Solving by Factoring


If \(ab=0\), then \(a=0\) or \(b=0\). (This includes the possibility that they are both \(0\).)

That is, if two factors multiply to \(0\), at least one of the factors must be equal to \(0\). 

Example 4

Solve \((3x+5)(7x-4)=0\). 

Solution

This is a quadratic equation, since it can be expanded and written in the form \(ax^2+bx+c=0\).  

However, the equation is easier to solve in its current, factored form.

If \((3x+5)(7x-4)=0\), then \(3x+5=0\) or \(7x-4=0\) (one of the factors must equal \(0\)).

Solve each equation separately.

\(\begin{align*}3x+5 &= 0 \\ 3x &= -5 \\ x &= -\frac53\end{align*}\)

or

\(\begin{align*} 7x-4&=0 \\ 7x &= 4 \\ x &= \frac47 \end{align*}\)

The two solutions to \((3x+5)(7x-4)=0\) or \(x=-\dfrac53\) and \(x=\dfrac47\). 

In this example, we saw another method for solving a quadratic equation: 

It is possible to solve a quadratic equation if one side of the equation is factored.  For this to be effective, the other side must equal \(0\).  

Once these conditions are met, set each of the factors equal to \(0\) and solve the resulting equations. These equations should be simpler to solve than the original one. 

Let's look at some more examples. 

 


Slide Notes

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Solving a Quadratic Equation by Factoring

Another way to solve a quadratic equation​​​: 

 

Example 5

Solve \(x^2+11x+24=0\).

 

 

Checking Example 5

Check that \(x=-3\) and \(x=-8\) are the solutions to  \(x^2+11x+24=0\).

 

Check Your Understanding 3

Identify the two solutions of the following equation by factoring.

\(((((((exp(r))*(e))*(s))*(s))*(i))*(o))*(n) = 0\)

Enter the two solutions separated by a semi-colon such as '\(a;b\)'.

The solutions are There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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The equation can be solved by factoring:

\(\begin{align*} ((((((exp(r))*(e))*(s))*(s))*(i))*(o))*(n) &= 0 \\ (((fact(o))*(r))*1.0)(((fact(o))*(r))*2.0) &= 0 \end{align*}\)

Either \(((fact(o))*(r))*1.0 = 0\) or \(((fact(o))*(r))*2.0 = 0\).

Therefore, \(x = (((s)*(o))*(l))*1.0\) or \(x = (((s)*(o))*(l))*2.0\).

Why One Side Needs to Equal 0

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Check Your Understanding 3


Identify the two solutions of the following equation by factoring.

\(((((((exp(r))*(e))*(s))*(s))*(i))*(o))*(n) = 0\)

Enter the two solutions separated by a semi-colon such as "\(a;b\)".

The solutions are There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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Slide Notes

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Example 6 — Part A

Solve by factoring.

  1. \(2x^2=9x+5\)
  2. \(4x^2-36x+56=0\)
  3. \(x^2+10x=0\)

 

Example 6 — Part B

Solve by factoring. 

  1. \(2x^2=9x+5\)
  2. \(4x^2-36x+56=0\)
  3. \(x^2+10x=0\)

 

Example 6 — Part B Continued

Solve by factoring. 

  1. \(2x^2=9x+5\)
  2. \(4x^2-36x+56=0\)
  3. \(x^2+10x=0\)

Solution — Part B

Check the solutions \(x=7\) or \(x=2\) by graphing

 

Example 6 — Part C

Solve by factoring. 

  1. \(2x^2=9x+5\)
  2. \(4x^2-36x+56=0\)
  3. \(x^2+10x=0\)

 

Example 7

Solve \(25x^2-64=0\) using inverse operations and by factoring. 

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Example 8

Solve \(16x^2+9=24x\).

Solution

To factor, ensure one side of the equation is equal to \(0\). In this case, subtract \(24x\) from both sides.

\(16x^2-24x+9=0\)

This equation factors as \((4x-3)^2=0\).

The two factors are the same, so there will only be one solution to the equation. It will satisfy \(4x-3=0\). 

Therefore, the solution to the equation is \(x = \dfrac34\).

Solving \(16x^2-24x+9=0\) is equivalent to finding the zeros of the relation \(y=16x^2-24x+9\).

Here is the graph of that relation:

The vertex of the graph occurs at the point (0.75, 0).

There is only one zero, at \(x=\dfrac34\), which is consistent with our solution.


Check Your Understanding 4


Identify the two solutions of the following equation by factoring.

\(5x^2 - 40x = 45\)

Enter the two solutions separated by a semi-colon. For example, "\(a;b\)".

The solutions are There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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