Answer
Factoring $ x^{2}+bx+c, \quad $ (leading coefficient is 1)
we search for two factors (m and n) of $ c $, such that their sum is $ b .$
If we find two such factors, we write
$ x^{2}+bx+c =(x+m)(x+n)$
Here, $ x^{2}+6x+8=(x+4)(x+2)$
So the equation becomes $(x+4)(x+2)=0$
By the zero-product principle, the product on the LHS can be zero only when one (or both) of the factors is zero.
So, we conclude that $ x $ can be $-4$ or $-2$
(the solution set is $\{ -4,-2\}$. )
Work Step by Step
Given above.
