5. Factoring Quadratic Expressions With a Leading Coefficient of 1
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Chapter 7
5.
Factoring Quadratic Expressions With a Leading Coefficient of 1
This lesson delves into the process of factoring quadratic expressions, specifically those with a leading coefficient of 1. Through various examples and scenarios, such as determining the dimensions of a photo frame or the area of a trough, readers are guided on how to break down these expressions into products of binomials. The lesson emphasizes the importance of understanding the signs of the coefficients and how they influence the factoring process. With practical applications like determining the perimeter of a triangle or the area of a rectangular rug, the lesson bridges the gap between theoretical knowledge and real-world application, making the concept of factoring more relatable and understandable.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Factoring Quadratic Expressions With a Leading Coefficient of 1
Slide of 10
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.
Challenge
Finding a Pair of Numbers
Discussion
Factoring a Quadratic Trinomial With Leading Coefficient 1
A quadratic trinomial in the form x2+bx+c can be factored as (x+p)(x+q) if there exist p and q such that p+q=b and pq=c.
x2+bx+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.
Therefore, the quadratic trinomial and the product of the binomials are equal.
p+q=bandp⋅q=c
To show x2+bx+c=(x+p)(x+q), substitute the equivalent expressions for b and c.
x2+bx+c
SubstituteExpressions
Substitute expressions
x2+(p+q)x+pq
▼
Factor
Distr
Distribute x
x2+px+qx+pq
FactorOut
Factor out x
x(x+p)+qx+pq
FactorOut
Factor out q
x(x+p)+q(x+p)
FactorOut
Factor out x+p
(x+q)(x+p)
CommutativePropMult
Commutative Property of Multiplication
(x+p)(x+q)
x2+bx+c=(x+p)(x+q)
As an example, the trinomial below will be factored.
x2+7x+12
These three steps can be followed to factor it.
1
Analyze the Signs of b and c
expand_more For some p and q, the aim is to write the given trinomial as follows.
x2+7x+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.
x2+7x+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⋅q is positive.
- Since b is positive, both p and q must be positive so that p+q is positive.
2
Find the Pair of Factors of c That Has a Sum of b
expand_more 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.
3
Write in Factored Form
expand_more For the given trinomial, two integers with a sum of 7 and a product of 12 were found.
x2+7x+12⇒p=3q=4
Therefore, the trinomial can be written as the product of the binomials x+3 and x+4.
x2+7x+12=(x+3)(x+4)
Example
Factoring x2+bx+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 x2+13x+42 and its length as x+7. Help Tearrik answer the following questions.
a Write a binomial that represents the width of the photo.
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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.
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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.
Ax2+13x+42=ℓ⋅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.
x2+13x+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⋅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 |
x2+13x+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⇒x=3
Before finding the perimeter of the photo, its length x+7 must also be computed.
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.
The photo's perimeter is 38 inches.
Example
Factoring x2+bx+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 x2+3x−10 is twice the area of the triangle and that the length of BC is greater than the length of AB.
a What are the possible leg lengths of the triangle when x=10?
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b Write an expression for the perimeter of the triangle to find the amount of fencing Vincenzo will need.
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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.
x2+3x−10=2⋅2AB⋅BC
To find AB and BC, the quadratic trinomial needs to be written as a product of two binomials.
x2+3x−10=(x+p)(x+q)
Now identify b and c to get an idea about the signs of p and q.
x2+3x−10⇓x2+3x+(-10)
For this expression, b=3 and c=-10, so b is positive but c is negative. - Since c is negative, the factors p and q must have opposite signs so that p⋅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 |
x2+3x−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−2x+5 x=10 8 x=10 15
This means that the longer side is 15 and the shorter side is 8. Since BC is greater than AB, BC=15 and AB=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)=3x+10
Example
Factoring x2+bx+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.
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.
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.
x2−9x−90⇓x2+(-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⋅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 |
x2−9x−90=(x−15)(x+6)
Example
Factoring x2+bx+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.
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Solution
Davontay needs to write a quadratic trinomial in the form x2+bx+c. It is also given that b is negative and c=60.
Of these possible values, -16 is the greatest, which meets Paulina's hint. Finally, the trinomial can be written.
x2+bx+c⇓x2+bx+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.
x2+bx+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⋅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 |
x2−16x+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.
x2−16x+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.
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
Solutions: w=70 and w=-175
b Only the positive solution makes sense because measures cannot be negative.
Width: 70 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
A=w⋅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=w2+105w
SubEqn
LHS−12250=RHS−12250
0=w2+105w−12250
RearrangeEqn
Rearrange equation
w2+105w−12250=0
w2+105w+(-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 |
w2+105w−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=0w+175=0(I)(II)
AddEqn
(I): LHS+70=RHS+70
w=70w+175=0
SubEqn
(II): LHS−175=RHS−175
w=70w=-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=11p⋅q=28
Help Jordan find the pair of numbers.
Answer
4 and 7
Solution
The first step is to isolate q in the first equation.
To solve this equation, the quadratic trinomial should be factored.
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+q=11⇒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⋅q=28
Substitute
q=11−p
p⋅(11−p)=28
▼
Rewrite
Distr
Distribute p
11p−p2=28
SubEqn
LHS−28=RHS−28
11p−p2−28=0
MultEqn
LHS⋅(-1)=RHS⋅(-1)
-11p+p2+28=0
CommutativePropAdd
Commutative Property of Addition
p2−11p+28=0
p2−11p+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 |
p2−11p+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 or 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.
Factoring Quadratic Expressions With a Leading Coefficient of 1
Exercise 2.1
A map of the USA is displayed on a whiteboard. The area of the map, in square centimeters, is represented by x2+3x−54. Its length is represented by (x+9).
External credits: TUBS
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We are given the area of a rectangle and one of its sides. We need to find the other side of the rectangle. First, let's look at just this part of the picture.
Let's substitute these pieces into the formula for the area of a rectangle. cc A&=& l &*& w ↓&& ↓ && ↓ x^2+3x-54 & = & (x+9)& * & w The area is represented by a quadratic trinomial with a leading coefficient of 1 and it is factorable. Therefore, it can be written as a product of two binomials x+p and x+q. x^2+3x-54 = (x+p)(x+q) ⇓ x^2+3x-54 = (x+9)(x+q) The number q in the factor (x+q), when multiplied by 9, should give - 54. Let's calculate the value of q. 9 * q = - 54 ⇒ q = - 6 We found that q= - 6. We can now say that the width of the map is x-6 cm.
We are given the width of the map as 70 centimeters. We also know that the expression x-6 represents the width.
x-6 = 70 ⇒ x = 76
Therefore, x is equal to 76 centimeters. Let's know write an expression for the perimeter of the map.
Now we can find the perimeter by substituting x=76 into our formula and simplifying.
When the width of the map is 70 centimeters, its perimeter is 310 centimeters.
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A lawyer's office and garden are on a rectangular piece of land. The area of the land, in square meters, is represented by x2+7x−30.
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We are given the area of a rectangle and one of its sides. We need to find the other side of the rectangle. First, let's look at just this part of the diagram.
Let's substitute these pieces into the formula for the area of a rectangle. cc A&=& l &*& w ↓&& ↓ && ↓ x^2+7x-30 & = & (x+10)& * & w The area is represented by a quadratic trinomial with a leading coefficient of 1 and it is factorable. Therefore, it can be written as a product of two binomials. x^2+7x-30 = (x+10)(x+q) The number q in the factor x+q, when multiplied by 10, should give - 30. Let's calculate its value. 10 * q = - 30 ⇒ q = - 3 We found that q=-3. We can now say that the width of the land is x-3 meters.
We are told that the width of the land is 15 meters. We also know that the expression x-3 represents the width of the land. Let's use this information to find the valuse of x.
x-3 = 15 ⇒ x = 18
Therefore, x is equal to 18 meters. Let's now write an expression for the area of the lawyer's office.
Now we can find the area by substituting x=18 into our equation for the area of the office and simplifying.
When the width of the land is 15 meters, the area of the lawyer's office is 273 square meters.
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The graph of y=x2−5x−14 is shown.
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We are given the graph of y=x^2-5x-14 and asked to factor the quadratic trinomial. To do so, we first need to remember that the curve crosses the x -axis when y=0. Let's look at those points.
We can see that the function crosses the x -axis at x=-2 and x=7. Since these are the values when y=0, we can look at our solution process as the reverse of using the Zero Product Property to find the zeros of the function.
| Solutions | x= -2 | x = 7 |
|---|---|---|
| Set Equal to Zero | x + 2=0 | x-7=0 |
| Zero Product Property | (x+2)(x-7)=0 | |
Therefore, the quadratic trinomial can be factored by looking at the zeros on the graph. x^2-5x-14 = (x-7)(x+2) Notice that the signs inside the parentheses are the opposite of the solutions because of the Properties of Equality while using the Zero Product Property.
We can also factor the given trinomial by looking for the factors of c that have a sum of b.
x^2+ bx+ c
⇓
x^2 +( - 5)x+( - 14)
For this expression, b= - 5 and c= - 14.
- Since c is negative, we will look for the factors that have opposite signs.
- Since b is negative, the factor with a greater absolute value must be negative.
Considering these conditions, let's lists the possible factor pairs of - 14.
| Factors of - 14 | Sum |
|---|---|
| - 14 and 1 | - 13 |
| - 7 and 2 | - 5 |
From this process, we found that the factor pair - 7 and 2 seems to fit our conditions. The factors of the given quadratic trinomial are, therefore, (x-7) and (x+2). x^2-5x-14 = (x-7)(x+2)
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Write an equation of the form x2+bx+c=0 that has the solutions x=6 and x=-7.
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Recall that the product of two binomials (x-p)(x-q) results in a trinomial of the form x^2+bx+c. (x-p)(x-q) = x^2 + (- p-q)^b x + p q^c Furthermore, according to the Zero Product Property, an equation of the form (x-p)(x-q)=0 holds true if either x-p=0 or x-q=0.
| (x-p)(x-q)=0 | |
|---|---|
| Solution I | Solution II |
| x-p=0 | x-q=0 |
| x=p | x=q |
As we can see, the solutions to the equation (x-p)(x-q) = 0 are x=p and x=q. Therefore, we can substitute the given solutions into this formula and expand the product in order to write the requested equation. Since it does not matter which solution we substitute for the variables, let's say that p=6 and q= - 7.
Finally, to find the equation in the form x^2+bx+c=0, we can use the FOIL method to multiply the binomials.
The equation x^2 +x-42=0 has the solutions x=- 7 and x=6.
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Factoring Quadratic Expressions With a Leading Coefficient of 1
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