Question

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

Answer 1

The solutions are

`color(blue)(n=-2`

`color(blue)(n=-5/2`

Explanation:

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

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

`n+2=0, color(blue)(n=-2`
`2n+5=0, color(blue)(n=-5/2`