How do you factor 9z ^ { 2} + 42z + 49?

Mar 13, 2017

$9 {z}^{2} + 42 z + 49 = {\left(3 z + 7\right)}^{2}$

Explanation:

First notice that since $9 = {3}^{2}$ and $49 = {7}^{2}$

we can say

$9 {z}^{2} + 42 z + 49 = {\left(3 z\right)}^{2} + 42 z + {7}^{2}$

also since $42 = 21 + 21$ and $21 = 3 \left(7\right)$

we can say

$9 {z}^{2} + 42 z + 49 = {\left(3 z\right)}^{2} + 3 z \left(7\right) + 3 z \left(7\right) + {7}^{2}$

$= {\left(3 z\right)}^{2} + 2 \left(3 z \left(7\right)\right) + {7}^{2}$

and since

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

we can say

${\left(3 z\right)}^{2} + 2 \left(3 z \left(7\right)\right) + {7}^{2} = \underline{{\left(3 z + 7\right)}^{2}}$