Question

factor completely
$n^2 - 12n + 35$
$4(n + 9)(n + 10)$
not factorable
$(n - 7)(n + 5)$

Explanation:

Step1: Find pair summing to -12

We need two numbers that add to $-12$ and multiply to $35$. These numbers are $-7$ and $-5$.

Step2: Factor the quadratic

Rewrite the quadratic using the pair, then factor by grouping.
$n^2 -7n -5n +35 = n(n-7) -5(n-7) = (n-7)(n-5)$
*(Note: The option $(n-7)(n+5)$ is incorrect, as $(-7)(+5)=-35$, not $+35$; the correct factored form is $(n-7)(n-5)$, but since it's not shown, we recheck: wait, correction: sum of $-7$ and $-5$ is $-12$, product is $35$. So the correct factorization is $(n-7)(n-5)$. However, if we evaluate the given options, none match exactly, but wait: $(n-7)(n+5)$ gives $n^2-2n-35$, which is wrong. The green option is $4(n+9)(n+10)=4n^2+76n+360$, wrong. "Not Factorable" is wrong because the quadratic is factorable. There may be a missing option, but based on the math, the correct factored form is $(n-7)(n-5)$.)

Answer:

(Note: The correct factored form is $(n-7)(n-5)$, which is not displayed in the visible options. Among the shown options, none are correct, but the quadratic is factorable, so the error is in the provided options. If forced to choose from visible options, there is no correct one, but mathematically the complete factorization is $\boldsymbol{(n-7)(n-5)}$)