How do you solve $(2 x + 8) (3 x - 6) = 0$ $(2x+8)(3x-6)=0$ using any method?

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How do you solve $(2 x + 8) (3 x - 6) = 0$ $(2x+8)(3x-6)=0$ using any method?

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Answer 1 · 2017-05-04T01:42:56

$x = - 4 or x = 2$ $x=-4 or x=2$

Explanation:

Expand the equation from the brackets

$(2 x + 8) (3 x - 6) = 0$ $(2x+8)(3x-6)=0$
$(2 x$ $(color(red)(2x)$ +8)( $3 x - 6)$ $color(red)(3x)-6)$ Multiply the highest power of $x$ $x$ first
$(2 x$ $(color(red)(2x)$ +8)(3x $- 6$ $color(red)(-6)$ ) Next multiply by the next highest power

(2x $+ 8$ $color(blue)(+8$ )( $3 x$ $color(blue)(3x)$ -6) Multiply by the highest power
(2x $+ 8$ $color(blue)(+8$ )(3x $- 6$ $color(blue)(-6)$ ) Multiply by next highest power.

Therefore,
$6 x^{2} - 12 x + 24 x - 48 = 0$ $6x^2-12x+24x-48=0$
$6 x^{2} + 12 x - 48$ $6x^2+12x-48$ =0 ------------------Divide the equation by 6
$x^{2} + 2 x - 8 = 0$ $x^2+2x-8=0$ --------------------------Simplify the equation
$(x + 4) (x - 2) = 0$ $(x+4)(x-2)=0$
$x + 4 = 0 or x - 2 = 0$ $x+4=0 or x-2=0$
$x = - 4 or x = 2$ $x=-4 or x=2$

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How do you solve $(2 x + 8) (3 x - 6) = 0$ $(2x+8)(3x-6)=0$ using any method?

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How do you solve (2x+8)(3x-6)=0(2x+8)(3x−6)=0(2x+8)(3x-6)=0 using any method?

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How do you solve $(2 x + 8) (3 x - 6) = 0$ $(2x+8)(3x-6)=0$ using any method?