How do you use the rational roots theorem to find all possible zeros of f(x) = x^3 - 12x - 16f(x)=x3−12x−16?

1 Answer
George C.
Jun 18, 2016

The zeros of this f(x)f(x) are all rational:

x=-2x=−2 with multiplicity 22

x=4x=4

Explanation:

f(x) = x^3-12x-16f(x)=x3−12x−16

By the rational roots theorem, any rational zeros of f(x)f(x) must be expressible in the form p/qpq for integers p, qp,q with pp a divisor of the constant term -16−16 and qq a divisor of the coefficient 11 of the leading term.

That means that the only possible rational zeros are:

+-1±1, +-2±2, +-4±4, +-8±8, +-16±16

Also note that the signs of the coefficients of f(x)f(x) follow the pattern + - -+−−, with one change of sign. Hence f(x)f(x) has exactly one positive zero.

So start trying the positive possibilities first:

f(1) = 1-12-16 = -27f(1)=1−12−16=−27

f(2) = 8-24-16 = -32f(2)=8−24−16=−32

f(4) = 64-48-16 = 0f(4)=64−48−16=0

So x=4x=4 is a zero and (x-4)(x−4) a factor:

x^3-12x-16 = (x-4)(x^2+4x+4) = (x-4)(x+2)(x+2)x3−12x−16=(x−4)(x2+4x+4)=(x−4)(x+2)(x+2)

So x=-2x=−2 is the other zero of f(x)f(x), with multiplicity 22

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