How do you factor 9z ^ { 2} + 42z + 499z2+42z+49?

1 Answer
Mar 13, 2017

9z^2+42z+49=(3z+7)^29z2+42z+49=(3z+7)2

Explanation:

First notice that since 9=3^29=32 and 49=7^249=72

we can say

9z^2+42z+49=(3z)^2+42z+7^29z2+42z+49=(3z)2+42z+72

also since 42=21+2142=21+21 and 21=3(7)21=3(7)

we can say

9z^2+42z+49=(3z)^2+3z(7)+3z(7)+7^29z2+42z+49=(3z)2+3z(7)+3z(7)+72

=(3z)^2+2(3z(7))+7^2=(3z)2+2(3z(7))+72

and since

(a+b)^2=a^2+2ab+b^2(a+b)2=a2+2ab+b2

we can say

(3z)^2+2(3z(7))+7^2=ul((3z+7)^2)