How do you find all the zeros of x^3+2x^2-2x-3x3+2x2−2x−3?

1 Answer
George C.
Feb 27, 2016

Use the rational root theorem to help find the first zero x=-1x=−1, then divide f(x)f(x) by (x+1)(x+1) to find a quadratic and hence the other two zeros:

x = (-1+-sqrt(13))/2x=−1±√132

Explanation:

f(x) = x^3+2x^2-2x-3f(x)=x3+2x2−2x−3

By the rational root theorem, any rational zeros of f(x)f(x) will be expressible in the form p/qpq for some integers pp and qq with pp a divisor of the constant term -3−3 and qq a divisor of the coefficient 11 of the leading term.

That means that the only possible rational zeros are:

+-1±1, +-3±3

Trying each in turn, we find:

f(1) = 1+2-2-3 = -2f(1)=1+2−2−3=−2

f(-1) = -1+2+2-3 = 0f(−1)=−1+2+2−3=0

So x=-1x=−1 is a zero and (x+1)(x+1) a factor:

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

The remaining quadratic factor is of the form ax^2+bx+cax2+bx+c with a=1a=1, b=1b=1 and c = -3c=−3. This has zeros given by the quadratic formula:

x = (-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a

=(-1+-sqrt(1-(4*1*-3)))/(2*1)=−1±√1−(4⋅1⋅−3)2⋅1

=(-1+-sqrt(13))/2=−1±√132

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