How do you factor the trinomial $6 d^{2} + 21 d = 10 d + 35$ $6d^2+21d= 10d+35$ ?

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How do you factor the trinomial $6 d^{2} + 21 d = 10 d + 35$ $6d^2+21d= 10d+35$ ?

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Answer 1 · 2015-11-19T08:44:23

Rearrange all the factors to be on one side of the expression and then factorize as normal

Explanation:

$6 d^{2} + 21 d - 10 d - 35 = 0$ $6d^2 +21d - 10d -35 = 0$
$6 d^{2} + 11 d - 35 = 0$ $6d^2 +11d -35 = 0$
35 is $7 \cdot 5$ $7*5$ and 6 is $3 \cdot 2$ $3*2$
By studying the numbers 11 can be obtained from 21 ( $7 \cdot 3$ $7*3$ ) minus 10 ( $5 \cdot 2$ $5*2$ )
$(3 d - 5) (2 d + 7) = 0$ $(3d - 5)(2d + 7) = 0$
$d = \frac{5}{3}$ $d = 5/3$ or $d = - \frac{7}{2}$ $d = -7/2$

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How do you factor the trinomial $6 d^{2} + 21 d = 10 d + 35$ $6d^2+21d= 10d+35$ ?

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How do you factor the trinomial 6d2+21d=10d+356d^2+21d= 10d+35 ?

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Explanation:

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How do you factor the trinomial $6 d^{2} + 21 d = 10 d + 35$ $6d^2+21d= 10d+35$ ?