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Introduction to Factoring

After studying this section, you will be able to:

1. Factor polynomials containing a common factor in each term.

Common Factors

Recall that when two or more numbers, variables, or algebraic expressions are multiplied, each is called a factor.

          

When you are asked to factor a number or an algebraic expression, you are being asked, "What, when multiplied, will give that number or expression?’’ For example, you can factor 6 as 3*2 since 3*2 = 6. You can factor 15x^5 as 3x^2*5x^3 since 3x^2*5x^3 = 15x^5. Factoring is simply the reverse of multiplying. 6 and 15x^5 are simple expressions to factor and can be factored in different ways. The factors of the polynomial 2x^2 + x -12 are not so easy to recognize. In this chapter we will be learning techniques to find the factors of a polynomial. We will begin with common factors.

EXAMPLE 1 Factor. (a) 3x-6y   (b) 9x + 2xy

Begin by looking for a common factor, the factor that both terms have in common. Then rewrite the expression as a product.

(a) 3x-6y = 3(x-2y)   This is true because 3(x-2y) = 3x-6y.

(b) 9x + 2xy = x(9 + 2y)   This is true because x(9 + 2y) = 9x + 2xy.

  Some people find it helpful to think of factoring as the distributive property in reverse. When we write 3x-6y = 3(x-2y), we are ‘‘undistributing’’ the 3.

  When we factor, we are looking for the largest common factor. For example, in the polynomial 48x-16y, a common factor is 2. We could factor 48-16y as 2(24-8y). However, this is not complete. To factor 48-16y completely, we look for the greatest common factor of 48 and of 16.

            48x-16y = 16(3x-y)

EXAMPLE 2 Factor. 24xy + 12x^2 + 36x^3

Find the greatest common factor of 24, 12, and 36. You may want to factor each number or you may notice that 12 is a common factor. 12 is the greatest common factor. Notice also that x is a factor of each term.

            24xy + 12x^2 + 36x^3 = 12x(2y + x + 3x^2)

Common Factors of a Polynomial

  1. You can determine the numerical common factor by asking, ‘‘What is the largest integer that will divide into the coefficient of each term?’’

  2. You can determine the variable common factor by asking, ‘‘What variables are common to each term, and what is the largest exponent of those variables that is common?”

EXAMPLE 3 Factor. (a) 12x^2 + 18y^2   (b) x^2y^2 + 3xy^2 + y^3

(a) Note that the largest integer that is common to both terms is 6 (not 3 or 2).

            12x^2 + 18y^2 = 6(2x^2 + 3y^2)

(b) Although y is common to all terms, we factor out y^2 since 2 is the largest exponent of y that is common to all terms. We do not factor out x since x is not common to all terms.

            x^2y^2 + 3xy^2 + y^3=y^2(x^2+3x+y


EXAMPLE 4 Factor. 9a^3b^2 + 9a^2b^2

We observe that both terms contain a common factor of 9. We can remove a^2 and b^2 from each term.

            9a^3b^2 + 9a^2b^2=9a^2b^2(a+1)

WARNING Don’t forget to include the 1 inside the parentheses in the above example. The solution is wrong without it. You will see why if you try to check a result written without the 1.

EXAMPLE 5 Factor. 3x(x-4y) + 2(x-4y)

Be sure you understand what are terms and what are factors in this example. There are two terms. The expression 3x(x-4y) is one term. The expression 2(x-4y) is the second term. Each term is made up of two factors. Observe that the binomial (x-4y) is a common factor in each term. A common factor may be a set of parentheses containing any type of polynomial. Thus we can remove the common factor (x-4y).

            3x(x-4y) + 2(x-4y) = (x-4y)(3x + 2)

EXAMPLE 6 Factor. 7x^2(2x-3y)-(2x-3y)

The common factor for each term is (2x-3y). What happens when we factor out (2x-3y)? What are we left with in the second term? Recall that (2x-3y) = 1(2x-3y). Thus

7x^2(2x-3y)-(2x-3y) = 7x^2(2x-3y)-1(2x-3y)   Rewrite the original expression.

            = (7x^2-1)(2x - 3y)   Factor out (2x-3y).

Factor by Grouping

After studying this section, you will be able to:

1. Factor problems with four terms by grouping.

Factoring by Grouping

A common factor can be a number, a variable, or an algebraic expression. Sometimes the polynomial is written so that it is easy to recognize the common factor. This is especially true when the common factor is enclosed by parentheses.

EXAMPLE 1 Factor. x(x-3) + 2(x-3)

Observe each term:

            

The common factor for both the first and second terms is the quantity (x-3), so we have
          x(x-3) + 2(x-3) = (x-3)(x + 2)

  In many cases these types of problems do not have any parentheses and we are asked to factor the polynomial. In such cases we remove a common factor from the first two terms and a different common factor from the second two terms. If the two parentheses contain the same expressions at this step, the problem may then be completed as we did in Example 1. This procedure of factoring is often called factoring by grouping.

EXAMPLE 2 Factor. 2x^2 + 3x + 6x +9


Now we finish the factoring.

            x(2x +3) +3(2x +3) =(2x +3)(x + 3)

EXAMPLE 3 Factor. 4x + 8y + ax + 2ay

Both sets of parentheses contain the same expression. The common factor of each term is the expression in parentheses, (x + 2y).

4(x+2y)+a(x+2y)=(x+2y)(4+a)

In some problems the terms are out of order. We have to rearrange the order of the terms first so that the first two terms have a common factor.

EXAMPLE 5 Factor. bx + 4y + 4b + xy

= bx + 4b + xy + 4y     Rearrange the terms so that the first terms have a common factor.

= b(x+4)+ y(x+4)   Remove the common factor of b from the first two terms. Remove the common factor of y from                the second two terms.

= (x + 4)(b + y)      Since the two sets of parentheses above contain the same expression, we can complete the                  factoring.

Sometimes you will need to remove a negative common factor from the second two terms to obtain two sets of parentheses that contain the same expression.

EXAMPLE 6 Factor. 2x^2 + 5x -4x -10

Remove the common factor of x from the first two terms and a common factor of -2 from the second two terms.

= x(2x+5)-2(2x+5)

Since the two sets of parentheses above contain the same expression, we can complete the factoring.

= (2x + 5)(x-2)

Notice that if you removed a common factor of +2 in the first step, the two sets of parentheses would not contain the same expression. If the expressions inside the two sets of parentheses are not exactly the same, you cannot express the polynomial as a product of two factors!

Factoring Trinomials of the Form x^2 + bx + c

After studying this section, you will be able to:

1. Factor polynomials of the form x^2 + bx +c.

2. Factor polynomials that have a common factor and a factor of the form x^2 + bx +c.

Factoring Polynomials of the Form x^2 + bx +c

Suppose that you wanted to factor x^2 + 5x + 6. After some trial and error you might obtain (x + 2)(x + 3) or you might get discouraged and not get an answer. If you did get these factors, you could check this answer by the FOIL method.

(x + 2)(x + 3) = x^2 + 3x+2x+6

            = x^2+ 5x+6

But trial and error can be a long process. There is another way.

  Look at the original polynomial x^2 + 5x + 6. We know immediately that the answer will be of the form ( )( ). Since the first term of the polynomial is x^2, we can place an x in each of the parentheses, (x     )(x     ). Next look at the sign of the last term. The sign is positive. This means that the sign of each factor must be the same. Since the sign of the middle term is positive, the sign of each factor must be positive, and we can write (x +  )(x +  ). Look again at the last term. Think of factors of 6. Try 2 and 3 since 2 + 3 = 5, the coefficient of the middle term. We write (x + 2)(x + 3) and our factoring is complete. We can check this using FOIL as shown above. To summarize:

        

Let’s write the procedure we have observed and try a few examples.

Factoring Trinomials of the Form x^2 + bx +c

1. The answer will be of the form (x + __)(x + __).

2. The two numbers at the end of each set of parentheses are numbers such that:

  (a) When you multiply them, you get the last term, which is c.

  (b) When you add them, you get the coefficient of x, which is b.

EXAMPLE 1 Factor. x^2 + 7x + 12

The answer is of the form (x + __)(x + __). We want to find the two numbers we can multiply to get 12 but add to get 7. The numbers are 3 and 4.

          x^2 + 7x + 12 = (x + 3)(x + 4)

EXAMPLE 2 Factor. x^2 + 12x + 20

We want two numbers that have a product of 20 but a sum of 12. The numbers are 10 and 2.

          x^2 + 12x + 20 = (x + 10)(x + 2)

Note: If you cannot think of the numbers in your head, write down the possible factors whose product is 20.

  So far we have only factored trinomials of the form x^2 + bx + c, where b and c are positive numbers. The same procedure applies if b is a negative number and c is positive.

Remember, if c is positive, the sign of both factors will be the same.

EXAMPLE 3 Factor. x^2 - 8x + 15

We want two numbers that have a product of +15 but a sum of -8. They must be negative numbers since the sign of the middle term is negative.

Multiply using FOIL to check.

EXAMPLE 4 Factor. x^2 - 9x + 14

We want two numbers whose product is 14 but whose sum is -9. The numbers are -7 and -2. So

          x^2-9x+ 14=(x-7)(x-2) or (x-2)(x-7)

  All the examples so far have had a positive last term. What happens when the last term is negative? If the last term is negative, we will need one positive number and one negative number. Why? The product of a positive and a negative number is negative.

EXAMPLE 5 Factor. x^2 + 2x -8

We want two numbers whose product is -8 but whose sum is +2.

x^2 +2x-8 =(x + 4)(x - 2)   Think: (+4)(-2) = -8 and +4 + (-2) = +2.

EXAMPLE 6 Factor. x^2- 3x -10

We want two numbers whose product is -10 but whose sum is -3. The two numbers are -5 and +2.

          x^2-3x -10 = (x-5)(x + 2)

What if we made a sign error and incorrectly factored the trinomial x^2-3x-10 as (x + 5)(x-2)? We could detect the error immediately since the sum of +5 and -2 is 3. We need a sum of -3!

WARNING It is very easy to make a sign error in these problems. Make sure that you mentally multiply your answer back to obtain the original expression. Check each sign carefully.

Check: (x-5)(x + 2)=x^2+2x-5x-10=x^2-3x-10  ✔

EXAMPLE 7 Factor. x^2-16x -36

We want two numbers whose product is -36 and whose sum is -16.

List all the possible factors of 36 (without regard to sign). Find the pair that has a difference of 16. We are looking for a difference because the signs of the factors are different.

        

Once we have picked the pair of numbers (18 and 2), it is easy to find the signs. For the coefficient of the middle term to be -16, we will have to add the numbers -18 and +2.

          x^2-16x-36 = (x -18)(x + 2)

Feel a little confused about the signs? If you do, you may find these facts helpful.

Do not memorize these facts; rather, try to understand the pattern.

Sometimes the exponent of the first term will be greater than 2. If the exponent is an even power, it is a square. For example, x^4 = (x^2)(x^2). Likewise, x^6 = (x^3)(x^3).

EXAMPLE 8 Factor. y^4-2y^2-35

Factoring Polynomials That Have a Common Factor

Some factoring problems require two steps. Often we must first remove a common factor from each term of the polynomial. Once this is done, we may find that the other factor is a trinomial that can be factored using the methods previously discussed in this section.

EXAMPLE 9 Factor. 2x^2 + 36x + 160

First remove the common factor of 2 from each term of the polynomial.

2(x^2 + 18x + 80)

Then factor the remaining polynomial.

2(x + 8)(x + 10) 

The final answer is 2(x + 8)(x + 10).   Be sure to list all parts of the answer.

Check: 2(x + 8)(x + 10) = 2(x^2 + 18x + 80) = 2x^2 + 36x + 160   ✔

Thus we are sure that the answer is 2(x + 8)(x + 10)

EXAMPLE 10 Factor. 3x^2 + 9x-162

First remove the common factor of 3 from each term of the polynomial.

3(x^2 + 3x-54) 

Then factor the remaining polynomial.

3(x-6)(x + 9) 

The final answer is 3(x-6)(x + 9).

Check: 3(x-6)(x + 9) = 3(x^2 + 3x-54) = 3x^2 + 9x- 162   ✔

Thus we are sure that the answer is 3(x-6)(x + 9)

Factoring Trinomials of the Form ax^2 + bx+c

After studying this section, you will be able to:

1. Factor a trinomial of the form ax^2 + bx + c by the trial and error method.

2. Factor a trinomial of the form ax^2 + bx + c by the grouping number method.

3. Factor a trinomial of the form ax^2 + bx + c after a common factor has been removed for each term.

Using the Trial and Error Method to Factor

When the coefficient of the x^2 term is not 1, the trinomial is more difficult to factor. Several possibilities must be considered.

EXAMPLE 1 Factor. 2x^2 + 5x + 3

To get the coefficient of the first term to be 2, the factors would be 2 and 1. 2x^2 +5x+3 = (2x   )(x   ).

To get the last term to be 3, the factors would be 3 and 1.

Since all signs are positive, we know that each set of parentheses will contain only
positive signs. However, we still have two possibilities. They are

          (2x + 3)(x + 1)

          (2x + 1)(x + 3)

We check them by multiplying by the FOIL method:

      (2x + 1)(x + 3) = 2x^2 + 7x + 3   Wrong middle term.

      (2x + 3)(x + 1) = 2x^2 + 5x +3   Right middle term.

Thus the correct answer is

      (2x + 3)(x+1) or (x + 1)(2x + 3)

Some problems have many more possibilities.

EXAMPLE 2 Factor. 4x^2-13x + 3

        

Let us list the possible factoring combinations and compute the middle term by the FOIL method. Note that both signs will be negative. Why?

        

The correct answer is (4x-1)(x-3) or (x-3)(4x-1)

This method is called the trial-and-error method.

EXAMPLE 3 Factor. 6x^2 +x -5

Let’s consider the last number in each set of parentheses. Because -5 is negative, the two numbers will have opposite signs. Since the last number can be either positive or negative, we have more possibilities. In this case:

            

The correct answer is

          (6x-5)(x+1) or (x + 1)(6x-5)

Look back to Example 3. Let’s see if we can shorten our work. Notice that there are four pairs of possible factors. The only difference between the two parentheses in the pair is that the last signs in each set of parentheses are reversed. This suggests that we list only half as many possibilities. Then, if the coefficient of the middle term is the opposite in sign of what we want, we can just reverse signs. Let’s try this shorter method.

EXAMPLE 4 Factor. 3x^2-2x-8

              

Let us list only one-half of the possibilities. The last term of that first set of parentheses will always be positive.

      

So we just reverse the signs of the last term in each set of parentheses.

        

The correct answer is

            (x -2)(3x +4) or (3x + 4)(x-2)


Using the Grouping Number Method for Factoring

One way to factor a trinomial of the form ax^2 + bx + c is to write it with four terms and factor by grouping, as we did previously. For example, the trinomial 2x^2 + 13x + 20 can be written as 2x^2 + 5x + 8x + 20. Using the previous methods, we factor

2x^2 + 5x + 8x + 20 = x(2x + 5) + 4(2x + 5)

            = (2x + 5)(x + 4)

We can factor all factorable trinomials of the form ax^2 + bx + c in this way. We will use the following procedure:

Let’s try the same problem we did in Example 1.

EXAMPLE 5  Factor by grouping. 2x^2 + 5x + 3

1. The grouping number is (2)(3) = 6.

2. The factors of 6 are 6*1 and 3*2. We chose the factors whose sum is 5.

3. We write 5x as the sum 3x + 2x.

4. Factor by grouping.

2x^2 + 5x + 3=2x^2+2x+3x+3

        = 2x(x + 1) + 3(x+ 1)

        = (x + 1)(2x + 3)

5. Multiply to check.

(x + 1)(2x + 3) = 2x^2 + 3x +2x+3

            = 2x^2+5x+3  ✔

EXAMPLE 6 Factor by grouping. 4x^2 -13x + 3

1. The grouping number is (4)(3) = 12.

2. The factors of 12 are (12)(1) or (4)(3) or (6)(2). We choose the factors whose sum is 13. Both factors receive the sign of the middle term in the trinomial.

3. We write -13x as the sum -12x + (-1).

4. Factor by grouping.

4x^2-13x +3 = 4x^2-12x - 1x +3

Remember to factor out a -1 from the last two terms so that both parentheses contain the same expression.

= 4x(x-3) -1(x-3) 

= (x-3)(4x - 1)

EXAMPLE 7  Factor by grouping. 3x^2-2x-8

1. The grouping number is (3)(-8) = -24.

2. Since the last sign is negative we want two numbers whose product is 24 but whose difference is 2. They are 6, 4. The largest gets the middle sign; so we have -6 and +4.

3. We write -2x as a sum -6x + 4x.

4. Factor by grouping.

3x^2-6x + 4x-8 = 3x(x-2) + 4(x-2)

        = (x-2)(3x + 4)

Factoring Polynomials That Have a Common Factor

Some problems require first removing a common factor and then factoring the trinomial by one of the two methods of this section.

EXAMPLE 8 Factor. 9x^2 + 3x -30

We first remove the common factor of 3 from each term of the trinomial.

9x^2 + 3x -30 = 3(3x^2 + 1x -10)

We then factor the trinomial by the grouping method or by the trial and error method. 

= 3(3x-5)(x + 2) 

Special Cases of Factoring

After studying this section, you will be able to:

1. Recognize and factor problems of the type a^2-b^2.

2. Recognize and factor problems of the type a^2 + 2ab + b^2.

3. Factor problems that require removing a common factor and then using the special case formula.

As we proceed in this section you will be able to reduce the time it takes you to factor a problem by quickly recognizing and factoring two special types of factoring problems. By becoming very familiar with the appropriate formula, you will be able to factor the difference of two squares and perfect square trinomials.

Factoring the Difference of Two Squares

Recall the formula from previous section:

              (a + b)(a-b) = a^2-b^2

In reverse form we can use it for factoring.

We can state it in words in this way: ‘‘The difference of two squares can be factored into the sum and difference of those values that were squared.”

EXAMPLE 1 Factor. 9x^2-1

We see that the problem is in the form of the difference of two squares. 9x^2 is a square and 1 is a square. So using the formula we can write

9x^2-1 = (3x+1)(3x-1)   Because 9x^2 = (3x)^2 and 1 = (1)^2.

Sometimes the problem will contain two variables.

EXAMPLE 2 Factor. 4x^2-49y^2

We see that

4x^2-49y^2 = (2x + 7y)(2x-y)

Some problems may involve more than one step.

EXAMPLE 3  Factor. 81x^4-1

We see that

81x^4-1 = (9x^2 +1)(9x^2-1)   Because 81x^4 = (9x)^2 and 1 = (1)^2.

Is the factoring complete? We can factor 9x^2-1.

81x^4-1 = (9x^2 + 1)(3x+1)(3x-1)   Because (9x^2-1) = (3x +1)(3x - 1)]].

Factoring Perfect Square Trinomials

There is a formula that will help us to factor very quickly certain trinomials, called perfect square trinomials. Recall the formula for a binomial squared from previous section:

            (a + b)^2 = a^2 + 2ab + b^2

            (a-b)^2 = a^2 -2ab + b^2

In reverse form we can use these two equations for factoring.

A perfect square trinomial is a trinomial that is the result of squaring a binomial. How can we recognize a perfect square trinomial?

1. The first and last terms are perfect squares.

2. The middle term is twice the product of the values whose squares are the first and last terms.

EXAMPLE 4 Factor. x^2 + 6x + 9

This is a perfect square trinomial.

1. The first and last terms are perfect squares because x^2 = (x)^2 and 9 = (3)^2.

2. The middle term is twice the product of x and 3.

Since x^2 + 6x + 9 is a perfect square trinomial, we can use the formula

            a^2 + 2ab + b^2 = (a + b)^2

and we have

            x^2 + 6x +9 =(x + 3)^2

EXAMPLE 5 Factor. 4x^2-20x + 25

This is a perfect square trinomial. 20x = 2(2x*5) Note the negative sign. We have

4x^2-20x + 25 = (2x-5)^2 Since a^2-2ab + b^2 = (a-b)^2.

More than one variable may be involved and the exponents may be higher than 2. The same principles apply.

EXAMPLE 6 Factor. (a) 49x^2+42xy+9y^2   (b) 36x^4 -12x^2 + 1

(a) This is a perfect square trinomial. Why?

Because 49x^2 = (7x)^2 and 9y^2 = (3y)^2. Also, 42xy = 2(7x*3y)

49x^2+42xy+9y^2 = (7x + 3y)^2.

(b) This is a perfect square trinomial. Why?
36x^4-12x^2 + 1 = (6x^2-1)^2 Because 36x^4 = (6x^2)^2 and 1 = (1)^2. Also, 12x^2 = 2(6x^2*1).

Some problems appear to be perfect square trinomials but are not. They were factored as other trinomials in previous section.

EXAMPLE 7 Factor. 49x^2 + 35x + 4

This is not  a perfect square trinomial! Although the first and last terms are perfect squares since (7x)^2 = 49x^2 and (2)^2 = 4, the middle term is not double the product of 2 and 7x! 35x != 28x! So we must factor by trial and error or by grouping to obtain

49x^2 + 35x +4 = (7x + 4)(7x + 1)

Factoring Polynomials by First Removing a Common Factor and Then Using One of the Special Factoring Formulas

In some cases, we will first remove a common factor. Then we will find an opportunity to use the difference of two squares formula or one of the perfect square trinomial formulas.

EXAMPLE 8 Factor. 12x^2-48

12x^2-48 = 12(x^2-4)   First we remove the greatest common factor 12.

= 12(x + 2)(x-2)   Then we use the difference of two squares formula: a^2- b^2 = (a + b)(a-b).

EXAMPLE 9 Factor. 24x^2 + 72x +54

First we remove the greatest common factor 6.

24x^2 + 72x + 54 = 6(4x^2 + 12x + 9) 

Then we use the perfect square trinomial formula: a^2 + 2ab + b^2 = (a + b)^2.

= 6(2x + 3)^2

A Brief Review of Factoring

After studying this section, you will be able to:

1. Identify and factor any polynomial that can be factored. 1 , :

2. Determine if a polynomial is prime.

Review of Factoring

Often the various types of factoring problems are all mixed together. We need to be able to identify each type of factoring problem quickly. The following table may be helpful in summarizing your knowledge of factoring.

Many problems involve more than one step when factoring. When you are asked to factor a polynomial, it is expected that you will factor it completely. Usually, the first step of factoring is removing the common factor, then the next step will be more apparent.

EXAMPLE 1 Factor.

(a) 25x^3-10x^2 + x   (b) 20x^2y^2-45y^2

(a) Remove the common factor of x. The other factor is a perfect square trinomial.

25x^3-10x^2 + x = x(25x^2-10x +1) 

= x(5x-1)^2

(b) Remove the common factor of 5y^2. The other factor is the difference of squares.

20x^2y^2-45y^2 = 5y^2(4x^2-9) 

= 5y^2(2x+3)(2x -3)

Determining If a Polynomial Is Prime

Not all polynomials can be factored using the methods in this lesson . If we cannot
factor a polynomial by elementary methods, we will identify it as a prime polynomial.

EXAMPLE 2  Factor, if possible. x^2 + 6x + 12

Since we are only using rational numbers, we find this polynomial is prime. The rational factors of 12 are (1)(12) or (2)(6) or (3)(4)

None of these pairs will add up to 6, the coefficient of the middle term. Thus the problem cannot be factored by the methods of this lesson. It is prime.

EXAMPLE 3  Factor, if possible. 25x^2 + 4

We have a formula to factor the difference of two squares. There is no way to factor the sum of two squares. That is, a^2 + b^2 cannot be factored. Thus 25x^2 + 4 is prime.

Solving Quadratic Equations by Factoring

After studying this section, you will be able to:

1. Solve a quadratic equation by factoring.

2. Solve a variety of applied problems by using a quadratic equation.

Solving a Quadratic Equation by Factoring

In earlier tutorial we learned how to solve linear equations such as 3x + 5 = 0 to find the root (or value of x) that satisfied the equation. Now we turn to the question of how to solve equations like 3x^2 + 5x + 2 = 0. Such equations are called quadratic equations. A quadratic equation is a polynomial equation that contains the power 2 of a variable as the highest power in the equation.

In this lesson, we will study quadratic equations in standard form, where a, b, and c are integers.

Usually we can find two real roots that satisfy a quadratic equation. But how can we find them? The most direct approach is the factoring method. This method depends on a very powerful property.

When we make a statement in mathematics using the word or, we intend it to mean one or the other or both. Therefore, both a and b can equal zero if the product is zero. We can use this principle to solve a quadratic equation. Before you start, make sure that the equation is in standard form.

EXAMPLE 1 Find the two roots. 3x^2 + 5x +2 =0

3x^2 + 5x+2=0   The equation is in standard form.

(3x + 2)(x+1)=0   Factor the quadratic expression.

3x +2=0   x+1=0   Set each factor equal to 0.

3x =-2   x=-1   Solve each equation to find the two roots.

x = −2/3

The two roots are −2/3 and -1.

EXAMPLE 2 Find the two roots. 2x^2 + 13x -7 =0

2x^2 + 13x -7=0   The equation is in standard form.

(2x-1)(x + 7) =0   Factor.

2x-1=0   x+7=0   Set each factor equal to 0.

2x = 1   x=-7   Solve each equation to find the two roots.

x=1/2

The two roots are 1/2 and -7.

Check. If x = 1/2, then

(2)(1/2)^2 + 13(1/2)-7=2(1/4)+13(1/2)-7=1/2+13/2-14/2=0  ✔

If x = -7, then

2(-7)^2 + 13(-7) -7 = 2(49) + 13(-7)-7=98-91-7=0  ✔

Thus x = 1/2 and x = -7 are both roots for the equation 2x^2 + 13x -7 = 0.

If the quadratic equation ax^2 + bx + c = 0 has no visible constant term, then c = 0. All such quadratic equations can be factored by removing a common factor to obtain two solutions that are real numbers.

EXAMPLE 3 Find the two roots. 7x^2-3x = 0

7x^2-3x =0   The equation is in standard form. Here c = 0.

x(7x-3) =0   Factor by removing the common factor.

x=0   7x -3=0   Set each factor equal to 0 by the zero factor property.

7x = 3

x= 3/7   Solve each equation to find the two roots.

The two roots are x = 0 and x = 3/7.

If the quadratic equation is not in standard form, we use the same basic algebraic methods we studied previously to set the equation equal to zero so that we can use the zero factor property.

EXAMPLE 4 Find the two roots. x^2 = 12-x

x^2=12-x

The equation is not in standard form.

Add x and -12 to both sides of the equation so that the left side is equal to zero and can be factored.

x^2+x-12=0

(x-3)(x + 4) =0   Factor.

x-3=0   x+4=0   Set each factor equal to 0 by the zero factor property.

[x=3]]   x=-4   Solve each equation for x.

EXAMPLE 5 Solve. (x + 3)(x-2) = 50

Be careful here! You must have a zero on the right hand side to use the zero factor property.

(x + 3)(x-2) = 50

x^2 + 3x-2x-6 = 50   Remove the parentheses.

x^2+x-56=0   Place in standard form.

(x + 8)(x -7) =0   Factor.

x+8=0   x-7=0   Set each factor equal to zero.

x=-8   x=7   Solve each equation for x.

Using a Quadratic Equation to Solve Applied Problems

Certain types of word problems lead to quadratic equations. These problems may be number problems, problems about geometric figures, or applied problems. We'll show the solution to some model problems in this section.

  It is particularly important to check the apparent solutions to the quadratic equation with conditions stated in the word problem. Often a particular solution to the quadratic equation will be eliminated by the conditions of the word problem.

EXAMPLE 6 The length of a rectangle is 3 meters longer than twice the width. The area of the rectangle is 44 square meters. Find the length and width of the rectangle.

1. Understand the problem.

Draw a picture.

Let w = the width in meters

Then 2w + 3 = the length in meters

2. Write an equation.

area = (width)(length)

44 = w(2w + 3)

3. Solve the equation and state the answer.

44 = w(2w + 3)

44 =2w^2 + 3w   Remove parentheses.

0 = 2w^2 + 3w -44   Subtract 44 from both sides.

0 = (2w + 11)(w-4)   Factor.

2w+11=0   w-4=0   Set each factor equal to 0.

2w=-11   w=4    Simplify and solve.

w = −51/2

Note that w = −51/2 is not a valid solution. It would not make sense to have a rectangle with a negative number as a width.
Since w = 4, the width of the rectangle is 4 meters. The length is 2w + 3, so we have 2(4) + 3 = 8 + 3 = 11. Thus the length of the rectangle is 11 meters.

4. Check. Is the length 3 meters more than twice the width?

11 ≟  3+2(4)   11 = 3+8   ✔

Is the area of the rectangle 44 square meters?

4 × 11 ≟ 44  44 = 44   ✔

EXAMPLE 7 The area of a triangle is 49 square centimeters. The altitude of the triangle is 7 centimeters longer than the base. Find the altitude and the base of the triangle.

                

Let b = the length of the base in centimeters

b + 7 = the length of the altitude in centimeters

To find the area of a triangle, we use

area = 1/2(altitude)(base) = 1/2ab = (ab)/2

(ab)/2=49  Write an equation.

((b + 7)(b))/2=49  Substitute the expressions for altitude and base.

(b^2 + 7b)/2=49  Simplify.

b^2 + 7b = 98   Multiply each side of the equation by 2.

b^2 + 7b-98 =0   Place the quadratic equation in standard form.

(b -7)(b + 14) =0   Factor.

b-7=0   b+ 14=0   Set each factor equal to zero.

b=7   b=-14   Solve each equation for b.

We cannot have a base of -14 centimeters, so we reject the negative answer. The only possible answer is b=7. So the base is 7 centimeters. The altitude is b+7=7+7= 14. The altitude is 14 centimeters.