Solving Quadratic Equations and Inequalities

Factoring Quadratic Equations

Factoring with a Leading Coefficient of 1

Quadratic equations of the form x2+bx+c=0x^2 + bx + c = 0 may be solved by factoring the quadratic expression and setting each factor equal to zero.

Factoring is a process of writing an algebraic expression as a product. Factoring can be used with the zero product property to solve quadratic equations. The zero product property states that if the product of real numbers is zero, then one of its factors is equal to zero. In other words, if pp and qq are real numbers and pq=0pq=0, then p=0p=0 or q=0q=0.

To solve a quadratic equation by factoring:

1. Set one side of the equation equal to zero.

2. Factor the other side of the equation.

3. Set each factor equal to zero.

4. Solve the resulting equations.

To factor an expression of the form x2+bx+cx^2+bx+c, look for two factors of cc that have a sum of bb. When cc is positive, the two factors have the same sign as the sign of bb. When cc is negative, the two factors have opposite signs, and the factor with the greater absolute value (greater distance from zero on a number line) will have the same sign as bb.

Step-By-Step Example
Factor to Solve a Quadratic Equation with b>0b>0 and c>0c>0
Solve the quadratic equation:
x2+8x+15=0x^2+8x+15=0
Step 1
The given equation contains the expression:
x2+8x+15x^2+8x+15
The expression is in the form:
x2+bx+cx^2+bx+c
In the given expression, b=8b=8 and c=15c=15.

Next, look for two factors of cc, or 15, that have a sum of bb, or 8. When cc is positive, the factors will have the same sign as bb. So, the factors are +3+3 and +5+5 are positive.

Use the factors to write the factored form of the expression:
(x+3)(x+5)(x+3)(x+5)
Step 2
Use the factored form of the quadratic expression to write an equivalent equation.
(x+3)(x+5)=0(x+3)(x+5)=0
Step 3
Apply the zero product property by setting each factor equal to zero.
x+3=0orx+5=0x+3 = 0\;\;\;\;\; \text{or}\;\;\;\;\; x+5=0
Solution
Solve each equation.
x+3=0x=3orx+5=0x=5\begin{aligned}x+3&=0\\x&=-3\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned} x+5&=0\\x&=-5\end{aligned}
The solutions are x=3x=-3 and x=5x=-5.
Step-By-Step Example
Factor to Solve a Quadratic Equation with b<0b\lt0 and c>0c\gt0
Solve the quadratic equation:
x214x+48=0x^2-14x+48=0
Step 1
The given equation contains the expression:
x214x+48x^2-14x+48
The expression is in the form:
x2+bx+cx^2+bx+c
In the given expression, b=14b=-14 and c=48c=48.

Next, look for two factors of cc, or 48, that have a sum of bb, or –14. When cc, or 48, is positive, both factors will have the same sign as bb. So, the factors are –6 and –8.

Use the factors to write the factored form of the expression:
(x6)(x8)(x-6)(x-8)
Step 2
Use the factored form of the quadratic expression to write an equivalent equation.
(x6)(x8)=0(x-6)(x-8)=0
Step 3
Apply the zero product property by setting each factor equal to zero.
x6=0orx8=0x-6 = 0\;\;\;\;\;\text{or}\;\;\;\;\;x-8=0
Solution
Solve each equation.
x6=0x=6orx8=0x=8\begin{aligned}x-6&=0\\x&=6\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned}x-8&=0\\x&=8\end{aligned}
The solutions are x=6x=6 or x=8x=8.
Step-By-Step Example
Factor to Solve a Quadratic Equation with b>0b>0 and c<0c<0
Solve the quadratic equation:
x2+x2=0x^2+x-2=0
Step 1
The given equation contains the expression:
x2+x2x^2+x-2
The expression is in the form:
x2+bx+cx^2+bx+c
In the given expression, b=1b=1 and c=2c=-2.

Next, look for two factors of cc, or –2 that have a sum of bb, or 1. When cc is negative, the factors will have different signs. The factor with the greater absolute value will have the same sign as bb. The factors are +2+2 and 1-1.

Use the factors to write the factored form of the expression:
(x+2)(x1)(x+2)(x-1)
Step 2
Use the factored form of the quadratic expression to write an equivalent equation.
(x+2)(x1)=0(x+2)(x-1)=0
Step 3
Apply the zero product property by setting each factor equal to zero.
x+2=0orx1=0x+2 = 0\;\;\;\;\;\text{or}\;\;\;\;\;x-1=0
Solution
Solve each equation.
x+2=0x=2orx1=0x=1\begin{aligned}x+2&=0\\x&=-2\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned}x-1&=0\\x&=1\end{aligned}
The solutions are x=2x=-2 and x=1x=1.
Step-By-Step Example
Factor to Solve a Quadratic Equation with b<0b<0 and c<0c<0
Solve the quadratic equation:
x24x21=0x^2-4x-21=0
Step 1
The given equation contains the expression:
x24x21x^2-4x-21
The given expression is in the form:
x2+bx+cx^2+bx+c
In the given expression, b=21b=-21 and c=4c=-4.

Next, look for two factors of bb, or –21, that have a sum of cc, or –4. When cc is negative, the factors will have different signs. The factor with the greater absolute value will have the same sign as bb. The factors are 7-7 and +3+3.

Use the factors to write the factored form of the expression:
(x7)(x+3)(x-7)(x+3)
Step 2
Use the factored form of the quadratic expression to write an equivalent equation.
(x7)(x+3)=0(x-7)(x+3)=0
Step 3
Apply the zero product property by setting each factor equal to zero.
x7=0orx+3=0x-7 = 0 \;\;\;\;\;\text{or}\;\;\;\;\; x+3=0
Solution
Solve each equation.
x7=0x=7orx+3=0x=3\begin{aligned}x-7&=0\\x&=7\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned}x+3&=0\\x&=-3\end{aligned}
The solutions are x=7x=7 and x=3x=-3.

Factoring with a Leading Coefficient

Quadratic equations of the form ax2+bx+c=0ax^2 + bx + c = 0 may be solved by factoring the quadratic expression and setting each factor equal to zero.
The expression has three terms: ax2ax^2, bxbx, and cc.
ax2+bx+cax^2+bx+c
The leading coefficient is aa because it is the coefficient of the first term when the expression is written in standard form.

To factor the expression, factor out –1 from each term if necessary so that aa is positive. Then look for two factors of acac that have a sum of bb. When acac is positive, the two factors have the same sign as determined by the sign of bb. When acac is negative, the two factors have opposite signs, and the factor with the greater absolute value, or the distance from zero on a number line, will have the same sign as bb. Use the two factors to write the middle term of the quadratic expression as a sum, and then factor by grouping.

To solve a quadratic equation by factoring by grouping:

1. Set one side of the equation equal to zero.

2. Factor the other side of the equation by grouping.

3. Set each factor equal to zero.

4. Solve the resulting equations.

Step-By-Step Example
Factor to Solve a Quadratic Equation with a1a\neq1
Solve the quadratic equation:
3x2+13x+4=03x^2+13x+4=0
Step 1
The given equation contains the expression:
3x2+13x+43x^2+13x+4
The expression is in the form:
ax2+bx+cax^2+bx+c
In the given expression, a=3a=3, b=13b=13, and c=4c=4. So, look for two factors of ac=12ac=12 that have a sum of, bb, or 13. The factors are +1+1 and +12+12. Use the factors to write the middle term of the expression as a sum:
3x2+13x+43x2+1x+12x+4\begin{gathered}3x^2+{\color{#c42126}{13x}}+4\\3x^2{\color{#c42126}{+1x+12x}}+4\end{gathered}
Step 2
Use parentheses to group the terms.
(3x2+1x)+(12x+4)(3x^2+1x)+(12x+4)
Then factor out the GCF (greatest common factor) from each group. In the first group, both terms have a GCF of xx. In the second group, both terms have a GCF of 4.
x(3x+1)+4(3x+1)x(3x+1)+4(3x+1)
The resulting terms have a common factor of 3x+13x+1. Factor out the common factor from each term. Group the remaining factors.
(3x+1)(x+4)(3x+1)(x+4)
Step 3
Use the factored form of the quadratic expression to write an equivalent equation.
(3x+1)(x+4)=0(3x+1)(x+4)=0
Step 4
Apply the zero product property by setting each factor equal to zero.
3x+1=0orx+4=03x+1=0\;\;\;\;\;\text{or}\;\;\;\;\;x+4=0
Solution
Solve each equation.
3x+1=03x=1x=13orx+4=0x=4\begin{aligned}3x+1&=0\\3x&=-1\\x&=-\frac{1}{3}\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned}x+4&=0\\x&=-4\end{aligned}
The solutions are x=13x=-\frac{1}{3} and x=4x=-4.
Step-By-Step Example
Factor to Solve a Quadratic Equation with a1a\neq1 and a<0a<0
Solve the quadratic equation:
2x2+15x18=0-2x^2+15x-18=0
Step 1
In the given equation, the coefficient of x2x^2 is negative. Multiply both sides of the equation by –1 to get a positive value for aa.
2x215x+18=01(2x215x+18)=1(0)2x215x+18=0\begin{aligned}2x^2-15x+18&=0\\-1(2x^2-15x+18)&=-1(0)\\{2x^2}-15x+18&=0\end{aligned}
The given equation contains the expression:
2x215x+18{2x^2}-15x+18
The expression is in the form:
ax2+bx+cax^2+bx+c
In the given expression, a=2a=2, b=15b=-15, and c=18c=18. Look for two factors of ac=36ac=36 that have a sum of bb, or –15. The factors are –3 and –12. Use the factors to write the middle term of the expression as a sum:
2x215x+182x23x12x+18\begin{gathered}-2x^2{\color{#c42126}{-15x}}+{18}\\{2x^2}{\color{#c42126}{-3x-12x}}+18\end{gathered}
Step 2
Use parentheses to group the terms.
(2x23x)+(12x+18)(2x^2-3x)+(-12x+18)
Then factor out the GCF (greatest common factor) from each group. In the first group, both terms have a GCF of xx. In the second group, both terms have a GCF of 6.
x(2x3)+6(2x+3)x(2x-3)+6(-2x+3)
The first term in the second group is negative, so factor out –1 from the second group.
x(2x3)6(2x3)x(2x-3)-6(2x-3)
The resulting terms have a common factor of 2x32x-3. Factor out the common factor from each term. Group the remaining factors.
(2x3)(x6)(2x-3)(x-6)
Step 3
Use the factored form of the quadratic expression to write an equivalent equation.
(2x3)(x6)=0(2x-3)(x-6)=0
Step 4
Apply the zero product property by setting each factor equal to zero.
2x3=0orx6=0{2x-3=0}\;\;\;\;\;\text{or}\;\;\;\;\;{x-6=0}
Solution
Solve each equation.
2x3=02x=3x=32orx6=0x=6\begin{aligned}{2x-3}&=0\\{2x}&=3\\{x}&=\frac{3}{2}\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned}{x-6}&=0\\{x}&=6\end{aligned}
The solutions are x=32x=\frac{3}{2} and x=6x=6.

Special Products

Recognizing special patterns of quadratic expressions, such as perfect square trinomials and the difference of squares, can help with factoring.
Some quadratic expressions fit a pattern that can be used for factoring. Some special products involving quadratic expressions include a perfect square trinomial or a difference of squares. A perfect square trinomial is a trinomial of the form a2+2ab+b2a^2+2ab+b^2 or a22ab+b2a^2-2ab+b^2, which can be written as the square of a binomial. A difference of squares is a binomial of the form a2b2a^2-b^2.

Special Products of Quadratic Expressions

Perfect Square Trinomial Difference of Squares
a2+2ab+b2=(a+b)(a+b)=(a+b)2a^2+2ab+b^2=(a+b)(a+b)=(a+b)^2
a22ab+b2=(ab)(ab)=(ab)2a^2-2ab+b^2=(a-b)(a-b)=(a-b)^2
a2b2=(a+b)(ab)a^2-b^2=(a+b)(a-b)

Step-By-Step Example
Factor to Solve a Quadratic Equation with a Perfect Square Trinomial Involving a+ba+b
Solve the equation:
4x2+12x+9=04x^2+12x+9=0
Step 1
The given equation contains the expression:
4x2+12x+94x^2+12x+9
It fits the pattern of a perfect square trinomial:
a2+2ab+b2=(a+b)(a+b)=(a+b)2a^2+2ab+b^2=(a+b)(a+b)=(a+b)^2
In the given expression, a=2xa=2x and b=3b=3. So, the factored form of the quadratic expression is:
(2x+3)2(2x+3)^2
Step 2
Use the factored form of the quadratic expression to write an equivalent equation.
(2x+3)2=0(2x+3)^2=0
Step 3
Apply the zero product property by setting the factor equal to zero. Note that the factor 2x+32x+3 will be repeated. So only one equation can be solved.
2x+3=02x+3=0
Solution
Solve the equation.
2x+3=02x=3x=32\begin{aligned}2x+3&=0\\2x&=-3\\x&=-\frac{3}{2}\end{aligned}
The solution is x=32x=-\frac{3}{2}.
Step-By-Step Example
Factor to Solve a Quadratic Equation with a Perfect Square Trinomial Involving aba-b
Solve the quadratic equation:
9x224x+16=09x^2-24x+16=0
Step 1
The given equation contains the expression:
9x224x+169x^2-24x+16
It fits the pattern of a perfect square trinomial:
a22ab+b2=(ab)(ab)=(ab)2a^2-2ab+b^2=(a-b)(a-b)=(a-b)^2
In the given expression, a=3xa=3x and b=4b=4. So, the factored form of the quadratic expression is:
(3x4)2(3x-4)^2
Step 2
Use the factored form of the quadratic expression to write an equivalent equation.
(3x4)2=0(3x-4)^2=0
Step 3
Apply the zero product property by setting the factor equal to zero. The factor 3x43x-4 will be repeated, so only one equation can be solved.
3x4=03x-4=0
Solution
Solve the equation.
3x4=03x=4x=43\begin{aligned}3x-4&=0\\3x&=4\\x&=\frac{4}{3}\end{aligned}
The solution is x=43x=\frac{4}{3}.
Step-By-Step Example
Factor to Solve a Quadratic Equation with a Difference of Squares
Solve the given quadratic equation:
5x280=05x^2-80=0
Step 1
The given equation contains the expression:
5x2805x^2-80
Factor the quadratic expression. First, factor out the greatest common factor (GCF) of 5 from both terms.
5(x216)5(x^2-16)
The expression inside the parentheses fits the pattern of the difference of squares, where a=xa=x and b=4b=4:
a2b2=(a+b)(ab)x216=(x+4)(x4)\begin{aligned}a^2-b^2&=(a+b)(a-b)\\x^2-16&=(x+4)(x-4)\end{aligned}
The factored form of the entire expression is:
5(x216)5(x+4)(x4)\begin{gathered}5(x^2-16)\\5(x+4)(x-4)\end{gathered}
Step 2
Use the factored form of the quadratic expression to write an equivalent equation.
5(x+4)(x4)=05(x+4)(x-4)=0
Step 3
Apply the zero product property by setting the factors in parentheses equal to zero.
x+4=0orx4=0x+4=0\;\;\;\;\;\text{or}\;\;\;\;\;x-4=0
Solution
Solve each equation.
x+4=0x=4orx4=0x=4\begin{aligned}x+4&=0\\x&=-4\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned}x-4&=0\\x&=4\end{aligned}
The solutions are x=4x=-4 and x=4x=4.

Substitution Methods

Equations that are not written in standard form of a quadratic equation may be solved using quadratic methods by substituting a temporary variable for part of the expression.
An equation may not appear as though it can be solved using methods for solving quadratic equations. However, it may be possible to substitute a temporary variable for part of the expression so that the new equation is factorable by quadratic methods.
Step-By-Step Example
Use Substitution and Factoring to Solve a Quadratic Equation
Solve the quadratic equation:
3(x+1)22(x+1)1=03(x+1)^2-2(x+1)-1=0
Step 1
The expression x+1x+1 is included in two terms of the equation. In the first term, it is squared, so the form of the equation is quadratic. To write the equation in standard form, substitute the variable uu for the expression x+1x+1.
3(x+1)22(x+1)13u22u1\begin{gathered}3(x+1)^2-2(x+1)-1\\3u^2-2u-1\end{gathered}
Step 2
Factor the quadratic expression, which is in the form:
ax2+bx+c3u22u1\begin{gathered}ax^2+bx+c\\3u^2-2u-1\end{gathered}
In the given quadratic expression, a=3a=3, b=2b=-2, and c=1c=-1. So, look for two factors of ac=3ac=-3 that have a sum of bb, or 2-2. The factors are 3-3 and +1+1. Use the factors to write the middle term of the expression as a sum:
3u22u13u23u+u1\begin{gathered}3u^2{\color{#c42126}{-2u}}-1\\3u^2{\color{#c42126}{-3u+u}}-1\end{gathered}
Step 3
Use parentheses to group the terms.
(3u23u)+(u1)(3u^2-3u)+(u-1)
Then factor out the greatest common factor (GCF) from each group.
3u(u1)+1(u1)3u(u-1)+1(u-1)
The terms have a common factor of u1u-1. Factor out the common factor (u1)(u-1) from each term. Group the remaining factors.
(u1)(3u+1)(u-1)(3u+1)
Step 4
Use the factored form of the quadratic expression to write an equivalent equation.
(u1)(3u+1)=0(u-1)(3u+1)=0
Step 5
Apply the zero product property by setting each factor equal to zero.
u1=0or3u+1=0u-1=0\;\;\;\;\;\text{or}\;\;\;\;\;3u+1=0
Step 6
Solve each equation.
u1=0u=1or3u+1=03u=1u=13\begin{aligned}u-1&=0\\u&=1\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned}3u+1&=0\\3u&=-1\\u&=-\frac{1}{3}\end{aligned}
Solution
Replace the temporary variable uu in each solution with the expression it represents, x+1x+1. Then solve for xx.
x+1=1x=0orx+1=13x=43\begin{aligned}x+1&=1\\x&=0\end{aligned}\;\;\;\;\;\text{or}\;\;\;\;\;\begin{aligned}x+1&=-\frac{1}{3}\\x&=-\frac{4}{3}\end{aligned}
The solutions are x=0x=0 and x=43x=-\frac{4}{3}.
Step-By-Step Example
Using Substitution and Factoring with a Difference of Squares
Solve the quadratic equation:
x46x2+5=0x^4-6x^2+5=0
Step 1
Substitute the variable uu for the expression x2x^2.
x46x2+5u26u+5\begin{gathered}x^4-6x^2+5\\u^2-6u+5\end{gathered}
Step 2
Factor the quadratic expression, which is in the form:
ax2+bx+cu26u+5\begin{gathered}ax^2+bx+c\\u^2-6u+5\end{gathered}
In the given expression, a=1a=1, b=6b=-6, and c=5c=5. So, look for two factors of cc, or 5, that have a sum of bb, or –6. The factors are –5 and –1. Use the factors to write the factored form of the expression:
(u5)(u1)(u-5)(u-1)
Step 3
Use the factored form of the quadratic expression to write an equivalent equation.
(u5)(u1)=0(u-5)(u-1)=0
Step 4
Replace the temporary variable uu with the expression it represents, x2x^2.
(x25)(x21)=0(x^2-5)(x^2-1)=0
Each factor fits the pattern for a difference of squares:
a2b2=(a+b)(ab)a^2-b^2=(a+b)(a-b)
Use the difference of squares to rewrite each factor.
(x25)=(x+5)(x5)(x21)=(x+1)(x1)(x^2-5)=(x+\sqrt{5})(x-\sqrt{5})\;\;\;\;\;(x^2-1)=(x+1)(x-1)
Rewrite the equation to include the rewritten factors.
(x25)(x21)=0(x+5)(x5)(x+1)(x1)=0\begin{aligned}(x^2-5)(x^2-1)&=0\\(x+\sqrt{5})(x-\sqrt{5})(x+1)(x-1)&=0\end{aligned}
Step 5
Apply the zero product property by setting each factor equal to zero.
x+5=0orx5=0x+1=0orx1=0\begin{aligned}x+\sqrt{5}=0\;\;\;\;\;&\text{or}\;\;\;\;\;x-\sqrt{5}=0\\\\x+1=0\;\;\;\;\;&\text{or}\;\;\;\;\;x-1=0\end{aligned}
Solution
Solve each equation.
x+5=0x=5orx5=0x=5\boxed{\begin{aligned}x+\sqrt{5}&=0\\x&=-\sqrt{5}\end{aligned}}\;\;\;\;\;\text{or}\;\;\;\;\;\boxed{\begin{aligned}x-\sqrt{5}&=0\\x&=\sqrt{5}\end{aligned}}
x+1=0x=1orx1=0x=1\boxed{\begin{aligned}x+1&=0\\x&=-1\end{aligned}}\;\;\;\;\;\text{or}\;\;\;\;\;\boxed{\begin{aligned}x-1&=0\\x&=1\end{aligned}}
The solutions are x=±5x=\pm \sqrt{5} and x=±1x=\pm1.

Factoring Methods

Method When to Use
Greatest Common Factor (GCF) Use with any number of terms. Look for a greatest common factor of the terms, and factor out the GCF if possible.
ac+bc=a(b+c)ac+bc=a(b+c)
Difference of squares Use when there are two terms that match the pattern for a difference of squares.
a2b2=(a+b)(ab)a^2-b^2=(a+b)(a-b)
Perfect square trinomial Use when there are three terms that match the pattern for a perfect square trinomial.
a2+2ab+b2=(a+b)(a+b)=(a+b)2a22ab+b2=(ab)(ab)=(ab)2\begin{aligned}a^2+2ab+b^2&=(a+b)(a+b)=(a+b)^2\\a^2-2ab+b^2&=(a-b)(a-b)=(a-b)^2\end{aligned}
Trinomial factoring with a=1a=1 Use when there are three terms in the form x2+bx+cx^2+bx+c. Look for factors of cc that have a sum of bb. Use those values to write factors of the expression.
x2+bx+c=(x+)(x+)x^2+bx+c=(x+\square)(x+\square)
Trinomial factoring with a1a\neq1 Use when there are three terms in the form ax2+bx+cax^2+bx+c with a1a\neq1. Look for factors of the product of aa and cc that have a sum of bb. Use those values to write the middle term as a sum. Then factor by grouping.
ax2+bx+c=ax2+x+x+c=(x+)(x+)ax^2+bx+c=ax^2+\square x+\square x+c=(\square x+\square)(\square x + \square)
Grouping Use when a quadratic trinomial has been written with four terms, such as when the middle term is written as a sum.
ax+bx+ay+by=x(a+b)+y(a+b)=(a+b)(x+y)\begin{aligned}ax+bx+ay+by&=x(a+b)+y(a+b)\\&=(a+b)(x+y)\end{aligned}
Substitution Use when a part of the expression can be replaced by a temporary variable so that another factoring method applies.
x4+5x236=u2+5u36=(u4)(u+9)=(x24)(x2+9)=(x+4)(x4)(x2+9)\begin{aligned}x^4+5x^2-36&=u^2+5u-36\\&=(u-4)(u+9)\\&=(x^2-4)(x^2+9)\\&=(x+4)(x-4)(x^2+9)\end{aligned}

There are many strategies for factoring quadratic expressions. The number of terms in the expression can help determine the factoring method to use.