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Study Guides > College Algebra

Factoring and the Square Root Property

Learning Objectives

  • Factor a quadratic equation to solve it
  • Use the square root property to solve a quadratic equation
  • Use the pythagorean theorem and the square root property to find the unknown length of a triangle leg
An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as 2x2+3x−1=02{x}^{2}+3x - 1=0 and x2−4=0{x}^{2}-4=0 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 a⋅b=0a\cdot b=0, then a=0a=0 or b=0b=0, 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 (x−2)(x+3)\left(x - 2\right)\left(x+3\right) by multiplying the two factors together.
(x−2)(x+3)=x2+3x−2x−6=x2+x−6\begin{array}{l}\left(x - 2\right)\left(x+3\right)\hfill&={x}^{2}+3x - 2x - 6\hfill \\ \hfill&={x}^{2}+x - 6\hfill \end{array}
The product is a quadratic expression. Set equal to zero, x2+x−6=0{x}^{2}+x - 6=0 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, ax2+bx+c=0a{x}^{2}+bx+c=0, where a, b, and c are real numbers, and a≠0a\ne 0. The equation x2+x−6=0{x}^{2}+x - 6=0 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.

A General Note: The Zero-Product Property and Quadratic Equations

The zero-product property states
If a⋅b=0, then a=0 or b=0\text{If }a\cdot b=0,\text{ then }a=0\text{ or }b=0,
where a and b are real numbers or algebraic expressions. A quadratic equation is an equation containing a second-degree polynomial; for example
ax2+bx+c=0a{x}^{2}+bx+c=0
where a, b, and c are real numbers, and if a≠0a\ne 0, it is in standard form.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation x2+x−6=0{x}^{2}+x - 6=0, the leading coefficient, or the coefficient of x2{x}^{2}, is 1. We have one method of factoring quadratic equations in this form.

How To: Given a quadratic equation with the leading coefficient of 1, factor it.

  1. Find two numbers whose product equals c and whose sum equals b.
  2. Use those numbers to write two factors of the form (x+k) or (x−k)\left(x+k\right)\text{ or }\left(x-k\right), where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and −2-2, the factors are (x+1)(x−2)\left(x+1\right)\left(x - 2\right).
  3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Example: Factoring and Solving a Quadratic with Leading Coefficient of 1

Factor and solve the equation: x2+x−6=0{x}^{2}+x - 6=0.

Answer: To factor x2+x−6=0{x}^{2}+x - 6=0, we look for two numbers whose product equals −6-6 and whose sum equals 1. Begin by looking at the possible factors of −6-6.

1⋅(−6)(−6)⋅12⋅(−3)3⋅(−2)\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}
The last pair, 3⋅(−2)3\cdot \left(-2\right) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.
(x−2)(x+3)=0\left(x - 2\right)\left(x+3\right)=0
To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
(x−2)(x+3)=0(x−2)=0x=2(x+3)=0x=−3\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}
The two solutions are x=2x=2 and x=−3x=-3. We can see how the solutions relate to the graph below. The solutions are the x-intercepts of x2+x−6=0{x}^{2}+x - 6=0. Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 7 to 7. The function x squared plus x minus six equals zero is graphed, with the x-intercepts (-3,0) and (2,0), plotted as well.

Try It

Factor and solve the quadratic equation: x2−5x−6=0{x}^{2}-5x - 6=0.

Answer: (x−6)(x+1)=0;x=6,x=−1\left(x - 6\right)\left(x+1\right)=0;x=6,x=-1

Using the Pythagorean Theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as a2+b2=c2{a}^{2}+{b}^{2}={c}^{2}, where aa and bb refer to the legs of a right triangle adjacent to the 90∘90^\circ angle, and cc refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications. We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side. The Pythagorean Theorem is given as
a2+b2=c2{a}^{2}+{b}^{2}={c}^{2}
where aa and bb refer to the legs of a right triangle adjacent to the 90∘{90}^{\circ } angle, and cc refers to the hypotenuse.
Right triangle with the base labeled: a, the height labeled: b, and the hypotenuse labeled: c Figure 4

Example: Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle. Right triangle with the base labeled: a, the height labeled: 4, and the hypotenuse labeled 12.

Answer: As we have measurements for side b and the hypotenuse, the missing side is a.

a2+b2=c2a2+(4)2=(12)2a2+16=144a2=128a=128=82\begin{array}{l}{a}^{2}+{b}^{2}\hfill&={c}^{2}\hfill \\ {a}^{2}+{\left(4\right)}^{2}\hfill&={\left(12\right)}^{2}\hfill \\ {a}^{2}+16\hfill&=144\hfill \\ {a}^{2}\hfill&=128\hfill \\ a\hfill&=\sqrt{128}\hfill \\ \hfill&=8\sqrt{2}\hfill \end{array}

 

Try It

Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.

Answer: 55 units

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