Factor the polynomial completely. $1 - 36 n ^ { 2 }$
A) $( 1 + 6 n ) ( 1 - 6 n )$
B) $( 1 + 36 n ) ( 1 - n )$
C) $( 6 + 1 n ) ( 1 - 6 n )$
D) $( 1 + n ) ( 1 - 36 n )$
E) not factorable

Question

Quizplus · Accepted Answer

We can use the difference of squares formula: $$a^2 - b^2 = (a+b)(a-b)$$ In this case, let $a=1$ and $b=6n$. Then we have: $$1-36n^2 = (1)^2 - (6n)^2 = (1+6n)(1-6n)$$ So the polynomial is completely factored as $(1+6n)(1-6n)$, which corresponds to choice letter A.

Factor the Polynomial Completely 1−36n21 - 36 n ^ { 2 }1−36n2

Factor the Polynomial Completely $1 - 36 n ^ { 2 }$