# Polynomial equations in factored form

All equations are composed of polynomials. Earlier we've only shown you how to solve equations containing polynomials of the first degree, but it is of course possible to solve equations of a higher degree.

One way to solve a polynomial equation is to use the zero-product property. If you remember from earlier chapters the property of zero tells us that the product of any real number and zero is zero. This means that for any real numbers x and y

$\\ if\: x=0\: or\: y=0,\: \: then\: xy=0 \\$

The zero-power property can be used to solve an equation when one side of the equation is a product of polynomial factors and the other side is zero. The solutions to a polynomial equation are called roots.

Example:

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

To satisfy this equation either

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

or

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

This means that the roots of the equation are 3 and -2.

This method can only work if your polynomial is in their factored form. The following sections will show you how to factor different polynomial.

We begin by looking at the following example:

$\\ x\cdot \left ( x+4 \right )=12 \\$

We multiply as usual:

$\\ x^{2}+4x=12 \\$

We may also do the inverse. By identifying the greatest common factor (GCF) in all terms we may then rewrite the polynomial into a product of the GCF and the remaining terms. Examples of this would be:

$\\ 3x+2x=15\Rightarrow \left \{ both\: 3x\: and\: 2x\: are\: divisible\: by\: x\right \} \\\Rightarrow x\left ( 3+2 \right )=15 \\$

$\\ 6x^{2}-x=9\Rightarrow \left \{ both\: terms\: are\: divisible\: by\: x \right \} \\\Rightarrow x\left ( 6x-1 \right )=9 \\$

$\\ 4x^{2}-2x^{3}=9\Rightarrow \left \{ both\: terms\: are\: divisible\: by\: 2x^{2} \right \} \\\Rightarrow 2x^{2}\left ( 2-x \right )=9 \\$

Video lesson: Solve the equation

$\\6x^{2}+18x=0$