What's wrong with my factoring?

Question

Khan Academy tells me I have to factor $-28$ into $2 \times k \times (-14)$ but I'm pretty sure $-2 \times k \times 14$ is acceptable.

Answer 1

Your answer and Khan Academy's are equivalent. You can rewrite the $-28k$ as $(-2)(14)(k)$ or $(2)(-14)(k)$.

However, I think you ran into a problem when you factored $\dfrac{3(k^2-2(14)(k)+14^2)}{-4(k+14)(k-14)}$. Essentially when factoring the numerator, you are finding two numbers that multiply to $14^2$ and add to the coefficient of the middle term, $(-2)(14)=-28$. This would be $-14$ and $-14$, as $-14-14=-28$ and $(-14)^2=(14^2)$. So the numerator would factor to $3(k-14)(k-14)$, which would cancel with the $(k-14)$ in the denominator, not the $(k+14)$. This gives you $-\dfrac{3(k-14)}{4(k+14)}$.