How do you find all the zeros of 2x3+9x2+6x−8?

1 Answer
George C.
Aug 13, 2016

This cubic has zeros −2 and 14(−5±√57)

Explanation:

f(x)=2x3+9x2+6x−8

By the rational roots theorem, any rational zeros of f(x) are expressible in the form pq for integers p,q with p a divisor of the constant term −8 and q a divisor of the coefficient 2 of the leading term.

That means that the only possible rational zeros are:

±12,±1,±2,±4,±8

We find:

f(−2)=2(−8)+9(4)+6(−2)−8=−16+36−12−8=0

So x=−2 is a zero and (x+2) a factor:

2x3+9x2+6x−8=(x+2)(2x2+5x−4)

The remaining quadratic is in the form ax2+bx+c with a=2, b=5, c=−4.

We can solve this using the quadratic formula:

x=−b±√b2−4ac2a

=−5±√52−4(2)(−4)2⋅2

=14(−5±√25+32)

=14(−5±√57)

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