understanding factors in a polynomial

A factor of a polynomial is one of the pieces that multiply together to build it.
In $(x + 2)(x - 1)$, both $(x + 2)$ and $(x - 1)$ are factors.
A factor divides the polynomial evenly, leaving no remainder.

Here is the central idea, called the Factor Theorem.
If $(x - a)$ is a factor of $f(x)$, then $f(a) = 0$.
The reason is simple: substituting $x = a$ makes that factor $(a - a) = 0$, and zero times anything is zero.

A value that makes $f(x) = 0$ is called a zero (or root) of the polynomial.
On a graph, a zero is exactly where the curve crosses the $x$-axis.
So factors, zeros, and $x$-intercepts are three ways of describing the same thing.

This works in reverse too.
If you know the zeros, you can write the factored form.
A polynomial with zeros at $x = -2$ and $x = 4$ has the form $a(x + 2)(x - 4)$, where $a$ is a constant.

A factor can repeat, like $(x - 2)^2$ in $(x - 1)(x - 2)^2$.
A repeated factor still gives the same zero, $x = 2$, but the graph touches the $x$-axis there instead of crossing it.

If $(x - 3)$ is a factor of a polynomial $f(x)$, what is $f(3)$?
By the Factor Theorem, a factor $(x - a)$ means $f(a) = 0$.
Here $a = 3$, so $f(3) = 0$.

A polynomial has zeros at $x = 1$ and $x = -2$. Write a possible factored form.
Each zero $x = c$ gives a factor $(x - c)$.
So the factors are $(x - 1)$ and $(x + 2)$, giving $f(x) = a(x - 1)(x + 2)$ for some constant $a$.

If $(x - 2)$ is a factor of $x^3 + ax^2 + bx + 6$, what equation must $a$ and $b$ satisfy?
By the Factor Theorem, $f(2) = 0$.
Substituting $x = 2$: $2^3 + a(2)^2 + b(2) + 6 = 0$, so $8 + 4a + 2b + 6 = 0$, giving $4a + 2b + 14 = 0$.