79. Without performing the operations, represent the product as a polynomial in standard form

$$
(x-1)(x+3)(x+5)
$$
Solution. The product is a polynomial of the third degree, the coefficient of the leading term (it is not difficult to verify) is 1, and the free term is -15, then we write:

$$
(x-1)(x+3)(x+5)=x^{3}+a x^{2}+b x-15
$$

where $a$ and $b$ are unknown coefficients. To calculate them, let $x=1$ and $x=-3$, then we get:

$$
\left\{\begin{aligned}
a+b-14 & =0 \\
9 a-3 b-42 & =0
\end{aligned}\right.
$$

From which $\quad a=7, \quad b=7 . \quad$ Therefore, $\quad(x-1)(x+3)(x+5)=x^{3}+$ $+7 x^{2}+7 x-15$

## Factorization of a Polynomial

For a quadratic trinomial $f(x)=a x^{2}+b x+c$, from the algebra course, the factorization is known:

$$
a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), \text { where } x_{1} \text { and } x_{2} \text { are the }
$$

roots of the trinomial.

For a polynomial of degree $n$ $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$, we have:

$$
a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}=a_{n}\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n}\right)\left(a_{n} \neq 0\right)
$$

$$
\text { where } x_{1}, x_{2}, \ldots, x_{n} \text { are the roots of the polynomial. }
$$

If a polynomial of degree $n$ has $k$ roots $(k<n)$, then

$$
f(x)=a_{n}\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{k}\right) g(x)
$$

where $g(x)$ is a polynomial of degree $(n-k)$.