 ## Solving Quadratic Equations and Inequalities

### Factoring with a Leading Coefficient of 1 Quadratic equations of the form $x^2 + bx + c = 0$ 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 $p$ and $q$ are real numbers and $pq=0$, then $p=0$ or $q=0$.

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 $x^2+bx+c$, look for two factors of $c$ that have a sum of $b$. When $c$ is positive, the two factors have the same sign as the sign of $b$. When $c$ 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 $b$.

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

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

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

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

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

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

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

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

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

### Factoring with a Leading Coefficient Quadratic equations of the form $ax^2 + bx + c = 0$ may be solved by factoring the quadratic expression and setting each factor equal to zero.
The expression has three terms: $ax^2$, $bx$, and $c$.
$ax^2+bx+c$
The leading coefficient is $a$ 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 $a$ is positive. Then look for two factors of $ac$ that have a sum of $b$. When $ac$ is positive, the two factors have the same sign as determined by the sign of $b$. When $ac$ 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 $b$. 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 $a\neq1$
$3x^2+13x+4=0$
Step 1
The given equation contains the expression:
$3x^2+13x+4$
The expression is in the form:
$ax^2+bx+c$
In the given expression, $a=3$, $b=13$, and $c=4$. So, look for two factors of $ac=12$ that have a sum of, $b$, or 13. The factors are $+1$ and $+12$. Use the factors to write the middle term of the expression as a sum:
$\begin{gathered}3x^2+{\color{#c42126}{13x}}+4\\3x^2{\color{#c42126}{+1x+12x}}+4\end{gathered}$
Step 2
Use parentheses to group the terms.
$(3x^2+1x)+(12x+4)$
Then factor out the GCF (greatest common factor) from each group. In the first group, both terms have a GCF of $x$. In the second group, both terms have a GCF of 4.
$x(3x+1)+4(3x+1)$
The resulting terms have a common factor of $3x+1$. Factor out the common factor from each term. Group the remaining factors.
$(3x+1)(x+4)$
Step 3
Use the factored form of the quadratic expression to write an equivalent equation.
$(3x+1)(x+4)=0$
Step 4
Apply the zero product property by setting each factor equal to zero.
$3x+1=0\;\;\;\;\;\text{or}\;\;\;\;\;x+4=0$
Solution
Solve each equation.
\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 $x=-\frac{1}{3}$ and $x=-4$.
Step-By-Step Example
Factor to Solve a Quadratic Equation with $a\neq1$ and $a<0$
$-2x^2+15x-18=0$
Step 1
In the given equation, the coefficient of $x^2$ is negative. Multiply both sides of the equation by –1 to get a positive value for $a$.
\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:
${2x^2}-15x+18$
The expression is in the form:
$ax^2+bx+c$
In the given expression, $a=2$, $b=-15$, and $c=18$. Look for two factors of $ac=36$ that have a sum of $b$, or –15. The factors are –3 and –12. Use the factors to write the middle term of the expression as a sum:
$\begin{gathered}-2x^2{\color{#c42126}{-15x}}+{18}\\{2x^2}{\color{#c42126}{-3x-12x}}+18\end{gathered}$
Step 2
Use parentheses to group the terms.
$(2x^2-3x)+(-12x+18)$
Then factor out the GCF (greatest common factor) from each group. In the first group, both terms have a GCF of $x$. In the second group, both terms have a GCF of 6.
$x(2x-3)+6(-2x+3)$
The first term in the second group is negative, so factor out –1 from the second group.
$x(2x-3)-6(2x-3)$
The resulting terms have a common factor of $2x-3$. Factor out the common factor from each term. Group the remaining factors.
$(2x-3)(x-6)$
Step 3
Use the factored form of the quadratic expression to write an equivalent equation.
$(2x-3)(x-6)=0$
Step 4
Apply the zero product property by setting each factor equal to zero.
${2x-3=0}\;\;\;\;\;\text{or}\;\;\;\;\;{x-6=0}$
Solution
Solve each equation.
\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 $x=\frac{3}{2}$ and $x=6$.

### 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 $a^2+2ab+b^2$ or $a^2-2ab+b^2$, which can be written as the square of a binomial. A difference of squares is a binomial of the form $a^2-b^2$.

### Special Products of Quadratic Expressions

Perfect Square Trinomial Difference of Squares
$a^2+2ab+b^2=(a+b)(a+b)=(a+b)^2$
$a^2-2ab+b^2=(a-b)(a-b)=(a-b)^2$
$a^2-b^2=(a+b)(a-b)$

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

### 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
$3(x+1)^2-2(x+1)-1=0$
Step 1
The expression $x+1$ 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 $u$ for the expression $x+1$.
$\begin{gathered}3(x+1)^2-2(x+1)-1\\3u^2-2u-1\end{gathered}$
Step 2
Factor the quadratic expression, which is in the form:
$\begin{gathered}ax^2+bx+c\\3u^2-2u-1\end{gathered}$
In the given quadratic expression, $a=3$, $b=-2$, and $c=-1$. So, look for two factors of $ac=-3$ that have a sum of $b$, or $-2$. The factors are $-3$ and $+1$. Use the factors to write the middle term of the expression as a sum:
$\begin{gathered}3u^2{\color{#c42126}{-2u}}-1\\3u^2{\color{#c42126}{-3u+u}}-1\end{gathered}$
Step 3
Use parentheses to group the terms.
$(3u^2-3u)+(u-1)$
Then factor out the greatest common factor (GCF) from each group.
$3u(u-1)+1(u-1)$
The terms have a common factor of $u-1$. Factor out the common factor $(u-1)$ from each term. Group the remaining factors.
$(u-1)(3u+1)$
Step 4
Use the factored form of the quadratic expression to write an equivalent equation.
$(u-1)(3u+1)=0$
Step 5
Apply the zero product property by setting each factor equal to zero.
$u-1=0\;\;\;\;\;\text{or}\;\;\;\;\;3u+1=0$
Step 6
Solve each equation.
\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 $u$ in each solution with the expression it represents, $x+1$. Then solve for $x$.
\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 $x=0$ and $x=-\frac{4}{3}$.
Step-By-Step Example
Using Substitution and Factoring with a Difference of Squares
$x^4-6x^2+5=0$
Step 1
Substitute the variable $u$ for the expression $x^2$.
$\begin{gathered}x^4-6x^2+5\\u^2-6u+5\end{gathered}$
Step 2
Factor the quadratic expression, which is in the form:
$\begin{gathered}ax^2+bx+c\\u^2-6u+5\end{gathered}$
In the given expression, $a=1$, $b=-6$, and $c=5$. So, look for two factors of $c$, or 5, that have a sum of $b$, or –6. The factors are –5 and –1. Use the factors to write the factored form of the expression:
$(u-5)(u-1)$
Step 3
Use the factored form of the quadratic expression to write an equivalent equation.
$(u-5)(u-1)=0$
Step 4
Replace the temporary variable $u$ with the expression it represents, $x^2$.
$(x^2-5)(x^2-1)=0$
Each factor fits the pattern for a difference of squares:
$a^2-b^2=(a+b)(a-b)$
Use the difference of squares to rewrite each factor.
$(x^2-5)=(x+\sqrt{5})(x-\sqrt{5})\;\;\;\;\;(x^2-1)=(x+1)(x-1)$
Rewrite the equation to include the rewritten factors.
\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.
\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.
\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}}
\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 $x=\pm \sqrt{5}$ and $x=\pm1$.

### 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.
$ac+bc=a(b+c)$
Difference of squares Use when there are two terms that match the pattern for a difference of squares.
$a^2-b^2=(a+b)(a-b)$
Perfect square trinomial Use when there are three terms that match the pattern for a perfect square trinomial.
\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 $a=1$ Use when there are three terms in the form $x^2+bx+c$. Look for factors of $c$ that have a sum of $b$. Use those values to write factors of the expression.
$x^2+bx+c=(x+\square)(x+\square)$
Trinomial factoring with $a\neq1$ Use when there are three terms in the form $ax^2+bx+c$ with $a\neq1$. Look for factors of the product of $a$ and $c$ that have a sum of $b$. Use those values to write the middle term as a sum. Then factor by grouping.
$ax^2+bx+c=ax^2+\square x+\square x+c=(\square x+\square)(\square x + \square)$
Grouping Use when a quadratic trinomial has been written with four terms, such as when the middle term is written as a sum.
\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.
\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.