Factoring Quadratic Expressions with Leading Coefficient 1: Master the Technique

One way to analyze a quadratic expression is to factor it — writing it as the product of two binomials. In this lesson, quadratic expressions with a leading coefficient of

1

will be factored.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Distributive Property
Polynomial
Standard Form of a Polynomial
Leading Coefficient
Degree of a Polynomial

Challenge

Finding a Pair of Numbers

Jordan wants to find a pair of integers with a sum of

11

and a product of

28 .

p + q = 11 p \cdot q = 28

Help Jordan find the pair of numbers.

Discussion

Factoring a Quadratic Trinomial With Leading Coefficient $1$

A quadratic trinomial in the form $x^{2} + b x + c$ can be factored as $(x + p) (x + q)$ if there exist $p$ and $q$ such that $p + q = b$ and $p q = c .$

x^{2} + b x + c = (x + p) (x + q)

Proof

Suppose that the sum of two numbers

p

and

q

is equal to

b

and their product is equal to

c .

p + q = b and p \cdot q = c

To show

x^{2} + b x + c = (x + p) (x + q),

substitute the equivalent expressions for

b

and

c .

x^{2} + b x + c

SubstituteExpressions

Substitute expressions

x^{2} + (p + q) x + p q

Factor

Distr

Distribute $x$

x^{2} + p x + q x + p q

FactorOut

Factor out $x$

x (x + p) + q x + p q

FactorOut

Factor out $q$

x (x + p) + q (x + p)

FactorOut

Factor out $x + p$

(x + q) (x + p)

CommutativePropMult

\CommutativePropMult

(x + p) (x + q)

Therefore, the quadratic trinomial and the product of the binomials are equal.

x^{2} + b x + c = (x + p) (x + q)

As an example, the trinomial below will be factored.

x^{2} + 7 x + 12

These three steps can be followed to factor it.

Analyze the Signs of

b

and

c

For some

p

and

q,

the aim is to write the given trinomial as follows.

x^{2} + 7 x + 12 = (x + p) (x + q)

To do so, the signs of

b

and

c

will be used to determine the signs of

p

and

q .

x^{2} + 7 x + 12

Here,

b = 7

and

c = 12,

so both

b

and

c

are positive.

Since $c$ is positive, the factors $p$ and $q$ must have the same sign so that $p \cdot q$ is positive.
Since $b$ is positive, both $p$ and $q$ must be positive so that $p + q$ is positive.

As a result,

p

and

q

are positive. To determine the signs in other cases, the following table can be used.

Find the Pair of Factors of

c

That Has a Sum of

b

It is known that $b = 7,$ $c = 12$ and that $p$ and $q$ are positive integers. Therefore, two positive factors of $12$ whose sum is $7$ need to be found. The positive factor pairs of $12$ will be listed and the pair with a sum of $7$ identified.

Positive Factors of $12$	Sum
$1$ and $12$	$1 + 12 = 13$
$2$ and $6$	$2 + 6 = 8$
$3$ and $4$	$3 + 4 = 7$

As seen, the factor pair of $3$ and $4$ meet these requirements, so the values of $p$ and $q$ are $3$ and $4 .$

Write in Factored Form

For the given trinomial, two integers with a sum of

7

and a product of

12

were found.

x^{2} + 7 x + 12 \Rightarrow p = 3 q = 4

Therefore, the trinomial can be written as the product of the binomials

x + 3

and

x + 4 .

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

Example

Factoring $x^{2} + b x + c$ When $b > 0$ and $c > 0$

It's a three-day weekend and Tearrik has the day off from school. He wants to use this time to make a present for his brother's birthday. He bought a nice frame and then chose a photo of the two of them. However, the photo does not fit in the frame, so Tearrik needs to edit it.

The editing program represents the area of the photo as $x^{2} + 13 x + 42$ and its length as $x + 7 .$ Help Tearrik answer the following questions.

a Write a binomial that represents the width of the photo.

b Tearrik is considering making a new picture frame himself. He needs to know approximately how much material he would need to make the frame. Find the perimeter of the photo if it is

9

inches wide.

Hint

a Start by factoring the given quadratic expression.

b Use the given information to find

x .

Then, use the formula for the perimeter of a rectangle.

Solution

a The given quadratic expression represents the area of the photo. Therefore, it should be equal to the product of its length

x + 7

and width.

A x^{2} + 13 x + 42 = ℓ \cdot w ⇓ = (x + 7) w

To find

w,

the trinomial must be factored. To do so, identify

b

and

c,

and determine the signs of

p

and

q .

x^{2} + 13 x + 42

Here,

b = 13

and

c = 42,

so both

b

and

c

are positive.

Since $c$ is positive, the factors $p$ and $q$ must have the same sign so that $p \cdot q$ is positive.
Since $b$ is positive, both $p$ and $q$ must be positive so that $p + q$ is positive.

Two positive factors of $42$ whose sum is $13$ need to be found. Now, list the positive factor pairs of $42$ and identify the pair with a sum of $13 .$

Positive Factors of $42$	Sum
$1$ and $42$	$1 + 42 = 43$
$2$ and $21$	$2 + 21 = 23$
$3$ and $14$	$3 + 14 = 17$
$6$ and $7$	$6 + 7 = 13$

As seen, the factor pair

6

and

7

satisfies the conditions, meaning that the values of

p

and

q

are

6

and

7 .

Therefore, the trinomial can be written as the product of the binomials

x + 7

and

x + 6 .

x^{2} + 13 x + 42 = (x + 7) (x + 6)

Since

x + 7

represents the length of the photo, the

x + 6

must represent the width of the photo.

b The width of the photo is given as

9

inches. Since the binomial

x + 6

represents the width of the photo, the value of

x

can be found.

x + 6 = 9 \Rightarrow x = 3

Before finding the perimeter of the photo, its length

x + 7

must also be computed.

x + 7

Substitute

$x = 3$

3 + 7

AddTerms

Add terms

10

Now, the length and width of the photo are known.

w = 9 and ℓ = 10

The perimeter can now be calculated by using the perimeter formula.

P = 2 (w + ℓ)

SubstituteII

$w = 9$ , $ℓ = 10$

P = 2 (9 + 10)

AddTerms

Add terms

P = 2 (19)

Multiply

P = 38

The photo's perimeter is

38

inches.

Example

Factoring $x^{2} + b x + c$ When $b > 0$ and $c < 0$

Vincenzo is using the extra day off school to begin building a kennel for his dog. In order to use the garden area in the most effective way, he has to build it on a triangular base. He draws a plan of a right triangle whose hypotenuse is represented by the binomial $x + 7 .$

It is known that the trinomial $x^{2} + 3 x - 10$ is twice the area of the triangle and that the length of $\overline{B C}$ is greater than the length of $\overline{A B} .$

a What are the possible leg lengths of the triangle when

x = 10 ?

b Write an expression for the perimeter of the triangle to find the amount of fencing Vincenzo will need.

Hint

a The given trinomial is equal to the product of the leg lengths of the triangle. Factor the given quadratic trinomial.

b The perimeter of a triangle is the sum of all of its side lengths.

Solution

a Since the given trinomial is twice the area of the triangle, the trinomial is equal to the product of leg lengths.

x^{2} + 3 x - 10 = 2 \cdot \frac{A B \cdot B C}{2}

To find

A B

and

B C,

the quadratic trinomial needs to be written as a product of two binomials.

x^{2} + 3 x - 10 = (x + p) (x + q)

Now identify

b

and

c

to get an idea about the signs of

p

and

q .

x^{2} + 3 x - 10 ⇓ x^{2} + 3 x + (- 10)

For this expression,

b = 3

and

c = - 10,

b

is positive but

c

is negative.

Since $c$ is negative, the factors $p$ and $q$ must have opposite signs so that $p \cdot q$ is negative.
Since $b$ is positive, the factor with a greater absolute value must be positive.

As such, the factor pairs of $- 10$ where one factor is negative should be listed. Then, look for the pair with a sum of $3 .$

Positive Factors of $- 10$	Sum
$- 1$ and $10$	$- 1 + 10 = 9$
$1$ and $- 10$	$1 + (- 10) = - 9$
$- 2$ and $5$	$- 2 + 5 = 3$
$2$ and $- 5$	$2 + (- 5) = - 3$

The factors that satisfy the conditions are

- 2

and

5 .

Therefore, the trinomial can be written as the product of the binomials

x - 2

and

x + 5 .

x^{2} + 3 x - 10 = (x - 2) (x + 5)

These two binomials represent the lengths of the legs. When

x = 10,

the expressions will be

8

and

15 .

x - 2 x + 5 x = 10 8 x = 10 15

This means that the longer side is

15

and the shorter side is

8 .

Since

B C

is greater than

A B,

B C = 15

and

A B = 8 .

b The binomials representing the lengths of the legs are

x - 2

and

x + 5 .

Additionally,

x + 7

represents the hypotenuse. To find the perimeter in terms of

x,

all these binomials should be added.

P = (x - 2) + (x + 5) + (x + 7) = 3 x + 10

Example

Factoring $x^{2} + b x + c$ When $b < 0$ and $c < 0$

Tadeo is using the extra day off school to catch up on homework. He is almost finished with his math homework but is stuck on one problem. He reviews the notes he wrote about factoring the given quadratic trinomial with a leading coefficient of $1 .$

Describe the error in his notes and help him factor the trinomial correctly so he can spend the rest of the weekend doing something more fun.

Answer

Error: When $∣ p ∣ > ∣ q ∣,$ $q$ must be a positive integer and $p$ must be a negative integer so that their sum is negative.
Factored Form: $(x - 15) (x + 6)$

Hint

Start by identifying the values of $b$ and $c$ for the given quadratic trinomial.

Solution

First,

b

and

c

will be identified.

x^{2} - 9 x - 90 ⇓ x^{2} + (- 9) x + (- 90)

Here,

b = - 9

and

c = - 90,

meaning that both

b

and

c

are negative.

Since $c$ is negative, the factors $p$ and $q$ must have opposite signs so that $p \cdot q$ is negative.
Since $b$ is negative, the factor with a greater absolute value must be negative so that their sum is negative.

Therefore, Tadeo's first point is correct. If it is assumed that $∣ p ∣ > ∣ q ∣,$ then $p$ should be negative. In other words, Tadeo's second point is incorrect.

The given trinomial can be factored using this information. To do so, the factor pairs of $- 90$ where one factor is negative and its absolute value is greater than the other factor are listed. Then, the pair with a sum of $- 9$ should be looked for.

Factors of $- 90$	Sum of Factors
$- 90$ and $1$	$- 90 + 1 = - 89$
$- 45$ and $2$	$- 45 + 2 = - 43$
$- 30$ and $3$	$- 30 + 3 = - 27$
$- 18$ and $5$	$- 18 + 5 = - 13$
$- 15$ and $6$	$- 15 + 6 = - 9$

The factors that satisfy the conditions are

- 15

and

6,

so the trinomial can be written as the product of the binomials

x - 15

and

x + 6 .

x^{2} - 9 x - 90 = (x - 15) (x + 6)

Example

Factoring $x^{2} + b x + c$ When $b < 0$ and $c > 0$

An inter-class quiz game is being held at Davontay's school this weekend, and he is his class's champion. The quizmaster Paulina asks Davontay to write a quadratic trinomial and then factor it. The conversation between the quizmaster and Dovantoy is shown in the diagram.

Conversation between quizmaster and Davontay

Help Davontay find the value of

b

and write the trinomial in factored form to win the quiz game.

Hint

If $b$ is negative, the sum of two factors of $60$ should be negative.

Solution

Davontay needs to write a quadratic trinomial in the form

x^{2} + b x + c .

It is also given that

b

is negative and

c = 60 .

x^{2} + b x + c ⇓ x^{2} + b x + 60

Since the above trinomial can be factored, the value of

b

should be the sum of the two factors,

p

and

q

, of

60 .

x^{2} + b x + 60 = (x + p) (x + q)

Now, two facts can be inferred from this trinomial.

Since $c$ is positive, the factors $p$ and $q$ must have the same sign so that $p \cdot q$ is positive.
Since $b$ is negative, both $p$ and $q$ should be negative so that $p + q$ is negative.

Based on this information, only negative factor pairs of $60$ need to be listed.

Negative Factors of $60$
$- 1$ and $- 60$
$- 2$ and $- 30$
$- 3$ and $- 20$
$- 4$ and $- 15$
$- 5$ and $- 12$
$- 6$ and $- 10$

The sum of the factors in each pair could be the value of $b,$ so Davontay's concern is valid. There are six values for $b .$ These values can be found as follows.

Factors of $60$	Sum
$- 1$ and $- 60$	$- 1 + (- 60) = - 61$
$- 2$ and $- 30$	$- 2 + (- 30) = - 32$
$- 3$ and $- 20$	$- 3 + (- 20) = - 23$
$- 4$ and $- 15$	$- 4 + (- 15) = - 19$
$- 5$ and $- 12$	$- 5 + (- 12) = - 17$
$- 6$ and $- 10$	$- 6 + (- 10) = - 16$

Of these possible values,

- 16

is the greatest, which meets Paulina's hint. Finally, the trinomial can be written.

x^{2} - 16 x + 60

The factors

- 6

and

- 10

have a product of

60

and a sum of

- 16 .

Therefore, the trinomial can be written as the product of the binomials

x - 6

and

x - 10 .

x^{2} - 16 x + 60 = (x - 6) (x - 10)

Pop Quiz

Practice Factoring Quadratic Trinomials

Factor each quadratic expression with a leading coefficient $1 .$ Write the answer in such a way that the value of $q$ is the greater factor.

Example

Finding the Dimensions of a Trough by Factoring

Maya and her father spent the long weekend building a trough for the animals on their farm. Her father knows that Maya has a math test coming up soon, so he decides to help her prepare for it by quizzing her about the trough they just built.

The trough's length is $105$ centimeters longer than its width. The area covered by the trough is $12250$ square centimeters.

a Write an equation for the area of the trough in term of its width

w

and solve the equation by factoring.

b State if the solutions make sense. What are the width and length of the trough?

Answer

a Equation:

w (w + 105) = 12250

Solutions:

w = 70

and

w = - 175

b Only the positive solution makes sense because measures cannot be negative.

Width: $70$ centimeters

Length:

175

centimeters

Hint

a Use the formula for the area of a rectangle to write an equation.

b Length cannot be negative.

Solution

a The width of the trough is represented by

w .

Since the length of the rectangular trough is

105

centimeters longer than its width,

w + 105

represents the length.

External credits: Colin Smith

Given that the area of the trough is

12250

square centimeters, the following equation can be written using the formula for the area of a rectangle.

A = w \cdot ℓ ⇓ 12250 = w (w + 105)

Now, the equation needs to be rewritten so that a quadratic trinomial is formed on one side of the equation. Then, it will be solved by factoring.

12250 = w (w + 105)

Rewrite

Distr

Distribute $w$

12250 = w^{2} + 105 w

SubEqn

$LHS - 12250 = RHS - 12250$

0 = w^{2} + 105 w - 12250

RearrangeEqn

Rearrange equation

w^{2} + 105 w - 12250 = 0

The quadratic trinomial should be factored.

w^{2} + 105 w + (- 12250)

For this trinomial,

b = 105

and

c = - 12250 .

Since the value of

c

is negative, only factor pairs of

- 12250

that have opposite signs will be considered. Since

b

is positive, the factor with a greater absolute value must be positive.

Factors of $- 12250$	Sum of Factors
$1225$ and $- 10$	$1225 + (- 10) = 1215$
$490$ and $- 25$	$490 + (- 25) = 465$
$350$ and $- 35$	$350 + (- 35) = 315$
$245$ and $- 50$	$245 + (- 50) = 195$
$175$ and $- 70$	$175 + (- 70) = 105$

The factors

- 175

and

70

satisfy the conditions. Therefore, the trinomial can be written as the product of the binomials

x - 70

and

x + 175 .

w^{2} + 105 w - 12250 = 0 ⇓ (w - 70) (w + 175) = 0

The left-hand side of the equation is a product of two factors. One of them must be zero so that the product is equal to zero. This is known as the Zero Product Property.

(w - 70) (w + 175) = 0

ZeroProdProp

Use the Zero Product Property

w - 70 = 0 w + 175 = 0 (I) (II)

AddEqn

$(I):$ $LHS + 70 = RHS + 70$

w = 70 w + 175 = 0

SubEqn

$(II):$ $LHS - 175 = RHS - 175$

w = 70 w = - 175

Therefore, the solutions of the equation are

70

and

- 175 .

b Recall the solutions of the equation written in Part A.

w = 70 and w = - 175

Since

w

represents a length, it cannot be negative. This means that only the positive value makes sense. The dimensions of the trough can be found by substituting

w = 75 .

	Expression	$w = 75$
Width	$w$	$75$
Length	$w + 105$	$75 + 105 = 175$

The width of the trough is $70$ centimeters and the length is $175$ centimeters.

Closure

Finding a Pair of Numbers

The challenge presented at the beginning can now be solved using the information covered in this lesson. Jordan wants to find a pair of integers with a sum of

11

and a product of

28 .

p + q = 11 p \cdot q = 28

Help Jordan find the pair of numbers.

Answer

$4$ and $7$

Hint

Isolate one of the variables and substitute it into the other equation. Then, solve the equation by factoring.

Solution

The first step is to isolate

q

in the first equation.

p + q = 11 \Rightarrow q = 11 - p

Now, the expression equivalent to

q

is substituted into the other equation. Then, the equation is rewritten so that a quadratic expression is formed on one side of the equation.

p \cdot q = 28

Substitute

$q = 11 - p$

p \cdot (11 - p) = 28

Rewrite

Distr

Distribute $p$

11 p - p^{2} = 28

SubEqn

$LHS - 28 = RHS - 28$

11 p - p^{2} - 28 = 0

MultEqn

$LHS \cdot (- 1) = RHS \cdot (- 1)$

- 11 p + p^{2} + 28 = 0

CommutativePropAdd

\CommutativePropAdd

p^{2} - 11 p + 28 = 0

To solve this equation, the quadratic trinomial should be factored.

p^{2} - 11 p + 28

In this trinomial,

b = - 11

and

c = 28 .

Since

c

is positive, factor pairs of

28

that have the same sign should be considered. Of those pairs, negative pairs are listed because

b

is negative.

Factors of $28$	Sum of Factors
$- 28$ and $- 1$	$- 28 + (- 1) = - 29$
$- 14$ and $- 2$	$- 14 + (- 2) = - 16$
$- 7$ and $- 4$	$- 7 + (- 4) = - 11$

The factors

- 7

and

- 4

has a sum of

- 11

and a product of

28 .

The trinomial can now be written as the product of the binomials

p - 7

and

p - 4 .

p^{2} - 11 p + 28 = 0 ⇓ (p - 7) (p - 4) = 0

The left-hand side of the equation is a product of two factors. One of them must be zero so that the product is equal to zero. Therefore,

p

is either

7

4

by the Zero Product Property.

(p - 7) (p - 4) = 0 ⇓ p = 7 or p = 4

When

p = 7,

q

will be

4,

as their sum is

11 .

Conversely, when

p = 4,

q = 7 .

Therefore, the numbers

7

and

4

are the numbers Jordan is looking for.

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount}}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount}}
	{{ 'ml-lesson-time-estimation' \| message }}

{{ article.displayTitle }}

Catch-Up and Review

Proof

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution