How do you find the zeros of f(x)=2x3−5x2−28x+15?

1 Answer
George C.
Jul 7, 2016

Zeros: x=12, x=5, x=−3

Explanation:

f(x)=2x3−5x2−28x+15

By the rational root theorem, since f(x) has integer coefficients, any rational zeros must be expressible in the form pq for integers p,q with p a divisor of the constant term 15 and q a divisor of the coefficient 2 of the leading term.

That means that the only possible rational zeros of f(x) are:

±12,±1,±32,±52,±3,±5,±152,±15

Trying the first one we find:

f(12)=28−54−282+15=1−5−56+604=0

So x=12 is a zero and (2x−1) a factor:

2x3−5x2−28x+15

=(2x−1)(x2−2x−15)

To factor the remaining quadratic, find a pair of factors of 15 which differ by 2. The pair 5,3 works, so:

x2−2x−15=(x−5)(x+3)

So the two remaining zeros are x=5 and x=−3

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