How do you solve $3 x^{2} + 5 x + 2 = 0$ $3x^2+5x+2=0$ by factoring?

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How do you solve $3 x^{2} + 5 x + 2 = 0$ $3x^2+5x+2=0$ by factoring?

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Answer 1 · 2015-03-27T16:27:33

Assuming $3 x^{2} + 5 x + 2$ $3x^2+5x+2$ can be factored into
$(a x + b) (c x + d)$ $(ax+b)(cx+d)$ with integer values for $a, b, c, d$ $a,b,c,d$
and noting that all of $a, b, c, d$ $a,b,c,d$ will be non-negative (from the form of the original quadratic)
we only have the possibilities
${a, c} = {1, 3}$ ${a,c} = {1,3}$ and
${b, d} = {1, 2}$ ${b,d} = {1,2}$

There are only 2 combinations to attempt:
$(1 x + 1) (3 x + 2)$ $(1x+1)(3x+2)$
and
$(1 x + 2) (3 x + 1)$ $(1x+2)(3x+1)$

revealing that $3 x^{2} + 5 x + 2 = 0$ $3x^2+5x+2 =0$ can be factored as
$(x + 1) (3 x + 2) = 0$ $(x+1)(3x+2)=0$

So either
$x + 1 = 0$ $x+1 = 0$ which implies $x = - 1$ $x = -1$
or
$3 x + 2 = 0$ $3x+2 = 0$ which implies $x = - \frac{2}{3}$ $x = -2/3$

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How do you solve $3 x^{2} + 5 x + 2 = 0$ $3x^2+5x+2=0$ by factoring?

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