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

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Answer 1 · 2017-03-13T01:39:13

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

First notice that since #9=3^2# and #49=7^2#

we can say

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

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

we can say

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

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

and since

#(a+b)^2=a^2+2ab+b^2#

we can say

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