This question is to clarify the confusion my middle schooler has.
During factoring we factor the negative sign of the leading coefficient so why is that not done when we factor a difference of squares like $81 - x^2$? if written in standard form it would be $- x^2 + 81.$ So why is this incorrect :
$$-(x^2-81) = -(x+9)(x-9)$$
His teacher expects the answer as $(9+x) ( 9-x)$
Would appreciate your help. Thank you.
There is a difference between finding "factors", and writing down a "factorization".
Factors are only determined up to sign. Consider integers, for instance: what are the factors (divisors) of $6$? Well, they are $2$ and $3$... and also $-2$, and $-3$, and $1$, and $-1$, and $6$, and $-6$. Same with $-6$. So $2$ and $3$ are both prime factors of $6$, and also of $-6$.
This does not mean that $2\times 3=-6$, though. If we want to write a factorization of $-6$, using positive primes, we need to add a $-1$: $-6 = (-1)\times2\times 3$.
Same with $x^2-81$ and $81-x^2$. They both have $x+9$ and $x-9$ as factors. But while $(x+9)(x-9)$ is equal to $x^2-81$, to get $81-x^2$ you need to add a sign. So either $$81 - x^2 = -(x+9)(x-9)$$ or distribute the $-1$ into either of the factors to get $$81 -x^2 = (x+9)(9-x) = (-x-9)(x-9) = -(x+9)(x-9).$$
With integers, we've decided that we will use the positive version of the primes ($2$ instead of $-2$, etc), and if we need to add a sign, we will just add the $-1$ at the beginning. That's why we usually write $-6$ as $(-1)\times2\times 3$, and not as $(-2)\times 3$ or $2\times(-3)$ (and why don't usually factor $6$ as $(-2)\times(-3)$, even though it is correct). That's a convention: an agreement we've made to make sure we come up with the same answers.
With polynomials, we've made a similar decision: we prefer monic factors, which are factors in which the leading coefficient is $1$. Because with polynomials it's even worse: it's not just the $-1$: you could also write, for example, $$x^2 -81 = (17x+153)\left(\frac{1}{17}x - \frac{9}{17}\right)$$ (because the extra factor of $17$ cancels out with the extra factor of $\frac{1}{17}$). So we will want to factor $81-x^2$ as $$81-x^2 = -(x+9)(x-9)$$ so that the factors, except for a single constant term at the beginning, are all monic.