17Calculus Precalculus - Zero Product Rule
The Zero Product Rule (also called Zero Product Property) is a simple yet powerful rule that you will use a lot in calculus. Here is how it works.
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If you have an equation like \( (a) (b) = 0 \), you can automatically assume that either \(a = 0 \) and/or \(b=0\). The reason this works is that when you multiply any number by zero, the result is always zero. For example, \( (2) 0 = 0 \) and \( 0 (621) = 0 \). Any number or variable multiplied by zero is always equal to zero.
This also works with more than one variable. For example, if you have \( (x) (y) (z) (a) = 0\), then you can write \( x=0 \), \(y=0\), \(z=0\) and/or \(a=0\). The key is that one or more of them have to be zero for this equation be true.
One common mistake students make is to assume this works when you have something other than zero on one side of the equation. For example, if you have \( (a) (b) = 1 \), you cannot assume that \( a=1\) or \( b=1\). This would hold if \( a=2\) and \( b=1/2\) neither of which are \(1\). So the key when using the zero product rule is that you need zero on one side of the equation.
Another thing to watch is if you have something other multiplication, i.e. \( (a) (b) + 1 = 0 \). Of course, you should know by now that you can't say \( a=0 \) and/or \(b=0\). The \( +1 \) term does not allow you to apply the zero product rule.
This rule is used mostly when solving for roots, as in this example.
Example
Find the roots (where the function crosses the x-axis) of \( f(x) = x^2 - 3x + 2 \).
Solution
When the function crosses the x-axis, the y-value is zero, so we need to solve \( x^2 - 3x +2 = 0 \). First, we factor.
\(
\begin{array}{rcl}
x^2 - 3x +2 & = & 0 \\
(x-1)(x-2) & = & 0
\end{array}
\)
Notice we have two expressions within parentheses multiplied together and we have zero on the other side of the equal sign. So we can apply the zero product rule.
\( x-1 = 0 \) or \( x-2 = 0 \)
Solving for \(x\) in both equations gives us
\( x=1 \) or \( x=2 \)
So we know that the roots of \( f(x) \) are \( x=1\) and \( x=2\).
Okay, time for some practice problems.
Practice
Instructions - - Unless otherwise instructed, use the zero product rule to solve these equations, i.e. find the x-values that make these equations true.
\( x(x-3) = 0 \)
Problem Statement |
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Solve \( x(x-3) = 0 \) using the zero product rule.
Final Answer |
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\(x = 0\) or \(x = 3\)
Problem Statement |
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Solve \( x(x-3) = 0 \) using the zero product rule.
Solution |
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2740 video
Final Answer |
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\(x = 0\) or \(x = 3\) |
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\( (x+2)(x-3) = 0 \)
Problem Statement |
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Solve \( (x+2)(x-3) = 0 \) using the zero product rule.
Final Answer |
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\(x = -2\) or \( x = 3 \)
Problem Statement |
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Solve \( (x+2)(x-3) = 0 \) using the zero product rule.
Solution |
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2741 video
Final Answer |
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\(x = -2\) or \( x = 3 \) |
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\( 4x(x+5)=0 \)
Problem Statement |
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Solve \( 4x(x+5)=0 \) using the zero product rule.
Final Answer |
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\( x = 0 \) or \( x = -5 \)
Problem Statement |
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Solve \( 4x(x+5)=0 \) using the zero product rule.
Solution |
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2742 video
Final Answer |
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\( x = 0 \) or \( x = -5 \) |
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\( (x-2)(x+7) = 0 \)
Problem Statement |
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Solve \( (x-2)(x+7) = 0 \) using the zero product rule.
Final Answer |
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\( x = 2 \) or \( x = -7 \)
Problem Statement |
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Solve \( (x-2)(x+7) = 0 \) using the zero product rule.
Solution |
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2743 video
Final Answer |
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\( x = 2 \) or \( x = -7 \) |
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\( (2x+3)(5x-1) = 0 \)
Problem Statement |
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Solve \( (2x+3)(5x-1) = 0 \) using the zero product rule.
Final Answer |
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\( x = -3/2 \) or \( x = 1/5 \)
Problem Statement |
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Solve \( (2x+3)(5x-1) = 0 \) using the zero product rule.
Solution |
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2744 video
Final Answer |
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\( x = -3/2 \) or \( x = 1/5 \) |
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\( x(x-1)(6x+11) = 0 \)
Problem Statement |
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Solve \( x(x-1)(6x+11) = 0 \) using the zero product rule.
Final Answer |
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\( x = 0 \) or \( x = -1 \) or \( x = -11/6 \)
Problem Statement |
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Solve \( x(x-1)(6x+11) = 0 \) using the zero product rule.
Solution |
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Final Answer |
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\( x = 0 \) or \( x = -1 \) or \( x = -11/6 \) |
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Problem Statement |
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\(-x+x^2=6\)
Solution |
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\( x^2 - x - 12 = 0 \)
Problem Statement |
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Solve \( x^2 - x - 12 = 0 \) by factoring and applying the zero product rule.
Final Answer |
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\( x = 4 \) or \( x = -3 \)
Problem Statement |
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Solve \( x^2 - x - 12 = 0 \) by factoring and applying the zero product rule.
Solution |
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2746 video
Final Answer |
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\( x = 4 \) or \( x = -3 \) |
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\( x^2 + 6x - 16 = 0 \)
Problem Statement |
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Solve by \( x^2 + 6x - 16 = 0 \) factoring and applying the zero product rule.
Solution |
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2747 video
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\( x^2 - 9x = 0 \)
Problem Statement |
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Solve \( x^2 - 9x = 0 \) by factoring and applying the zero product rule.
Final Answer |
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\( x = 0 \) or \( x = 9 \)
Problem Statement |
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Solve \( x^2 - 9x = 0 \) by factoring and applying the zero product rule.
Solution |
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2748 video
Final Answer |
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\( x = 0 \) or \( x = 9 \) |
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\( x^2 - 3x - 10 = 0 \)
Problem Statement |
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Solve \( x^2 - 3x - 10 = 0 \) by factoring and applying the zero product rule.
Final Answer |
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\( x = 5 \) or \( x = -2 \)
Problem Statement |
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Solve \( x^2 - 3x - 10 = 0 \) by factoring and applying the zero product rule.
Solution |
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2749 video
Final Answer |
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\( x = 5 \) or \( x = -2 \) |
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\( x^2 - 36 = 0 \)
Problem Statement |
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Solve \( x^2 - 36 = 0 \) by factoring and applying the zero product rule.
Final Answer |
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\( x = 6 \) or \( x = -6 \)
Problem Statement |
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Solve \( x^2 - 36 = 0 \) by factoring and applying the zero product rule.
Solution |
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2750 video
Final Answer |
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\( x = 6 \) or \( x = -6 \) |
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\( 3x^2 + 4x - 7 = 0 \)
Problem Statement |
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Solve \( 3x^2 + 4x - 7 = 0 \) by factoring and applying the zero product rule.
Final Answer |
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\( x = 0 \) or \( x = -7/3 \)
Problem Statement |
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Solve \( 3x^2 + 4x - 7 = 0 \) by factoring and applying the zero product rule.
Solution |
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2751 video
Final Answer |
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\( x = 0 \) or \( x = -7/3 \) |
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Really UNDERSTAND Precalculus
Topics You Need To Understand For This Page
algebra factoring |
Related Topics and Links
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
|
Set 2 - squared identities | ||
|---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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