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What are the x-intercepts of the equation f(x))=3x^2+10x-8 ?

1 Answer

Oct 26, 2017

x - intercepts are $(\frac{2}{3}, 0)$ $(2/3,0)$ and $(- 4, 0)$ $(-4,0)$

Explanation:

Given -

$f (x) = 3 x^{2} + 10 x - 8$ $f(x)=3x^2+10x-8$
$y = 3 x^{2} + 10 x - 8$ $y=3x^2+10x-8$

Put $y = 0$ $y=0$

$3 x^{2} + 10 x - 8 = 0$ $3x^2+10x-8=0$

$3 x^{2} - 2 x + 12 x - 8 = 0$ $3x^2-2x+12x-8=0$

$x (3 x - 2) + 4 (3 x - 2) = 0$ $x(3x-2)+4(3x-2)=0$

$(3 x - 2) (x + 4) = 0$ $(3x-2)(x+4)=0$

$3 x - 2 = 0$ $3x-2=0$

$x = \frac{2}{3}$ $x=2/3$

$x + 4) = 0$ $x+4)=0$

$x = - 4$ $x=-4$

x - intercepts are $(\frac{2}{3}, 0)$ $(2/3,0)$ and $(- 4, 0)$ $(-4,0)$

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