How do you solve $3 x^{3} - 2 x^{2} - 12 x + 8 = 0$ $3x^3 -2x^2 -12x + 8 = 0$ ?

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How do you solve $3 x^{3} - 2 x^{2} - 12 x + 8 = 0$ $3x^3 -2x^2 -12x + 8 = 0$ ?

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Answer 1 · 2017-07-06T23:42:50

With a large polynomial like this, you'll need to factor in order to isolate the x and solve for it.

Explanation:

This is the equation you're given: $3 x^{3} - 2 x^{2} - 12 x + 8 = 0$ $3x^3-2x^2-12x+8=0$

Separate the polynomial into two parts, and factor that way.
$(3 x^{3} - 2 x^{2}) + (- 12 x + 8) = 0$ $(3x^3-2x^2)+(-12x+8)=0$
$x^{2} (3 x - 2) - 4 (3 x - 2)$ $x^2(3x-2)-4(3x-2)$
$(3 x - 2) (x^{2} - 4) = 0$ $(3x-2)(x^2-4)=0$

You can factor out $(x^{2} - 4)$ $(x^2-4)$ to $(x + 2) (x - 2)$ $(x+2)(x-2)$ .

The fully factored polynomial is $(3 x - 2) (x + 2) (x - 2)$ $(3x-2)(x+2)(x-2)$ .

But remember... the question asks you to solve the equation.
Set the factored polynomial to zero and solve for x (you will end up with three values of x):
$(3 x - 2) (x + 2) (x - 2) = 0$ $(3x-2)(x+2)(x-2)=0$

$3 x - 2 = 0$ $3x-2=0$ ---> $x = \frac{2}{3}$ $x=2/3$
$x + 2 = 0$ $x+2=0$ ---> $x = - 2$ $x=-2$
$x - 2 = 0$ $x-2=0$ ---> $x = 2$ $x=2$

Your final answers are $x = \frac{2}{3}, - 2, 2$ $x=2/3,-2,2$ .

Answer 2 · 2017-07-06T23:46:40

$x = - 2$ $x=-2$

$x = 2$ $x=2$

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

Explanation:

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

Common factors

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

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

Three solutions (B.S $=$ $=$ both sides)

(1) $x + 2 = 0$ $" "x+2=0" "$ ( $- 2$ $-2$ B.S)

$x = - 2$ $x=-2$

(2) $x - 2 = 0$ $" "x-2=0" "$ ( $+ 2$ $+2$ B.S)

$x = 2$ $x=2$

(3) $3 x - 2 = 0$ $" "3x-2=0" "$ ( $+ 2$ $+2$ B.S)

$3 x = 2$ $3x=2" "$ ( $\div 3$ $div3$ B.S)

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

So, the problem has 3 solutions, $x = - 2$ $x=-2$ , $x = 2$ $x=2$ , and $x = \frac{2}{3}$ $x=2/3$ .

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How do you solve $3 x^{3} - 2 x^{2} - 12 x + 8 = 0$ $3x^3 -2x^2 -12x + 8 = 0$ ?

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