How do you solve $(n + 2) (2 n + 5) = 0$ $(n+2) (2n+5)= 0$ by factoring?

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Answer 1 · 2015-08-17T06:37:39

The solutions are

$n = - 2$ $color(blue)(n=-2$

$n = - \frac{5}{2}$ $color(blue)(n=-5/2$

$(n + 2) (2 n + 5) = 0$ $(n+2)(2n+5)=0$ is the already factorised form of equation
$2 n^{2} + 9 n + 10$ $2n^2+9n+10$ .

Since we have the factors we obtain the solution by equating these factors to the R.HS.

$n + 2 = 0, n = - 2$ $n+2=0, color(blue)(n=-2$
$2 n + 5 = 0, n = - \frac{5}{2}$ $2n+5=0, color(blue)(n=-5/2$

How do you solve (n+2)(2n+5)=0(n+2) (2n+5)= 0 by factoring?