7.7: Zero Product Property for Quadratic Equations
The area of a particular rectangle was found to be \begin{align*}A(w)=w^28w58\end{align*}
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Khan Academy Factoring Special Products
Guidance
Recall that when solving an equation, you are trying to determine the values of the variable that make the equation true. For the equation \begin{align*}2x^2+10x+8=0\end{align*}

\begin{align*}2(1)^2+10(1)+8=2(1)10+8=0\end{align*}
2(−1)2+10(−1)+8=2(1)−10+8=0 
\begin{align*}2(4)^2+10(4)+8=2(16)40+8=3240+8=0\end{align*}
2(−4)2+10(−4)+8=2(16)−40+8=32−40+8=0
Here you will focus on solving quadratic equations. One of the methods for quadratic equations utilizes your factoring skills and a property called the zero product property.
If \begin{align*}a\cdot b=0\end{align*}
The zero product property has to do with products being equal to zero. When you factor, you turn a quadratic expression into a product. If you have a quadratic expression equal to zero, you can factor it and then use the zero product property to solve. So, if you were given the equation \begin{align*}2x^2+5x3=0\end{align*}
\begin{align*}2x^2+5x3=(x+3)(2x1)\end{align*}
You can rewrite the equation you are trying to solve as \begin{align*}(x+3)(2x1)=0\end{align*}
Now, you have the product of two binomials equal to zero. This means at least one of those binomials must be equal to zero. So, you have two miniequations that you can solve to find the values of \begin{align*}x\end{align*}

\begin{align*}x+3=0\end{align*}
x+3=0 , which means \begin{align*}x=3\end{align*}x=−3 OR 
\begin{align*}2x1=0\end{align*}
2x−1=0 , which means \begin{align*}x=\frac{1}{2}\end{align*}x=12
The two solutions to the equation \begin{align*}2x^2+5x3=0\end{align*}
Keep in mind that you can only use the zero product property if your equation is set equal to zero! If you have an equation not set equal to zero, first rewrite it so that it is set equal to zero. Then factor and use the zero product property.
Example A
Solve for \begin{align*}x\end{align*}
Solution: First, change \begin{align*}x^2+5x+6\end{align*}
\begin{align*}x^2+5x+6=(x+3)(x+2)\end{align*}
Next, rewrite the equation you are trying to solve:
\begin{align*}x^2+5x+6=0\end{align*}
Finally, set up two miniequations to solve in order to find the values of \begin{align*}x\end{align*}

\begin{align*}x+3=0\end{align*}
x+3=0 , which means that \begin{align*}x=3\end{align*}x=−3 
\begin{align*}x+2=0\end{align*}
x+2=0 , which means that \begin{align*}x=2\end{align*}x=−2
The solutions are \begin{align*}x=3\end{align*}
Example B
Solve for \begin{align*}x\end{align*}
In order to solve for \begin{align*}x\end{align*}
Solution: First, change \begin{align*}6x^2+x15\end{align*}
\begin{align*}6x^2+x15=(3x+5)(2x3)\end{align*}
Next, rewrite the equation you are trying to solve:
\begin{align*}6x^2+x15=0\end{align*} becomes \begin{align*}(3x+5)(2x3)=0\end{align*}.
Finally, set up two miniequations to solve in order to find the values of \begin{align*}x\end{align*} that cause each binomial to be equal to zero.
 \begin{align*}3x+5=0\end{align*}, which means that \begin{align*}x=\frac{5}{3}\end{align*}
 \begin{align*}2x3=0\end{align*}, which means that \begin{align*}x=\frac{3}{2}\end{align*}
The solutions are \begin{align*}x=\frac{5}{3}\end{align*} or \begin{align*}x=\frac{3}{2}\end{align*}.
Example C
Solve for \begin{align*}x\end{align*}: \begin{align*}x^2+2x35=0\end{align*}.
Solution: First, change \begin{align*}x^2+2x35\end{align*} into a product so that you can use the zero product property. Change the expression into a product by factoring:
\begin{align*}x^2+2x35=(x+7)(x5)\end{align*}
Next, rewrite the equation you are trying to solve:
\begin{align*}x^2+2x35=0\end{align*} becomes \begin{align*}(x+7)(x5)=0\end{align*}.
Finally, set up two miniequations to solve in order to find the values of \begin{align*}x\end{align*} that cause each binomial to be equal to zero.
 \begin{align*}x+7=0\end{align*}, which means that \begin{align*}x=7\end{align*}
 \begin{align*}x5=0\end{align*}, which means that \begin{align*}x=5\end{align*}
The solutions are \begin{align*}x=7\end{align*} or \begin{align*}x=5\end{align*}.
Concept Problem Revisited
The area of a particular rectangle was found to be \begin{align*}A(w)=w^28w58\end{align*}. Determine the dimensions of the rectangle if the area was known to be 7 units.
In other words, you are being asked to solve the problem:
\begin{align*}& w^28w58=7\\ & OR\\ & w^28w65=0\end{align*}
You can solve this problem by factoring and using the zero product property.
\begin{align*}w^28w65=0\end{align*} becomes \begin{align*}(w+5)(w13)=0\end{align*}
\begin{align*}& \quad (w+5)(w13)=0\\ &\qquad \swarrow \qquad \quad \searrow\\ & w + 5 = 0 \qquad \quad w  13 = 0\\ &\quad \ \ w = 5 \quad or \qquad \ \ w = 13\end{align*}
Since you are asked for dimensions, a width of –5 units does not make sense. Therefore for the rectangle, the width would be 13 units.
Vocabulary
 Quadratic Equation
 A quadratic equation is an equation in which the highest power of a variable is 2. Standard form for a quadratic equation is \begin{align*}ax^2+bx+c=0\end{align*}.
 Zeroproduct property
 The zeroproduct property states that if two factors are multiplied together and their product is zero, then one of the factors must equal zero
Guided Practice
1. Solve for the variable in the polynomial: \begin{align*}x^2+4x21=0\end{align*}
2. Solve for the variable in the polynomial: \begin{align*}20m^2+11m4=0\end{align*}
3. Solve for the variable in the polynomial: \begin{align*}2e^2+7e+6=0\end{align*}
Answers:
1. \begin{align*}x^2+4x21=(x3)(x+7)\end{align*}

\begin{align*}& \qquad (x3)(x+7)=0\\
&\quad \qquad \swarrow \qquad \quad \searrow\\
&(x  3) = 0 \qquad (x+7) = 0\\
&\qquad \ x = 3 \quad \qquad \quad \ \ x = 7\end{align*}
2. \begin{align*}20m^2+11m4=(4m1)(5m+4)\end{align*}

\begin{align*}& \qquad (4m1)(5m+4)=0\\
&\quad \qquad \swarrow \qquad \quad \searrow\\
&4m  1 = 0 \qquad 5m+4 = 0\\
&\quad \ \ 4m = 1 \quad \qquad \ \ 5m = 4\\
&\qquad m = \frac{1}{4} \quad \qquad \quad m = \frac{4}{5}\end{align*}
3. \begin{align*}2e^2+7e+6=(2e+3)(e+2)\end{align*}

\begin{align*}& \qquad (2e+3)(e+2)=0\\
&\quad \qquad \swarrow \qquad \quad \searrow\\
& 2e+3 = 0 \qquad \quad e+2 = 0\\
&\quad \ \ 2e = 3 \quad \qquad \quad e = 2\\
&\qquad e = \frac{3}{2}\end{align*}
Practice
Solve for the variable in each of the following equations.
 \begin{align*}(x+1)(x3)=0\end{align*}
 \begin{align*}(a+3)(a+5)=0\end{align*}
 \begin{align*}(x5)(x+4)=0\end{align*}
 \begin{align*}(2t4)(t+3)=0\end{align*}
 \begin{align*}(x8)(3x7)=0\end{align*}
 \begin{align*}x^2+x12=0\end{align*}
 \begin{align*}b^2+2b24=0\end{align*}
 \begin{align*}t^2+3t18=0\end{align*}
 \begin{align*}w^2+3w108=0\end{align*}
 \begin{align*}e^22e99=0\end{align*}
 \begin{align*}6x^2x2=0\end{align*}
 \begin{align*}2d^2+14d16=0\end{align*}
 \begin{align*}3s^2+20s+12=0\end{align*}
 \begin{align*}18x^2+12x+2=0\end{align*}
 \begin{align*}3j^217j+10=0\end{align*}
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Description
Learning Objectives
Here you'll learn how to solve a quadratic equation by factoring and using the zero product property.