We have looked at many different ways to factor polynomials.
Recall that we can factor by:
Once we can factor, we can solve equations algebraically to find the unknown value(s). There is a great benefit to factoring quadratics in order to solve them, and in order to understand why, we need to think about zero.
The property of $0$0 is very special. The only way two things that are being multiplied can have a product of $0$0, is if one, or both of those things are $0$0 themselves.
So if we have two factors, like $a$a and $b$b, and we multiply them together so that they equal $0$0, then one of those factors ($a$a or $b$b) must be $0$0. A written solution to a question like this would be similar to the following:
If $a\times b=0$a×b=0 then $a=0$a=0 or $b=0$b=0. This is known as the zero product property.
The product of any number and $0$0 is $0$0. That is:
If $ab=0$ab=0, then $a=0$a=0 or $b=0$b=0
Solve $\left(x-2\right)\left(x+5\right)=0$(x−2)(x+5)=0.
Think: This quadratic equation is already in factored form, so we can go straight to using the zero product property to find the solutions.
Do: Using the zero product property, either
$\left(x-2\right)$(x−2) | $=$= | $0$0 | or | $\left(x+5\right)$(x+5) | $=$= | $0$0 |
$x$x | $=$= | $2$2 | $x$x | $=$= | $-5$−5 |
So the solutions to the equation $\left(x-2\right)\left(x+5\right)=0$(x−2)(x+5)=0 are $x=2$x=2 or $x=-5$x=−5.
What are the solutions to the equation $x^2-25=0$x2−25=0?
Think: Notice that the left-hand side of this equation is actually a difference of two squares polynomial, and so we can rewrite it first in factored form.
Do: Let's first factor the equation, then apply the zero product property to find the solutions.
$x^2-25$x2−25 | $=$= | $0$0 |
$\left(x-5\right)\left(x+5\right)$(x−5)(x+5) | $=$= | $0$0 |
Now we know from the zero product property that either $\left(x-5\right)=0$(x−5)=0 or $\left(x+5\right)=0$(x+5)=0, and so either $x=5$x=5 or $x=-5$x=−5.
Solve the equation $\left(2x-8\right)^2=0$(2x−8)2=0.
Think: This perfect square quadratic equation is also already in a form that can make use of the zero product property.
Do: To make it clear, we'll rewrite the equation using the fact that $\left(\editable{}\right)^2=\editable{}\times\editable{}$()2=×.
$\left(2x-8\right)^2$(2x−8)2 | $=$= | $0$0 |
$\left(2x-8\right)\left(2x-8\right)$(2x−8)(2x−8) | $=$= | $0$0 |
What does this last line mean? The zero product property tells us that either $\left(2x-8\right)=0$(2x−8)=0 or $\left(2x-8\right)=0$(2x−8)=0. But these are the same factor! This is actually an example of a double root, and we find the answer by considering the single factor $2x-8=0$2x−8=0 by itself, which gives $x=4$x=4.
Solve for the two possible values of $x$x:
$\left(x-7\right)\left(x-6\right)=0$(x−7)(x−6)=0
Write all solutions on the same line, separated by commas.
Solve $3y-15y^2=0$3y−15y2=0
Write all solutions on the same line, separated by commas.
Solve the following equation by first factoring the left hand side of the equation.
$4x^2-17x+15=0$4x2−17x+15=0
Write all solutions on the same line, separated by commas.