# Zero Factor Property

## Key Questions

• You use the zero factor property after you have factored the quadratic to find the solutions.

It is best to look at an example: ${x}^{2} + x - 6 = 0$

This factors into:

$\left(x + 3\right) \left(x - 2\right) = 0$

We find our solutions by setting each factor to zero and solve:

$x + 3 = 0$
$x = - 3$

or

$x - 2 = 0$
$x = 2$

Previous answer (I was thinking some more complicated before):

You are not using the words precisely. You use the factor theorem with the factor property. The factor theorem states that if you find a $k$ such that $P \left(k\right) = 0$, then $x - k$ is a factor of the polynomial. The factor property states that $k$ must a factor of the constant term in $P \left(x\right)$.

Having said all that, you wouldn't normally use the factor theorem or factor property to solve a quadratic; they are many used to find factors of higher order polynomials. Once you reduce the higher order polynomial to a quadratic, you use regular factoring methods such as FPS or PFS: Factors, Product, and Sum.

$P \left(x\right) = a {x}^{2} + b x + c$

The problem with the factor theorem and factor property is that it's not as easy to use when $a \ne 1$.

• #### Answer:

Well, if you a polynomial is factorable then its roots/zeroes can be easily found by setting it to zero and using the zero factor property. Please see explanation below.

#### Explanation:

The Zero Product Property:
A product of factors is zero if and only if one or more of the factors is zero. Or:
if $a \cdot b = 0$, then either $a = 0$ or $b = 0$ or both.
Example: Find the roots of the polynomial by factoring:
$P \left(x\right) = {x}^{3} - {x}^{2} - x + 1$, set to zero:
${x}^{3} - {x}^{2} - x + 1 = 0$, factor by grouping:
${x}^{2} \left(x - 1\right) - 1 \left(x - 1\right) = 0$
$\left({x}^{2} - 1\right) \left(x - 1\right) = 0$, use difference of squares to factor further:
$\left(x + 1\right) \left(x - 1\right) \left(x - 1\right) = 0$, use the zero factor property:
$x + 1 = 0 \implies x = - 1$
$x - 1 = 0 \implies x = 1$
Notice that $x = 1$ has a multiplicity of 2.

• The zero factor property states that if $a b = 0$, then either $a = 0$ or $b = 0$.

Example: find the roots of ${x}^{2} - x - 6$.

${x}^{2} - x - 6 = 0$

$\left(x - 3\right) \left(x + 2\right) = 0$

Now, the zero factor property can be applied, since two thing are being multiplied and equal zero.

We know that either

$x - 3 = 0 \textcolor{w h i t e}{s s s s}$ or $\textcolor{w h i t e}{s s s s} x + 2 = 0$

Solve both to find that $x = 3$ or $x = - 2$.