How do you find all the zeros of f(x)=x3+x2−7x+2?

1 Answer
George C.
Apr 16, 2016

f(x) has zeros x=2 and x=−32±√132

Explanation:

By the rational root theorem, any rational zeros of f(x) must be expressible in the form pq for integers p,q with p a divisor of the constant term 2 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational zeros are:

±1, ±2

We find:

f(2)=8+4−14+2=0

So x=2 is a zero and (x−2) a factor:

x3+x2−7x+2=(x−2)(x2+3x−1)

We can factor the remaining quadratic expression by completing the square. I will multiply by 4 first to cut down on the fractions involved:

4(x2+3x−1)

=4x2+12x−4

=(2x+3)2−9−4

=(2x+3)2−(√13)2

=((2x+3)−√13)((2x+3)+√13)

=(2x+3−√13)(2x+3+√13)

Hence x=−32±√132

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