10.3: Solving Quadratic Equations by Factoring

Factoring Method

To solve quadratic equations by factoring, we must make use of the zero-factor property.

Factoring Method

1. 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.
   
   $ax^2 + bx + c = 0$
2. Factor the quadratic expression.
   
   $()() = 0$
3. 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.)

Example $\PageIndex{1}$

$\begin{array}{flushleft} x^2 - 7x + 12 &= 0 & & & \text{The equation is already set equal to } 0\\ (x-3)(x-4) &= 0 & & & \text{Factor. Set each factor equal to } 0.\\ x-3 &= 0 & \text{ or } x - 4 &= 0\\ x &= 3 & \text{ or } x &= 4 \end{array}$

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 $x = 3, 4$.

Example $\PageIndex{2}$

$\begin{array}{flushleft} x^2 &= 25 & \text{Set the equation equal to } 0\\ x^2 - 25 &= 0 & \text{Factor.}\\ (x+5)(x-5) &= 0 & \text{Set each factor equal to } 0\\ x+5=0 & \text{ or } & x - 5 = 0\\ x = -5 & \text{ or } & x=5\\ \end{array}$

Thus, the solutions to this equation are $x = 5, -5$.

Example $\PageIndex{3}$

$\begin{array}{flushleft} x^2 &= 2x & \text{Set the equation equal to } 0\\ x^2 - 2x &= 0 & \text{Factor.}\\ x(x-2) && \text{Set each factor equal to} 0\\ x=0 & \text{ or } & x-2=0\\ && x=2 \end{array}$

Thus, the solutions to this equation are $x = 0, 2$

Example $\PageIndex{4}$

$\begin{array}{flushleft} 2x^2 + 7x - 15 &= 0 & \text{Factor.}\\ (2x - 3)(x+5) &= 0 & \text{Set each factor equal to } 0\\ 2x-3=0 & \text{ or } & x + 5 = 0\\ 2x=3 & \text{ or } & x=-5\\ x=\dfrac{3}{2} \end{array}$

Thus, the solutions to this equation are $x = \dfrac{3}{2}, -5$.

Example $\PageIndex{5}$

$63x^2 = 13x + 6$

$\begin{array}{flushleft} 63x^2 - 13x - 6 &= 0\\ (9x + 2)(7x - 3) &= 0\\ 9x + 2 = 0 & \text{ or } & 7x - 3 = 0\\ 9x = -2 & \text{ or } & 7x = 3\\ x=\dfrac{-2}{9} & \text{ or } & x = \dfrac{3}{7} \end{array}$

Thus, the solutions to this equiation are $x = \dfrac{-2}{9}, \dfrac{3}{7}$

Practice Set A

Solve the following equations, if possible.

Practice Problem $\PageIndex{1}$

$(x−7)(x+4)=0$

Answer

$x=7, −4$

Practice Problem $\PageIndex{2}$

$(2x+5)(5x−7)=0$

Answer

$x = \dfrac{-5}{2}, \dfrac{7}{5}$

Practice Problem $\PageIndex{3}$

$x^2 + 2x - 24 = 0$

Answer

$x=4, −6$

Practice Problem $\PageIndex{4}$

$6x^2 + 13x - 5 = 0$

Answer

$x = \dfrac{1}{3}, \dfrac{-5}{2}$

Practice Problem $\PageIndex{5}$

$5y^2 + 2y = 3$

Answer

$y = \dfrac{3}{5}, -1$

Practice Problem $\PageIndex{6}$

$m(2m - 11) = 0$

Answer

$m = 0, \dfrac{11}{2}$

Practice Problem $\PageIndex{7}$

$6p^2 = -(5p + 1)$

Answer

$p = \dfrac{-1}{3}, \dfrac{-1}{2}$

Practice Problem $\PageIndex{8}$

$r^2 - 49 = 0$

Answer

$r=7,−7$

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

Example $\PageIndex{6}$

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

Practice Problem $\PageIndex{9}$

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.

Exercise $\PageIndex{1}$

$(x+1)(x+3)=0$

Answer

$x=−1, −3$

Exercise $\PageIndex{2}$

$(x+4)(x+9)=0$

Exercise $\PageIndex{3}$

$(x−5)(x−1)=0$

Answer

$x=1,  5$

Exercise $\PageIndex{4}$

$(x−6)(x−3)=0$

Exercise $\PageIndex{5}$

$(x−4)(x+2)=0$

Answer

$x=−2, 4$

Exercise $\PageIndex{6}$

$(x+6)(x−1)=0$

Exercise $\PageIndex{7}$

$(2x+1)(x−7)=0$

Answer

$x = -\dfrac{1}{2}, 7$

Exercise $\PageIndex{8}$

$(3x+2)(x−1)=0$

Exercise $\PageIndex{9}$

$(4x+3)(3x−2)=0$

Answer

$x = -\dfrac{3}{4}, \dfrac{2}{3}$

Exercise $\PageIndex{10}$

$(5x−1)(4x+7)=0$

Exercise $\PageIndex{11}$

$(6x+5)(9x−4)=0$

Answer

$x = -\dfrac{5}{6}, \dfrac{4}{9}$

Exercise $\PageIndex{12}$

$(3a+1)(3a−1)=0$

Exercise $\PageIndex{13}$

$x(x+4)=0$

Answer

$x=−4, 0$

Exercise $\PageIndex{14}$

$y(y−5)=0$

Exercise $\PageIndex{15}$

$y(3y−4)=0$

Answer

$y = 0, \dfrac{4}{3}$

Exercise $\PageIndex{16}$

$b(4b+5)=0$

Exercise $\PageIndex{17}$

$x(2x+1)(2x+8)=0$

Answer

$x = -4, -\dfrac{1}{2}, 0$

Exercise $\PageIndex{18}$

$y(5y+2)(2y−1)=0$

Exercise $\PageIndex{19}$

$(x-8)^2 = 0$

Answer

$x=8$

Exercise $\PageIndex{20}$

$(x-2)^2 = 0$

Exercise $\PageIndex{21}$

$(b + 7)^2 = 0$

Answer

$b=−7$

Exercise $\PageIndex{22}$

$(a + 1)^2$

Exercise $\PageIndex{23}$

$(x(x-4)^2 = 0$

Answer

$x=0, 4$

Exercise $\PageIndex{24}$

$y(y + 9)^2 = 0$

Exercise $\PageIndex{25}$

$y(y-7)^2 = 0$

Answer

$y=0, 7$

Exercise $\PageIndex{26}$

$y(y + 5)^2 = 0$

Exercise $\PageIndex{27}$

$x^2 - 4 = 0$

Answer

$x=−2, 2$

Exercise $\PageIndex{28}$

$x^2 + 9 = 0$

Exercise $\PageIndex{29}$

$x^2 + 36$

Answer

no solution

Exercise $\PageIndex{30}$

$x^2 - 25 = 0$

Exercise $\PageIndex{31}$

$a^2 - 100 = 0$

Answer

$a=−10, 10$

Exercise $\PageIndex{32}$

$a^2 - 81 = 0$

Exercise $\PageIndex{33}$

$b^2 - 49 = 0$

Answer

$b=7, −7$

Exercise $\PageIndex{34}$

$y^2 - 1 = 0$

Exercise $\PageIndex{35}$

$3a^2 - 75 = 0$

Answer

$a=5, −5$

Exercise $\PageIndex{36}$

$5b^2 - 20 = 0$

Exercise $\PageIndex{37}$

$y^3 - y = 0$

Answer

$y=0, 1, −1$

Exercise $\PageIndex{38}$

$a^2 = 9$

Exercise $\PageIndex{39}$

$b^2 = 4$

Answer

$b=2, −2$

Exercise $\PageIndex{40}$

$b^2 = 1$

Exercise $\PageIndex{41}$

$a^2 = 36$

Answer

$a=6, −6$

Exercise $\PageIndex{42}$

$3a^2 = 12$

Exercise $\PageIndex{43}$

$-2x^2 = -4$

Answer

$x = \sqrt{2}, -\sqrt{2}$

Exercise $\PageIndex{44}$

$-2a^2 = -50$

Exercise $\PageIndex{45}$

$-7b^2 = -63$

Answer

$b=3,  −3$

Exercise $\PageIndex{46}$

$-2x^2 = -32$

Exercise $\PageIndex{47}$

$3b^2 = 48$

Answer

$b=4, −4$

Exercise $\PageIndex{48}$

$a^2 - 8a + 16 = 0$

Exercise $\PageIndex{49}$

$y^2 + 10y + 25 = 0$

Answer

$y=−5$

Exercise $\PageIndex{50}$

$y^2 + 9y + 16 = 0$

Exercise $\PageIndex{51}$

$x^2 - 2x - 1 = 0$

Answer

no solution

Exercise $\PageIndex{52}$

$a^2 + 6a + 9 = 0$

Exercise $\PageIndex{53}$

$a^2 + 4a + 4 = 0$

Answer

$a=−2$

Exercise $\PageIndex{54}$

$x^2 + 12x = -36$

Exercise $\PageIndex{55}$

$b^2 - 14b = -49$

Answer

$b=7$

Exercise $\PageIndex{56}$

$3a^2 + 18a + 27 = 0$

Exercise $\PageIndex{57}$

$2m^3 + 4m^2 + 2m = 0$

Answer

$m=0, −1$

Exercise $\PageIndex{58}$

$3mn^2 - 36mn + 36m = 0$

Exercise $\PageIndex{59}$

$a^2 + 2a - 3 = 0$

Answer

$a=−3, 1$

Exercise $\PageIndex{60}$

$a^2 + 3a - 10 = 0$

Exercise $\PageIndex{61}$

$x^2 + 9x + 14 = 0$

Answer

$x=−7, −2$

Exercise $\PageIndex{62}$

$x^2 - 7x + 12 = 3$

Exercise $\PageIndex{63}$

$b^2 + 12b + 27 = 0$

Answer

$b=−9,  −3$

Exercise $\PageIndex{64}$

$b^2 - 3b + 2 = 0$

Exercise $\PageIndex{65}$

$x^2 - 13x = -42$

Answer

$x=6, 7$

Exercise $\PageIndex{66}$

$a^3 = -8a^2 - 15a$

Exercise $\PageIndex{67}$

$6a^2 + 13a + 5 = 0$

Answer

$a = -\dfrac{5}{3}, -\dfrac{1}{2}$

Exercise $\PageIndex{68}$

$6x^2 - 4x - 2 = 0$

Exercise $\PageIndex{69}$

$12a^2 + 15a + 3 = 0$

Answer

$a = -\dfrac{1}{4}, -1$

Exercise $\PageIndex{70}$

$18b^2 + 24b + 6 = 0$

Exercise $\PageIndex{71}$

$12a^2 + 24a + 12 = 0$

Answer

$a=−1$

Exercise $\PageIndex{72}$

$4x^2 - 4x = -1$

Exercise $\PageIndex{73}$

$2x^2 = x + 15$

Answer

$x = -\dfrac{5}{2}, 3$

Exercise $\PageIndex{74}$

$4a^2 = 4a + 3$

Exercise $\PageIndex{75}$

$4y^2 = -4y - 2$

Answer

no solution

Exercise $\PageIndex{76}$

$9y^2 = 9y + 18$

Exercises For Review

Exercise $\PageIndex{77}$

Simplify $(x^4y^3)^2(xy^2)^4$

Answer

$x^{12}y^{14}$

Exercise $\PageIndex{78}$

Write $(x^{-2}y^3w^4)^{-2}$ so that only positive exponents appear.

Exercise $\PageIndex{79}$

Find the sum: $\dfrac{x}{x^2 - x - 2} + \dfrac{1}{x^2 - 3x + 2}$

Answer

$\dfrac{x^2 + 1}{(x+1)(x-1)(x-2)}$

Exercise $\PageIndex{80}$

Simplify $\dfrac{\dfrac{1}{a} + \dfrac{1}{b}}{\dfrac{1}{a} - \dfrac{1}{b}}$

Exercise $\PageIndex{81}$

Solve $(x + 4)(3x + 1) = 0$

Answer

$x = -4, \dfrac{-1}{3}$