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How do you solve $3 x^{2} - 13 x - 10 = 0$ $3x^2-13x-10=0$ by factoring?

1 Answer

Oct 10, 2015

$(3 x + 2) (x - 5)$ $(3x+2)(x-5)$

Explanation:

Because $3$ $3$ can only have factors of $3$ $3$ and1
you know those will be your coefficients for $x$ $x$ .
That is your factors will be of the form
$XXX (3 x + a) (x + b)$ $color(white)("XXX") (3x +a)(x +b )$

You also know that one of $a$ $a$ or $b$ $b$ will have $+$ $+$ and one side will have $-$ $-$ due to the $- 10$ $color(red)(-)10$ .

The options for $10$ $10$ are
$2 \times 5$ $2xx5$ or $1 x 10$ $1x10$ .

Using $2 \times 5$ $2xx5$
if you multiply the $5$ $5$ by $3$ $3$ it gives you $15$ $15$
and the $2$ $2$ by $1$ $1$ 1 you get $2$ $2$ .

Because $13$ $13$ in the center (of the original form) is negative
you know the $5$ $5$ needs to be negative

and the factors are
$XXX (3 x + 2) (x - 5)$ $color(white)("XXX")(3x+2)(x-5)$

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