

x2−7x+12=(x−4)(x−3)

Putting it all together:

x3−5x2−2x+24=(x+2)(x−4)(x−3)

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