30 Solving Quadratic Equations by Factoring

An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as [latex]2{x}^{2}+3x - 1=0[/latex] and [latex]{x}^{2}-4=0[/latex] are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.

Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.

If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if [latex]a\cdot b=0[/latex], then [latex]a=0[/latex] or [latex]b=0[/latex], where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.

Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression [latex]\left(x - 2\right)\left(x+3\right)[/latex] by multiplying the two factors together.

The product is a quadratic expression. Set equal to zero, [latex]{x}^{2}+x - 6=0[/latex] is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.

The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, [latex]a{x}^{2}+bx+c=0[/latex], where a, b, and c are real numbers, and [latex]a\ne 0[/latex]. The equation [latex]{x}^{2}+x - 6=0[/latex] is in standard form.

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation [latex]{x}^{2}+x - 6=0[/latex], the leading coefficient, or the coefficient of [latex]{x}^{2}[/latex], is 1. We have one method of factoring quadratic equations in this form.

Solution

To factor [latex]{x}^{2}+x - 6=0[/latex], we look for two numbers whose product equals [latex]-6[/latex] and whose sum equals 1. Begin by looking at the possible factors of [latex]-6[/latex].

[latex]\begin{array}{l}1\cdot \left(-6\right)\hfill \\ \left(-6\right)\cdot 1\hfill \\ 2\cdot \left(-3\right)\hfill \\ 3\cdot \left(-2\right)\hfill \end{array}[/latex]

The last pair, [latex]3\cdot \left(-2\right)[/latex] sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

[latex]\left(x - 2\right)\left(x+3\right)=0[/latex]

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

[latex]\begin{array}{l}\left(x - 2\right)\left(x+3\right)\hfill&=0\hfill \\ \left(x - 2\right)\hfill&=0\hfill \\ x\hfill&=2\hfill \\ \left(x+3\right)\hfill&=0\hfill \\ x\hfill&=-3\hfill \end{array}[/latex]

The two solutions are [latex]x=2[/latex] and [latex]x=-3[/latex]. We can see how the solutions relate to the graph in Figure 2. The solutions are the x-intercepts of [latex]{x}^{2}+x - 6=0[/latex].

Factoring and Solving a Quadratic Equation of Higher Order

When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:

1. With the quadratic in standard form, [latex]a{x}^{2}+bx+c=0[/latex], multiply [latex]a\cdot c[/latex].
2. Find two numbers whose product equals [latex]ac[/latex] and whose sum equals [latex]b[/latex].
3. Rewrite the equation replacing the [latex]bx[/latex] term with two terms using the numbers found in step 1 as coefficients of x.
4. Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
5. Factor out the expression in parentheses.
6. Set the expressions equal to zero and solve for the variable.

Solution

First, multiply [latex]ac:4\left(9\right)=36[/latex]. Then list the factors of [latex]36[/latex].

[latex]\begin{array}{l}1\cdot 36\hfill \\ 2\cdot 18\hfill \\ 3\cdot 12\hfill \\ 4\cdot 9\hfill \\ 6\cdot 6\hfill \end{array}[/latex]

The only pair of factors that sums to [latex]15[/latex] is [latex]3+12[/latex]. Rewrite the equation replacing the b term, [latex]15x[/latex], with two terms using 3 and 12 as coefficients of x. Factor the first two terms, and then factor the last two terms.

[latex]\begin{array}{l}4{x}^{2}+3x+12x+9=0\hfill \\ x\left(4x+3\right)+3\left(4x+3\right)=0\hfill \\ \left(4x+3\right)\left(x+3\right)=0\hfill \end{array}[/latex]

Solve using the zero-product property.

[latex]\begin{array}{l}\left(4x+3\right)\left(x+3\right)=0\hfill \\ \hfill \\ \left(4x+3\right)=0\hfill \\ x=-\frac{3}{4}\hfill \\ \hfill \\ \left(x+3\right)=0\hfill \\ x=-3\hfill \end{array}[/latex]

The solutions are [latex]x=-\frac{3}{4}[/latex], [latex]x=-3[/latex].