# 1.4 Solving Polynomial Equations

## Polynomial Equations

In MATH 110, there will be many situations where we are interested to solve polynomial equations.

One method which can be used to solve certain equations is called the zero-product property which states that if the product of two algebraic expressions is zero, then at least one of the factors is equal to zero.

This can be stated as:

If $A\times B = 0$, then $A = 0$ or $B = 0$

This method can be used to solve a quadratic equations using factoring:

1. Write the equation in the form $ax^2 + bx + c = 0$, where the equation is set equal to zero.
2. Factor the quadratic expression if possible.
3. Use the zero-product property to set each factor separately equal to zero.
4. Solve each of the separate equations.

Example:
Solve the equation: $x^2 + 3x = 40$

Solution:
First, get the equation set equal to zero by subtracting 40 from both sides of the equation.

$x2 + 3x - 40 = 0$

Next, factor the left side of the equation:
$(x - 5)(x + 8) = 0$

Next use the zero-product property to set each factor separately equal to zero.

$x - 5 = 0$  which results in $x = 5$
$x + 8 = 0$ which results in $x = -8$

The two solutions to the equation are $x = 5$, $x = -8$.

The zero-product property can be extended such that if the product of any number of algebraic expressions is zero, then at least one of the factors is equal to zero.

For example we can write this for three factors as:

If $A\times B\times C = 0$, then $A = 0$ or $B = 0$ or $C = 0$.

Example:
Solve the equation: $5x^4 -45x^2 = 0$

Solution:
First, note that the equation is set equal to zero.

Next, factor the left side of the equation.  A greatest common factor of $5x^2$ is noted:
$5x^2(x^2 - 9) = 0$

Notice that $(x^2 - 9)$ can be further factored as the difference of two squares.

$5x^2(x + 3)(x - 3) = 0$

Next use the zero-product property to set each factor separately equal to zero.

$5\times 2 = 0$ which results in $x = 0$
$x - 3 = 0$  which results in $x = 3$
$x + 3 = 0$ which results in $x = -3$

The three solutions to the equation are then $x = 0, x = 3, x = -3$.