10.3: Solving Quadratic Equations by Factoring
- Page ID
- 49403
Factoring Method
To solve quadratic equations by factoring, we must make use of the zero-factor property.
- Set the equation equal to zero, that is, get all the nonzero terms on one side of the equal sign and 0 on the other.
\(ax^2 + bx + c = 0\) - Factor the quadratic expression.
\(()() = 0\) - By the zero-factor property, at least one of the factors must be zero, so, set each of the factors equal to 0 and solve for the variable.
Sample Set A
Solve the following quadratic equations. (We will show the check for problem 1.)
\(\begin{array}{flushleft}
x^2 - 7x + 12 &= 0 & & & \text{The equation is already set equal to } 0\\
(x-3)(x-4) &= 0 & & & \text{Factor. Set each factor equal to } 0.\\
x-3 &= 0 & \text{ or } x - 4 &= 0\\
x &= 3 & \text{ or } x &= 4
\end{array}\)
Check:
If \(x = 3, x^2 - 7x + 12 = 0\)
\(\begin{array}{flushleft}
3^2 - 7 \cdot 3 + 12 & = 0 & \text{Is this correct?}\\
9 - 21 + 12 &= 0 & \text{Is this correct?}\\
0 &= 0 & \text{Yes, this is correct?}
\end{array}\)
Check: If \(x = 4, x^2 - 7x + 12 = 0\)
\(\begin{array}{flushleft}
4^2 - 7 \cdot 4 + 12 &= 0 & \text{Is this correct?}\\
16 - 28 + 12 &= 0 & \text{Is this correct?}\\
0 &= 0 & \text{Yes, this is correct}
\end{array}\)
Thus, the solutions to this equation are \(x = 3, 4\).
\(\begin{array}{flushleft}
x^2 &= 25 & \text{Set the equation equal to } 0\\
x^2 - 25 &= 0 & \text{Factor.}\\
(x+5)(x-5) &= 0 & \text{Set each factor equal to } 0\\
x+5=0 & \text{ or } & x - 5 = 0\\
x = -5 & \text{ or } & x=5\\
\end{array}\)
Thus, the solutions to this equation are \(x = 5, -5\).
\(\begin{array}{flushleft}
x^2 &= 2x & \text{Set the equation equal to } 0\\
x^2 - 2x &= 0 & \text{Factor.}\\
x(x-2) && \text{Set each factor equal to} 0\\
x=0 & \text{ or } & x-2=0\\
&& x=2
\end{array}\)
Thus, the solutions to this equation are \(x = 0, 2\)
\(\begin{array}{flushleft}
2x^2 + 7x - 15 &= 0 & \text{Factor.}\\
(2x - 3)(x+5) &= 0 & \text{Set each factor equal to } 0\\
2x-3=0 & \text{ or } & x + 5 = 0\\
2x=3 & \text{ or } & x=-5\\
x=\dfrac{3}{2}
\end{array}\)
Thus, the solutions to this equation are \(x = \dfrac{3}{2}, -5\).
\(63x^2 = 13x + 6\)
\(\begin{array}{flushleft}
63x^2 - 13x - 6 &= 0\\
(9x + 2)(7x - 3) &= 0\\
9x + 2 = 0 & \text{ or } & 7x - 3 = 0\\
9x = -2 & \text{ or } & 7x = 3\\
x=\dfrac{-2}{9} & \text{ or } & x = \dfrac{3}{7}
\end{array}\)
Thus, the solutions to this equiation are \(x = \dfrac{-2}{9}, \dfrac{3}{7}\)
Practice Set A
Solve the following equations, if possible.
\((x−7)(x+4)=0\)
- Answer
-
\(x=7, −4\)
\((2x+5)(5x−7)=0\)
- Answer
-
\(x = \dfrac{-5}{2}, \dfrac{7}{5}\)
\(x^2 + 2x - 24 = 0\)
- Answer
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\(x=4, −6\)
\(6x^2 + 13x - 5 = 0\)
- Answer
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\(x = \dfrac{1}{3}, \dfrac{-5}{2}\)
\(5y^2 + 2y = 3\)
- Answer
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\(y = \dfrac{3}{5}, -1\)
\(m(2m - 11) = 0\)
- Answer
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\(m = 0, \dfrac{11}{2}\)
\(6p^2 = -(5p + 1)\)
- Answer
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\(p = \dfrac{-1}{3}, \dfrac{-1}{2}\)
\(r^2 - 49 = 0\)
- Answer
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\(r=7,−7\)
Solving Mentally After Factoring
Let's consider problems 4 and 5 of Sample Set A in more detail. Let's look particularly at the factorizations \((2x-3)(x + 5) = 0\) and \((9x + 2)(7x - 3) = 0\)/ The next step is to set each factor equal to zero and solve. We can solve mentally if we understand how to solve linear equations: we transpose the constant from the variable term and then divide by the coefficient of the variable.
Sample Set B
Solve the following equation mentally.
\((2x - 3)(x + 5) = 0\)
\(\begin{array}{flushleft}
2x - 3 &= 0 & \text{Mentally add } 3 \text{ to both sides. The constant changes sign.}\\
2x &= 3 & \text{Divide by } 2 \text{ the coefficient of } x \text{. The } 2 \text{ divides the cosntant } 3 \text{ into } \dfrac{3}{2}\\
& & \text{The coefficient becomes the denominator.}\\
x &= \dfrac{3}{2}\\
x + 5 &= 0 & \text{ Mentally subtract } 5 \text{ from both sides. The constant changes sign.}\\
x &= -5 & \text{Divide by the coefficient of } x, 1. \text{ The coefficient becomes the denominator}\\
x = \dfrac{-5}{1} &= -5\\
x &= -5
\end{array}\)
Now, we can immediately write the solution to the equation after factoring by looking at each factor, changing the sign of the constant, then divide by the coefficient.
Practice Set B
Solve \((9x + 2)(7x - 3) = 0\) using this mental method.
- Answer
-
\(x = -\dfrac{2}{9}, \dfrac{3}{7}\)
Exercises
For the following problems, solve the equations, if possible.
\((x+1)(x+3)=0\)
- Answer
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\(x=−1, −3\)
\((x+4)(x+9)=0\)
\((x−5)(x−1)=0\)
- Answer
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\(x=1, 5\)
\((x−6)(x−3)=0\)
\((x−4)(x+2)=0\)
- Answer
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\(x=−2, 4\)
\((x+6)(x−1)=0\)
\((2x+1)(x−7)=0\)
- Answer
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\(x = -\dfrac{1}{2}, 7\)
\((3x+2)(x−1)=0\)
\((4x+3)(3x−2)=0\)
- Answer
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\(x = -\dfrac{3}{4}, \dfrac{2}{3}\)
\((5x−1)(4x+7)=0\)
\((6x+5)(9x−4)=0\)
- Answer
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\(x = -\dfrac{5}{6}, \dfrac{4}{9}\)
\((3a+1)(3a−1)=0\)
\(x(x+4)=0\)
- Answer
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\(x=−4, 0\)
\(y(y−5)=0\)
\(y(3y−4)=0\)
- Answer
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\(y = 0, \dfrac{4}{3}\)
\(b(4b+5)=0\)
\(x(2x+1)(2x+8)=0\)
- Answer
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\(x = -4, -\dfrac{1}{2}, 0\)
\(y(5y+2)(2y−1)=0\)
\((x-8)^2 = 0\)
- Answer
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\(x=8\)
\((x-2)^2 = 0\)
\((b + 7)^2 = 0\)
- Answer
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\(b=−7\)
\((a + 1)^2\)
\((x(x-4)^2 = 0\)
- Answer
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\(x=0, 4\)
\(y(y + 9)^2 = 0\)
\(y(y-7)^2 = 0\)
- Answer
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\(y=0, 7\)
\(y(y + 5)^2 = 0\)
\(x^2 - 4 = 0\)
- Answer
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\(x=−2, 2\)
\(x^2 + 9 = 0\)
\(x^2 + 36\)
- Answer
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no solution
\(x^2 - 25 = 0\)
\(a^2 - 100 = 0\)
- Answer
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\(a=−10, 10\)
\(a^2 - 81 = 0\)
\(b^2 - 49 = 0\)
- Answer
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\(b=7, −7\)
\(y^2 - 1 = 0\)
\(3a^2 - 75 = 0\)
- Answer
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\(a=5, −5\)
\(5b^2 - 20 = 0\)
\(y^3 - y = 0\)
- Answer
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\(y=0, 1, −1\)
\(a^2 = 9\)
\(b^2 = 4\)
- Answer
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\(b=2, −2\)
\(b^2 = 1\)
\(a^2 = 36\)
- Answer
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\(a=6, −6\)
\(3a^2 = 12\)
\(-2x^2 = -4\)
- Answer
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\(x = \sqrt{2}, -\sqrt{2}\)
\(-2a^2 = -50\)
\(-7b^2 = -63\)
- Answer
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\(b=3, −3\)
\(-2x^2 = -32\)
\(3b^2 = 48\)
- Answer
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\(b=4, −4\)
\(a^2 - 8a + 16 = 0\)
\(y^2 + 10y + 25 = 0\)
- Answer
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\(y=−5\)
\(y^2 + 9y + 16 = 0\)
\(x^2 - 2x - 1 = 0\)
- Answer
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no solution
\(a^2 + 6a + 9 = 0\)
\(a^2 + 4a + 4 = 0\)
- Answer
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\(a=−2\)
\(x^2 + 12x = -36\)
\(b^2 - 14b = -49\)
- Answer
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\(b=7\)
\(3a^2 + 18a + 27 = 0\)
\(2m^3 + 4m^2 + 2m = 0\)
- Answer
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\(m=0, −1\)
\(3mn^2 - 36mn + 36m = 0\)
\(a^2 + 2a - 3 = 0\)
- Answer
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\(a=−3, 1\)
\(a^2 + 3a - 10 = 0\)
\(x^2 + 9x + 14 = 0\)
- Answer
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\(x=−7, −2\)
\(x^2 - 7x + 12 = 3\)
\(b^2 + 12b + 27 = 0\)
- Answer
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\(b=−9, −3\)
\(b^2 - 3b + 2 = 0\)
\(x^2 - 13x = -42\)
- Answer
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\(x=6, 7\)
\(a^3 = -8a^2 - 15a\)
\(6a^2 + 13a + 5 = 0\)
- Answer
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\(a = -\dfrac{5}{3}, -\dfrac{1}{2}\)
\(6x^2 - 4x - 2 = 0\)
\(12a^2 + 15a + 3 = 0\)
- Answer
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\(a = -\dfrac{1}{4}, -1\)
\(18b^2 + 24b + 6 = 0\)
\(12a^2 + 24a + 12 = 0\)
- Answer
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\(a=−1\)
\(4x^2 - 4x = -1\)
\(2x^2 = x + 15\)
- Answer
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\(x = -\dfrac{5}{2}, 3\)
\(4a^2 = 4a + 3\)
\(4y^2 = -4y - 2\)
- Answer
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no solution
\(9y^2 = 9y + 18\)
Exercises For Review
Simplify \((x^4y^3)^2(xy^2)^4\)
- Answer
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\(x^{12}y^{14}\)
Write \((x^{-2}y^3w^4)^{-2}\) so that only positive exponents appear.
Find the sum: \(\dfrac{x}{x^2 - x - 2} + \dfrac{1}{x^2 - 3x + 2}\)
- Answer
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\(\dfrac{x^2 + 1}{(x+1)(x-1)(x-2)}\)
Simplify \(\dfrac{\dfrac{1}{a} + \dfrac{1}{b}}{\dfrac{1}{a} - \dfrac{1}{b}}\)
Solve \((x + 4)(3x + 1) = 0\)
- Answer
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\(x = -4, \dfrac{-1}{3}\)