Unlocking the Secrets of Factoring Quadratic Expressions with Greatest Common Factor and Zero Product Property

Some of the easiest quadratic expressions to factor are those with a leading coefficient of

1 .

This lesson will explain how to factor quadratic expressions with leading coefficients other than

1 .

Catch-Up and Review

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

Challenge

Finding a Pair of Numbers

Consider the numbers $a = 2,$ $b = 12,$ and $c = 10 .$ Find two numbers $p$ and $q$ such that the following conditions are satisfied.

Condition 1 p q = \frac{b}{a} Condition 2 p + q = \frac{c}{a}

Discussion

Factoring a Quadratic Trinomial

If the greatest common factor (GCF) of a quadratic trinomial of the form

a x^{2} + b x + c

a,

then it can be rewritten by factoring out

a .

In this case, since

a

is a factor of both

b

and

c,

these two coefficients can be written in terms of

a .

b c = a \cdot b_{1}, for some number b_{1} = a \cdot c_{1}, for some number c_{1}

With this information in mind, the expression

a x^{2} + b x + c

can be factored.

a x^{2} + b x + c

SubstituteII

$b = a \cdot b_{1}$ , $c = a \cdot c_{1}$

a x^{2} + a \cdot b_{1} x + a \cdot c_{1}

FactorOut

Factor out $a$

a (x^{2} + b_{1} x + c_{1})

The leading coefficient of the quadratic expression in the parentheses is

1 .

Therefore, it can be factored by finding two numbers

p

and

q

such that

p + q = b_{1}

and

p q = c_{1} .

a (x^{2} + b_{1} x + c_{1}) = a (x + p) (x + q) ⇕ a x^{2} + b x + c = a (x + p) (x + q)

Example

Factoring the Face Area

Magdalena's class is planning a camping trip. Magdalena volunteered to research tents to purchase. When she asked her teacher what size they would buy, he instead told her that the base area of the tent should be $3 x^{2} + 12 x + 9$ square feet and its width $w$ is $x + 3$ feet.

To choose a suitable tent, Magdalena needs to know the dimensions of the tent's base.

a Write a binomial that represents the length by factoring the given area.

b What will be the dimensions of the tent when

x = 1

foot?

Hint

a Think about the greatest common factor of the given quadratic trinomial.

b Substitute the given value into the expressions that represent the width and length of the tent.

Solution

a Since the base of a tent is a rectangle, its area is the product of its width and its length.

A = w ℓ ⇓ 3 x^{2} + 12 x + 9 = (x + 3) ℓ

To find an expression for the length

ℓ,

the quadratic trinomial that represents the area will be factored. The first step to factoring the polynomial is to find its greatest common factor (GCF), if there is one. Note that the GCF of this expression is

3 .

3 x^{2} + 12 x + 9

SplitIntoFactors

Split into factors

3 x^{2} + 3 (4) x + 3 (3)

FactorOut

Factor out $3$

3 (x^{2} + 4 x + 3)

The simplified quadratic trinomial in parentheses now has a leading coefficient of

1 .

Therefore, it can be factored accordingly. The expression

x^{2} + 4 x + 3

can be written as

(x + p) (x + q),

where

p

and

q

are factor pairs of

3

whose sum is

4 .

Factors of $3$	Sum of the Factors
$- 1$ and $- 3$	$- 1 + (- 3) = - 4 \times$
$1$ and $3$	$1 + 3 = 4 ✓$

The table shows that

p = 1

and

q = 3 .

With this information in mind, the expression can be fully factored.

3 (x^{2} + 4 x + 3) ⇕ 3 (x + 1) (x + 3)

This expression represents the area of a tent's base.

A = w ℓ ⇓ 3 (x + 1) (x + 3) = (x + 3) ℓ

The expression

(x + 3)

on the right-hand side of the equation represents the width of the tent's base. A measurement for a width cannot be zero, which means that both sides of the equation can be divided by

(x + 3) .

3 (x + 1) (x + 3) = (x + 3) ℓ

DivEqn

$LHS / (x + 3) = RHS / (x + 3)$

\frac{3 ( x + 1 ) ( x + 3 )}{x + 3} = \frac{( x + 3 ) ℓ}{x + 3}

CancelCommonFac

Cancel out common factors

\frac{3 ( x + 1 ) ( x + 3 )}{x + 3} = \frac{( x + 3 ) ℓ}{x + 3}

SimpQuot

Simplify quotient

3 (x + 1) = ℓ

RearrangeEqn

Rearrange equation

ℓ = 3 (x + 1)

Distr

Distribute $3$

ℓ = 3 x + 3

The length of the tent's base is

3 x + 3

inches.

b Now that the equations for the width and length of the tent, substitute

x = 1

into each binomial to find the corresponding values.

	Substitution	Calculation
$w = x + 3$	$w = 1 + 3$	$w = 4$ feet
$ℓ = 3 x + 3$	$ℓ = 3 \cdot 1 + 3$	$ℓ = 6$ feet

The class will buy tents that are $4$ feet wide and $6$ feet long.

Discussion

Factoring a Quadratic Trinomial

When trying to factor a quadratic trinomial of the form

a x^{2} + b x + c,

it can be difficult to see its factors. Consider the following expression.

8 x^{2} + 26 x + 6

Here,

a = 8,

b = 26,

and

c = 6 .

There are six steps to factor this trinomial.

Factor Out the GCF of

a,

b,

and

c

To fully factor a quadratic trinomial, the Greatest Common Factor (GCF) of

a,

b,

and

c

has to be factored out first. To identify the GCF of these numbers, their prime factors will be listed.

8 = 26 = 6 = 2 \cdot 2 \cdot 2 2 \cdot 13 2 \cdot 3

It can be seen that

8,

26,

and

6

share exactly one factor,

2 .

GCF (8, 26, 6) = 2

Now,

2

can be factored out.

8 x^{2} + 26 x + 6

SplitIntoFactors

Split into factors

2 (4) x^{2} + 2 (13) x + 2 (3)

FactorOut

Factor out $2$

2 (4 x^{2} + 13 x + 3)

In the remaining steps, the factored coefficient

2

before the parentheses can be ignored. The new considered quadratic trinomial is

4 x^{2} + 13 x + 3 .

Therefore, the current values of

a,

b,

and

c,

are

4,

13,

and

3,

respectively. If the GCF of the coefficients is

1,

this step can be ignored.

Find the Factor Pair of

a c

Whose Sum Is

b

It is known that $a = 4$ and $c = 3,$ so $a c = 12 > 0 .$ Therefore, the factors must have the same sign. Also, $b = 13 .$ Since the sum of the factors is positive and they must have the same sign, both factors must be positive. All positive factor pairs of $12$ can now be listed and their sums checked.

Factors of $a c$	Sum of Factors
$1$ and $12$	$1 + 12 = 13 ✓$
$2$ and $6$	$2 + 6 = 8 \times$
$3$ and $4$	$3 + 4 = 7 \times$

In this case, the correct factor pair is $1$ and $12 .$ The following table sums up how to determine the signs of the factors based on the values of $a c$ and $b .$

$a c$	$b$	Factors
Positive	Positive	Both positive
Positive	Negative	Both negative
Negative	Positive	One positive and one negative. The absolute value of the positive factor is greater.
Negative	Negative	One positive and one negative. The absolute value of the negative factor is greater.

Such analysis makes the list of possible factor pairs shorter.

Write

b x

as a Sum

The factor pair obtained in the previous step will be used to rewrite the

x -

term — the linear term — of the quadratic trinomial as a sum. Remember that the factors are

1

and

12 .

13 x

WriteSum

Write as a sum

12 x + 1 x

IdPropMult

Identity Property of Multiplication

12 x + x

The linear term

13 x

can be rewritten in the original expression as

12 x + x .

4 x^{2} + 13 x + 3 ⇕ 4 x^{2} + 12 x + x + 3

Factor Out the GCF of the First Two Terms

The expression has four terms, which can be grouped into the

first

two

and the

last

two

terms .

Then, the GCF of each group can be factored out.

4 x^{2} + 12 x + x + 3

The first two terms,

4 x^{2}

and

12 x,

can be factored.

4 x^{2} = 12 x = 2 \cdot 2 \cdot x \cdot x 2 \cdot 2 \cdot 3 \cdot x

The GCF of

4 x^{2}

and

12 x

2 \cdot 2 \cdot x = 4 x .

4 x^{2} + 12 x + x + 3

Rewrite

Rewrite $4 x^{2}$ as $4 x \cdot x$

4 x \cdot x + 12 x + x + 3

Rewrite

Rewrite $12 x$ as $4 x \cdot 3$

4 x \cdot x + 4 x \cdot 3 + x + 3

FactorOut

Factor out $4 x$

4 x (x + 3) + x + 3

Factor Out the GCF of the Last Two Terms

The process used in Step

3

will be repeated for the last two terms. In this case,

x

and

3

cannot be factored, so their GCF is

1 .

4 x (x + 3) + x + 3

Rewrite

Rewrite $x$ as $1 \cdot x$

4 x (x + 3) + 1 \cdot x + 3

Rewrite

Rewrite $3$ as $1 \cdot 3$

4 x (x + 3) + 1 \cdot x + 1 \cdot 3

FactorOut

Factor out $1$

4 x (x + 3) + 1 (x + 3)

Factor Out the Common Factor

If all the previous steps have been performed correctly, there should now be two terms with a common factor.

4 x (x + 3) + 1 (x + 3)

The common factor will be factored out.

4 x (x + 3) + 1 (x + 3)

FactorOut

Factor out $(x + 3)$

(4 x + 1) (x + 3)

The factored form of

4 x^{2} + 13 x + 3

(4 x + 1) (x + 3) .

Remember that the original trinomial was

8 x^{2} + 26 x + 6

and that the GCF

2

was factored out in Step

1 .

This GCF has to be included in the final result.

8 x^{2} + 26 x + 6 = 2 (4 x + 1) (x + 3)

Example

Factoring $a x^{2} + b x + c$ When $a c$ is Positive

Heichi and Maya are reading a pamphlet about local campsites. They volunteered to find a suitable campground for their class field trip. Their math teacher, who goes camping often, gives them a challenge to help them decide where to go. If they answer his questions about the pamphlet correctly, he will tell them about his favorite local campsite.

Their teacher tells them that the area of the pamphlet is $3 x^{2} - 5 x + 2$ square inches and that it has a width of $x - 1$ inches.

a Write a binomial that represents the length of the pamphlet.

b Find the perimeter of the pamphlet if the width is

4

inches.

Hint

a Factor the trinomial that represents the area of the pamphlet.

b Substitute the given value into the binomial that represents the width. Then, find the value of

x .

Solution

a The area of the rectangular pamphlet is equal to the product of its width and length. This relationship will be used to find the length and perimeter of the pamphlet.

A = w ℓ ⇓ 3 x^{2} - 5 x + 2 = (x - 1) ℓ

To find an expression for the length

ℓ,

the trinomial that represents the area will be factored. First, the values of

a,

b,

and

c

for the trinomial will be determined.

3 x^{2} - 5 x + 2 ⇕ 3 x^{2} + (- 5) x + 2

Here,

a = 3,

b = - 5,

and

c = 2 .

With this in mind, the following conclusions can be made.

The Greatest Common Factor (GCF) of $a,$ $b,$ and $c$ is $1 .$
Since $a c = 6,$ which is greater than zero, the factors must have the same sign.
Since $b = - 5,$ which is negative, the sum of the factors will be negative. Therefore, both of the factors must be negative.

Now, all the negative factor pairs of $6$ will be listed and checked whether their sum is $- 5 .$

Factors of $a c = 6$	Sum of Factors
$- 1$ and $- 6$	$- 1 + (- 6) = - 7$
$- 2$ and $- 3$	$- 2 + (- 3) = - 5$

Therefore, the

x -

term of the trinomial will be rewritten by using the factors

- 2

and

- 3 .

3 x^{2} - 5 x + 2 ⇕ 3 x^{2} - 2 x - 3 x + 2

Next, the expression will be factored by grouping its terms.

3 x^{2} - 2 x - 3 x + 2

CommutativePropAdd

Commutative Property of Addition

3 x^{2} - 3 x - 2 x + 2

▼

Rewrite

Rewrite $3 x^{2}$ as $3 x \cdot x$

3 x \cdot x - 3 x - 2 x + 2

Rewrite

Rewrite $3 x$ as $3 x \cdot 1$

3 x \cdot x - 3 x \cdot 1 - 2 x + 2

FactorOut

Factor out $3 x$

3 x (x - 1) - 2 x + 2

Rewrite

Rewrite $2$ as $2 \cdot 1$

3 x (x - 1) - 2 x + 2 \cdot 1

FactorOut

Factor out $- 2$

3 x (x - 1) - 2 (x - 1)

FactorOut

Factor out $(x - 1)$

(3 x - 2) (x - 1)

Finally, the area can be written as the product of the binomials

x - 1

and

3 x - 2 .

3 x^{2} - 5 x + 2 = (x - 1) (3 x - 2)

Since

x - 1

represents the width of the pamphlet, the binomial

3 x - 2

represents its length.

b To find the perimeter of a rectangle, all four side lengths must be added. The width of the pamphlet is given as

4

inches. Since the binomial

x - 1

represents the width, the value of

x

can be found from here.

x - 1 = 4 \Leftrightarrow x = 5

Then, the length of the pamphlet will be found by substituting

x = 5

into the binomial that it was found in Part A.

ℓ = 3 x - 2

Substitute

$x = 5$

ℓ = 3 (5) - 2

Multiply

ℓ = 15 - 2

SubTerm

Subtract term

ℓ = 13

The length of the pamphlet is

13

inches. The perimeter of the pamphlet can finally be calculated now that both the length and width are known.

P = 2 (w + ℓ)

SubstituteII

$w = 4$ , $ℓ = 13$

P = 2 (4 + 13)

AddTerms

Add terms

P = 2 (17)

Multiply

P = 34

The perimeter of the pamphlet is

34

inches.

Example

Factoring $a x^{2} + b x + c$ When $a c$ is Negative

At the campsite, each camper is assigned a rectangular area of

28

square feet for their tent. Dominika wants to place a tarp on the ground below her tent to make sure it stays dry underneath. She does not remember the size of the tarp she has, but she knows that its length is

1

inch less than twice its width.

Dominika finds that the tarp fits exactly in the allocated area. Help her find the length and width of the tarp.

Hint

Start by writing a quadratic expression for the area of the tarp. Then, factor the obtained expression.

Solution

An expression for the area of the tarp can be written by using the expressions for its length and width. Let $w$ be the width of the tarp. Since the length is $1$ inch less than twice the width, it can be expressed as $2 w - 1 .$

Since the base is a rectangle, its area is equal to the product of its width and length.

A = w ℓ

SubstituteII

$A = 28$ , $ℓ = 2 w - 1$

28 = w (2 w - 1)

▼

Rewrite

Distr

Distribute $w$

28 = 2 w^{2} - w

SubEqn

$LHS - 28 = RHS - 28$

0 = 2 w^{2} - w - 28

RearrangeEqn

Rearrange equation

2 w^{2} - w - 28 = 0

Now, the obtained quadratic expression will be factored. First, its coefficients will be identified.

2 w^{2} - w - 28 = 0 ⇕ 2 w^{2} + (- 1) w + (- 28) = 0

Here,

a = 2,

b = - 1,

and

c = - 28 .

With this in mind, the following conclusions can be made.

The Greatest Common Factor (GCF) of $a,$ $b,$ and $c$ is $1 .$
Since $a c = - 56,$ which is less than zero, the factors must have the different signs. Therefore, there is a positive factor and a negative factor.
Since $b = - 1,$ the sum of the factors is negative. This means that the absolute value of the negative factor is greater than the absolute value of the positive factor.

Now, the factor pairs of $- 56$ with opposite signs and where the absolute value of the negative factor is greater than the absolute value of the positive number will be listed. Also, their sum will be calculated to see if it is $- 1 .$

Factors of $a c = - 56$	Sum of Factors
$- 56$ and $1$	$- 56 + 1 = - 55$
$- 28$ and $2$	$- 28 + 2 = - 26$
$- 14$ and $4$	$- 14 + 4 = - 10$
$- 8$ and $7$	$- 8 + 7 = - 1$

Of these factors,

- 8

and

7

satisfy the conditions. Therefore, the

x -

term of the trinomial on the left-hand side of the equation will be rewritten by using these numbers as a sum.

2 w^{2} - w - 28 = 0 ⇓ 2 w^{2} - 8 w + 7 w - 28 = 0

Then, the obtained expression will be factored by grouping.

2 w^{2} - 8 w + 7 w - 28 = 0

▼

Rewrite

Rewrite $2 w^{2}$ as $2 w \cdot w$

2 w \cdot w - 8 w + 7 w - 28 = 0

Rewrite

Rewrite $8 w$ as $2 w \cdot 4$

2 w \cdot w - 2 w \cdot 4 + 7 w - 28 = 0

FactorOut

Factor out $2 w$

2 w (w - 4) + 7 w - 28 = 0

SplitIntoFactors

Split into factors

2 w (w - 4) + 7 w - 7 \cdot 4 = 0

FactorOut

Factor out $7$

2 w (w - 4) + 7 (w - 4) = 0

FactorOut

Factor out $(w - 4)$

(2 w + 7) (w - 4) = 0

Since the product of the two binomials is zero, at least one binomial must equal zero. With this in mind, the Zero Product Property will be applied.

(w - 4) (2 w + 7) = 0

▼

Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

w - 4 = 0 2 w + 7 = 0 (I) (II)

AddEqn

$(I):$ $LHS + 4 = RHS + 4$

w = 4 2 w + 7 = 0

SubEqn

$(II):$ $LHS - 7 = RHS - 7$

w = 4 2 w = - 7

DivEqn

$(II):$ $LHS / 2 = RHS / 2$

w = 4 w = - 3.5

The solutions to the equation are

4

and

- 3.5 .

Since a negative length does not make sense,

w

cannot be negative. This means that the value of the width is

4

feet. With this information and knowing that the length is

1

foot less than twice the width, the length can be found.

ℓ = 2 w - 1

Substitute

$w = 4$

ℓ = 2 (4) - 1

▼

Evaluate right-hand side

Multiply

ℓ = 8 - 1

SubTerm

Subtract term

ℓ = 7

The base of the tarp has a length and a width of

7

and

4

feet, respectively.

Width : Length : 4 feet 7 feet

Example

Factoring $a x^{2} + b x + c$ When $a$ is Negative

Dylan is practicing on the basketball court at the campground. For the ball to fall through the hoop without hitting the rim or backboard, it needs to follow a specific path after being thrown at a

4 5^{\circ}

angle.

This path can be modeled with a quadratic function.

h (t) = - 5 t^{2} + 13 t + 6

Here,

h

is the height of the ball in feet

t

seconds after it has been thrown. Using this model, find the time it takes for the ball to hit the ground after Dylan throws it.

Hint

The value of $h$ is zero when the ball hits the ground.

Solution

The height of the ball is zero when the it hits the ground. Therefore, to find the total time it takes for the ball to hit the ground,

0

will be substituted for

h (t)

in the given equation.

h (t) = - 5 t^{2} + 13 t + 6 ⇓ 0 = - 5 t^{2} + 13 t + 6

Now, the trinomial will be factored. First, its coefficients will be identified.

- 5 t^{2} + 13 t + 6

Here,

a = - 5,

b = 13,

and

c = 6 .

With this information in mind, the following conclusions can be made.

The Greatest Common Factor (GCF) of $a,$ $b,$ and $c$ is $1 .$
Since $a c = - 30,$ which is less than zero, the factors must have different signs. Therefore, one of the factors is positive and one of the factors is negative.
Since $b = 13,$ the sum of the factors is positive. Therefore, the absolute value of the positive factor is greater than the absolute value of the negative factor.

Now, the factor pairs of $- 30$ with opposite signs and where the absolute value of the positive factor is greater than the absolute value of the negative factor will be listed. Also, their sum will be calculated to see whether it is $13 .$

Factors of $a c = - 30$	Sum of Factors
$- 1$ and $30$	$29$
$- 2$ and $15$	$13$
$- 3$ and $10$	$7$
$- 5$ and $6$	$1$

The factors of

- 30

whose sum is

13

are

- 2

and

15 .

The

t -

term of the right-hand side can be rewritten as a subtraction of these numbers.

0 = - 5 t^{2} + 13 t + 6 ⇓ 0 = - 5 t^{2} + 15 t - 2 t + 6

Now, the obtained expression will be factored by grouping.

0 = - 5 t^{2} + 15 t - 2 t + 6

RearrangeEqn

Rearrange equation

- 5 t^{2} + 15 t - 2 t + 6 = 0

▼

Rewrite

Rewrite $- 5 t^{2}$ as $- 5 t \cdot t$

- 5 t \cdot t + 15 t - 2 t + 6 = 0

Rewrite

Rewrite $15 t$ as $5 t \cdot 3$

- 5 t \cdot t + 5 t \cdot 3 - 2 t + 6 = 0

FactorOut

Factor out $- 5 t$

- 5 t (t - 3) - 2 t + 6 = 0

Rewrite

Rewrite $6$ as $2 \cdot 3$

- 5 t (t - 3) - 2 t + 2 \cdot 3 = 0

FactorOut

Factor out $- 2$

- 5 t (t - 3) + (- 2) (t - 3) = 0

AddNeg

$a + (- b) = a - b$

- 5 t (t - 3) - 2 (t - 3) = 0

FactorOut

Factor out $(t - 3)$

(- 5 t - 2) (t - 3) = 0

The factoring process is finally done. Now, the Zero Product Property can be used to find the

t -

values.

(- 5 t - 2) (t - 3) = 0

▼

Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

- 5 t - 2 = 0 t - 3 = 0 (I) (II)

AddEqn

$(I):$ $LHS + 2 = RHS + 2$

- 5 t = 2 t - 3 = 0

DivEqn

$(I):$ $LHS / (- 5) = RHS / (- 5)$

t = \frac{2}{- 5} t - 3 = 0

MoveNegDenomToFrac

$(I):$ Put minus sign in front of fraction

t = - \frac{2}{5} t - 3 = 0

CalcQuot

$(I):$ Calculate quotient

t = - 0.4 t - 3 = 0

AddEqn

$(II):$ $LHS + 3 = RHS + 3$

t = - 0.4 t = 3

Since a negative time does not make sense,

t = 3

is the only solution that satisfies the conditions. This means that the ball will hit the ground

3

seconds after it is thrown.

Example

The Width of a Path

There is a jogging track around the campground with the same width on every side. It is known that the campground has a rectangular shape with a width of $30$ meters and a length of $50$ meters.

a Write a trinomial that represents the total area of the campground and the jogging track.

b The total area for the campground and the jogging track altogether is

2204

square meters. What is the width of the jogging track?

Hint

a Start by writing binomials for the overall width and the overall length.

b Write an equation using the trinomial found in Part A and the total area for the campground and jogging track together.

Solution

a Let

x

be the width of the track. The dimensions of the figure can be shown on the diagram.

As seen above, the overall width is

x + 30 + x

meters. The overall length is

x + 50 + x

meters.

Width : Length : x + 30 + x \Leftrightarrow 2 x + 30 x + 50 + x \Leftrightarrow 2 x + 50

Since the given figure is a rectangle, its area is the product of its width and length.

A = w ℓ ⇓ A = (2 x + 30) (2 x + 50)

The product will be rewritten as a trinomial.

A = (2 x + 30) (2 x + 50)

▼

Multiply parentheses

Distr

Distribute $(2 x + 50)$

A = 2 x (2 x + 50) + 30 (2 x + 50)

Distr

Distribute $2 x$

A = 4 x^{2} + 100 x + 30 (2 x + 50)

Distr

Distribute $30$

A = 4 x^{2} + 100 x + 60 x + 1500

AddTerms

Add terms

A = 4 x^{2} + 160 x + 1500

The trinomial that represents the total area is

4 x^{2} + 160 x + 1500 .

b To find the width of the jogging track

x,

the trinomial found in Part A will be set equal to the total area of the jogging track and campground,

2204

square meters.

2204 = 4 x^{2} + 160 x + 1500

The equation will be rewritten to make the right-hand side equal to zero.

2204 = 4 x^{2} + 160 x + 1500

SubEqn

$LHS - 2204 = RHS - 2204$

0 = 4 x^{2} + 160 x - 704

RearrangeEqn

Rearrange equation

4 x^{2} + 160 x - 704 = 0

Next, the Greatest Common Factor (GCF) will be factored out.

4 x^{2} + 160 x - 704 = 0

SplitIntoFactors

Split into factors

4 \cdot x^{2} + 4 \cdot 40 x - 4 \cdot 176 = 0

FactorOut

Factor out $4$

4 (x^{2} + 40 x - 176) = 0

Now that the GCF has been factored out, the quadratic trinomial in parentheses will be factored. The expression

x^{2} + 40 x - 176

will be written as

(x + p) (x + q),

where

p

and

q

are factor pairs of

- 176

whose sum is

40 .

Therefore, all factor pairs of

- 176

will be listed and their sum will be calculated.

Factors of $- 176$	Sum of Factors
$- 1$ and $176$	$- 1 + 176 = 175$
$- 2$ and $88$	$- 2 + 88 = 86$
$- 4$ and $44$	$- 4 + 44 = 40$
$- 8$ and $22$	$- 8 + 22 = 14$
$- 11$ and $16$	$- 11 + 16 = 5$
$- 16$ and $11$	$- 16 + 11 = - 5$
$- 22$ and $8$	$- 22 + 8 = - 14$
$- 44$ and $4$	$- 44 + 4 = - 40$
$- 88$ and $2$	$- 88 + 2 = - 86$
$- 176$ and $1$	$- 176 + 1 = - 175$

As seen, the factors

- 4

and

44

satisfy the conditions, so

p = - 4

and

q = 44 .

With this information in mind, the equation can now be fully factored.

4 x^{2} + 160 x - 706 = 0 ⇕ 4 (x - 4) (x + 44) = 0

Finally, since the product of the two binomials is zero, at least one binomial must equal zero. With this in mind, the Zero Product Property can be used to find the value of

x .

4 (x - 4) (x + 44) = 0

DivEqn

$LHS / 4 = RHS / 4$

(x - 4) (x + 44) = 0

▼

Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

x - 4 = 0 x + 44 = 0 (I) (II)

AddEqn

$(I):$ $LHS + 4 = RHS + 4$

x = 4 x + 44 = 0

SubEqn

$(II):$ $LHS - 44 = RHS - 44$

x = 4 x = - 44

Since

x

represents the width of the track, it cannot be negative. Therefore, the only solution to the equation is

x = 4 .

This means that the width of the jogging track is

4

meters.

Closure

Finding a Pair of Numbers

The topics learned in this lesson can be applied to the challenge presented at the beginning. Three numbers

a = 2,

b = 10,

and

c = 12

are given. The challenge is to find two numbers

p

and

q

that satisfy the following conditions.

Condition 1 p q = \frac{c}{a} Condition 2 p + q = \frac{b}{a}

p

is less than

q,

find these two values.

Hint

Substitute the given numbers $a,$ $b,$ and $c$ into the equations of the conditions.

Solution

Start by substituting the given numbers

a = 2,

b = 10,

and

c = 12

into the equations of the conditions.

Condition 1 p \cdot q = \frac{1 2}{2} ⇓ p \cdot q = 6 Condition 2 p + q = \frac{1 0}{2} ⇓ p + q = 5

In this case,

p

and

q

are factors of

6

and they have a positive sum. To find these numbers, all positive factor pairs of

6

will be listed and their sum will be calculated.

Factors of $6$	Sum of Factors
$1$ and $6$	$7$
$2$ and $3$	$5$

The table shows that

2

and

3

satisfy the conditions. Since

p

is less than

q,

p = 2

and

q = 3 .

If the given numbers

a = 2,

b = 10,

and

c = 12

are the coefficients of a quadratic expression, then this trinomial can be factored by considering the methods and examples already discussed.

a x^{2} + b x + c ⇓ 2 x^{2} + 10 x + 12

First, the Greatest Common Factor of all therms,

2,

must be factored out.

2 x^{2} + 10 x + 12 ⇕ 2 (x^{2} + 5 x + 6)

The leading coefficient of the trinomial in parentheses is

1 .

Therefore, it can be factored as

(x + p) (x + q)

where

p

and

q

are factor pairs of

6

whose sum is

5 .

These values are already known and are

p = 2

and

q = 3 .

2 x^{2} + 10 x + 12 ⇕ 2 (x + 2) (x + 3)

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Catch-Up and Review

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