How do you find all the zeros of $f(x) = 9x^2 + 6x - 8$ ?

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Answer 1 · 2016-08-05T22:43:47

The zeros of $f(x)$ are $x=2/3$ and $x=-4/3$

$f(x) = 9x^2+6x-8$

Use an AC method to factor $f(x)$ :

Find a pair of factors of $AC=9*8=72$ with difference $B=6$ .

The pair $12, 6$ works.

Use this pair to split the middle term and factor by grouping:

$9x^2+6x-8$

$=9x^2+12x-6x-8$

$=(9x^2+12x)-(6x+8)$

$=3x(3x+4)-2(3x+4)$

$=(3x-2)(3x+4)$

Hence the zeros of $f(x)$ are $x=2/3$ and $x=-4/3$

How do you find all the zeros of f(x) = 9x^2 + 6x - 8f(x) = 9x^2 + 6x - 8?