How do you factor 6r² + 7r -10?

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Answer 1 · 2015-05-17T23:49:11

The answer is $(6 r - 5) (r + 2)$ $(6r-5)(r+2)$

Problem: Factor $6 r^{2} + 7 r - 10$ $6r^2+7r-10$ . The format of this equation is $a x + b x + c$ $ax+bx+c$ .

Multiply $a \cdot c$ $a*c$ .

$6 \cdot - 10 = - 60$ $6*-10=-60$ .

Determine two numbers that when added make $b$ $b$ , and when multiplied make $a \cdot c$ $a*c$ .

$b = 7$ $b=7$ ; $a \cdot c = - 60$ $a*c=-60$

The numbers $12$ $12$ and $- 5$ $-5$ fit the required pattern.

Rewrite the equation with $12 r$ $12r$ and $- 5 r$ $-5r$ in place of $7 r$ $7r$ .

$6 r^{2} + 12 r - 5 r - 10$ $6r^2+12r-5r-10$

Factor by grouping.

$(6 r^{2} + 12 r) - (5 r - 10)$ $(6r^2+12r)-(5r-10)$

Factor out common terms.

$6 r (r + 2)$ $6r(r+2)$

$- 5 (r + 2)$ $-5(r+2)$

Factor out the common term, $r + 2$ $r+2$ .

Answer: $(6 r - 5) (r + 2)$ $(6r-5)(r+2)$