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

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

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Answer 1 · 2016-08-05T22:43:47

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

Explanation:

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

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

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

The pair $12, 6$ $12, 6$ works.

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

$9 x^{2} + 6 x - 8$ $9x^2+6x-8$

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

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

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

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

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

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

1 Answer

Explanation:

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How do you find all the zeros of f(x)=9x2+6x−8f(x) = 9x^2 + 6x - 8?

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