Example
Solve:
−t2+t=0
Answer:
Each term has a common factor of t, so we can factor and use the zero product principle. Rewrite each term as the product of the GCF and the remaining terms.
−t2=t(−t)t=t(1)
Rewrite the polynomial equation using the factored terms in place of the original terms.
−t2+t=0t(−t)+t(1)t(−t+1)=0
Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.
t=0 OR −t+1=0−1−1−t=−1−1−t=−1−1t=1
Therefore,
t=0 OR t=1.
In the following video, we show two more examples of using both factoring and the principle of zero products to solve a polynomial equation.
https://youtu.be/gIwMkTAclw8
In our next example, we will solve a quadratic equation that has a leading coefficient of 1. In the quadratic equation
Example
Solve:
s2−4s=5
Answer:
First, move all the terms to one side. The goal is to try and see if we can use the zero product principle since that is the only tool we know for solving polynomial equations.
s2−4s=5s2−4s−5=0
We now have all the terms on the left side and zero on the other side. The polynomial s2−4s−5 factors nicely which makes this equation a good candidate for the zero product principle.
s2−4s−5=0(s+1)(s−5)=0
We separate our factors into two linear equations using the principle of zero products.
(s−5)=0s−5=0s=5
OR
(s+1)=0s+1=0s=−1
Therefore,
s=−1 OR s=5.
In our next example, we solve a quadratic equation with a leading coefficient that is not equal to 1. Recall that when the leading coefficient is not
Example
Solve the quadratic equation:
4x2+15x+9=0.
Answer:
We can see that this equation is already in standard for, so we will proceed with factoring.
Because this equation has a leading coefficient not equal to 1, we will factor by grouping:
multiply ac:4(9)=36. Then list the factors of 36.
1⋅362⋅183⋅124⋅96⋅6
The only pair of factors that sums to
15 is
3+12. Rewrite the equation replacing the
b term,
15x, with two terms using
3 and
12 as coefficients of
x. Factor the first two terms, and then factor the last two terms.
4x2+3x+12x+9=0x(4x+3)+3(4x+3)=0(4x+3)(x+3)=0
Solve using the zero-product property.
(4x+3)=0x=−43(4x+3)(x+3)=0(x+3)=0x=−3
The solutions are
x=−43,
x=−3.

Notice that when we graph the equation
y=4x2+15x+9 we can see that the x-intercepts are
(−3,0) and
(−43,0). This agrees with our solutions!
The example below shows a quadratic equation where neither side is originally equal to zero. Remember, we cannot solve quadratic equations unless one side is equal to zero because this allows us to use the Zero Product Principle. (Note that the factoring sequence has been shortened.)
Example
Solve
y2−5=−27y+25
Answer:
We can solve this in one of two ways. One way is to eliminate the fractions like you may have done when solving linear equations, and the second is to find a common denominator and factor fractions. Eliminating fractions is easier, so we will show that way.
Start by multiplying the whole equation by 2 to eliminate fractions:
2(y2−5=−27y+25)2(y2)+2(−5)=2(−27y)+2(25)2y2−10=−7y+5
Now we can move all the terms to one side and see if this will factor so we can use the principle of zero products.
2y2−10=−7y+52y2−10+7y−5=02y2−15+7y=02y2+7y−15=0
We can now check whether this polynomial will factor. Using a table we can list factors until we find two numbers with a product of
2⋅−15=−30 and a sum of 7.
Factors of 2⋅−15=−30 |
Sum of Factors |
1,−30 |
−29 |
−1,30 |
29 |
2,−15 |
−13 |
−2,15 |
13 |
3,−10 |
−7 |
−3,10 |
7 |
We have found the factors that will produce the middle term we want,
−3,10. We need to place the factors in a way that will lead to a middle term of
7y:
(2y−3)(y+5)=0
Now we can set each factor equal to zero and solve:
(2y−3)=0 OR (y+5)=02y=3 OR y=−5y=23 OR y=−5
You can always check your work to make sure your solutions are correct:
Check
y=23
(23)2−5=−27(23)+2549−5=−421+25 common denominator = 449−420=−421+410−411=−411
y=23[/latex]isindeedasolution,nowcheck [latex]y=−5
(−5)2−5=−27(−5)+2525−5=235+2520=24020=20
y=−5 is also a solution, so we must have done something right!
Therefore, y=23 OR y=−5.
In our last video, we show how to solve another quadratic equation that contains fractions.
https://youtu.be/kDj_qdKW-ls
The following video contains another example of solving a quadratic equation using factoring with grouping.
https://youtu.be/04zEXaOiO4U
It is useful to remember that, if you factor out a constant, the constant will never equal
Example
Solve for k:
−2k2+90=−8k
Answer:
We need to move all the terms to one side so we can use the zero product principle.
−2k2+90=−8k+8k+8k−2k2+8k+90=0
You will either need to try to factor out a
−2, or use the method where we multiply
−2⋅90 and find factors that sum to
8. Each term is divisible by
2, so we can factor out
−2.
−2(k2−4k−45)=0
Note how we changed the signs when we factored out a negative number. If we can factor the polynomial, we will be able to solve.
Using the shortcut for factoring we will start with the variable and place a plus and a minus sign in the binomials. We do this because
45 is negative and the only way to get a product that is negative is if one of the factors is negative.
−2(k−)(k+)=0
We want our factors to have a product of −45 and a sum of −4:
Factors whose product is -45 |
Sum of the factors |
1⋅−45=−45 |
1−45=−44 |
3⋅−15=−45 |
3−15=−12 |
5⋅−9=−45 |
5−9=−4 |
There are more factors that will give
−45, but we have found the ones that sum to
−4, so we will stop. Fill in the rest of the binomials with the factors we found.
−2(k−9)(k+5)=0
Now we can set each factor equal to zero using the zero product rule.
−2=0 This solution is nonsense so we discard it.
k−9=0,k=9
k+5=0,k=−5
Answers
k=9 and k=−5
We can use the zero-product property to solve quadratic equations in which we first have to factor out the Did you have an idea for improving this content? We’d love your input.