6. Factoring Any Quadratic Expression
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Chapter 7
6.
Factoring Any Quadratic Expression
This lesson delves into the techniques for breaking down quadratic expressions into simpler forms. It emphasizes the role of the greatest common factor (GCF) and the Zero Product Property in this process. Understanding these methods can be incredibly useful in various scenarios, such as calculating areas or determining the dimensions of objects. For instance, if you're planning a garden layout or trying to figure out the dimensions of a room, these techniques can help you find the most efficient arrangement. The lesson aims to make these mathematical concepts accessible and practical, helping you make informed decisions in both academic and everyday settings.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Factoring Any Quadratic Expression
Slide of 10
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.
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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.
cc Condition1 & Condition2 [1em] pq=b/a & p+q=c/a
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Discussion
Factoring a Quadratic Trinomial
If the greatest common factor (GCF) of a quadratic trinomial of the form ax^2+bx+c is 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&=a* b_1, for some numberb_1 c&=a* c_1, for some numberc_1 With this information in mind, the expression ax^2+bx+c can be factored.
ax^2+bx+c
SubstituteII
b= a* b_1, c= a* c_1
ax^2+ a* b_1x+ a* c_1
FactorOut
Factor out a
a(x^2+b_1x+c_1 )
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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 3x^2+12x+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.
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b What will be the dimensions of the tent when x=1 foot?
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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= w l ⇓ 3x^2+12x+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.
3x^2+12x+9
SplitIntoFactors
Split into factors
3 x^2+ 3(4) x+ 3(3)
FactorOut
Factor out 3
3 (x^2+ 4x+ 3)
The simplified quadratic trinomial in parentheses now has a leading coefficient of 1. Therefore, it can be factored accordingly. The expression x^2+ 4x+ 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 * |
| 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+4x+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).
3(x+1)(x+3)/x+3=(x+3)l/x+3
CancelCommonFac
Cancel out common factors
3(x+1)(x+3)/x+3=(x+3)l/x+3
SimpQuot
Simplify quotient
3(x+1)=l
RearrangeEqn
Rearrange equation
l=3(x+1)
Distr
Distribute 3
l=3x+3
The length of the tent's base is 3x+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=3x+3 | l=3 * 1+3 | l=6 feet |
The class will buy tents that are 4 feet wide and 6 feet long.
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Discussion
Factoring a Quadratic Trinomial
When trying to factor a quadratic trinomial of the form ax^2+ bx+ c, it can be difficult to see
its factors. Consider the following expression.
8x^2+ 26x+ 6
Here, a= 8, b= 26, and c= 6. There are six steps to factor this trinomial.
1
Factor Out the GCF of a, b, and c
expand_more 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 = & 2* 2* 2 26 = & 2* 13 6 = & 2* 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.
In the remaining steps, the factored coefficient 2 before the parentheses can be ignored. The new considered quadratic trinomial is 4x^2+ 13x+ 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.
2
Find the Factor Pair of ac Whose Sum Is b
expand_more 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 * |
| 3 and 4 | 3+4=7 * |
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 ac and b.
| ac | 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.
3
Write bx as a Sum
expand_more 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.
The linear term 13x can be rewritten in the original expression as 12x+x. 4x^2+ 13x+3 ⇕ 4x^2+ 12x+x+3
4
Factor Out the GCF of the First Two Terms
expand_more 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. 4x^2+12x+ x+ 3
The first two terms, 4x^2 and 12x, can be factored.
4x^2 = & 2* 2* x* x 12x = & 2* 2* 3* x
The GCF of 4x^2 and 12x is 2* 2*x=4x.
4x^2+12x+x+3
Rewrite
Rewrite 4x^2 as 4x* x
4x* x+12x+x+3
Rewrite
Rewrite 12x as 4x* 3
4x* x+4x* 3+x+3
FactorOut
Factor out 4x
4x(x+3)+x+3
5
Factor Out the GCF of the Last Two Terms
expand_more 6
Factor Out the Common Factor
expand_more If all the previous steps have been performed correctly, there should now be two terms with a common factor.
4x (x+3)+1 (x+3)
The common factor will be factored out.
The factored form of 4x^2+13x+3 is (4x+1)(x+3). Remember that the original trinomial was 8x^2+26x+6 and that the GCF 2 was factored out in Step 1. This GCF has to be included in the final result. 8x^2+26x+6 = 2(4x+1)(x+3)
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Example
Factoring ax^2+bx+c When ac 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 3x^2-5x+2 square inches and that it has a width of x-1 inches.
a Write a binomial that represents the length of the pamphlet.
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b Find the perimeter of the pamphlet if the width is 4 inches.
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Solution
A= w l ⇓ 3x^2-5x+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. 3x^2-5x+2 ⇕ 3x^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 ac=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. 3x^2 - 5x+2 ⇕ 3x^2 - 2x - 3x+2 Next, the expression will be factored by grouping its terms.
3x^2-2x-3x+2
CommutativePropAdd
Commutative Property of Addition
3x^2-3x-2x+2
3x* x-3x* 1-2x+2
FactorOut
Factor out 3x
3x(x-1)-2x+2
Rewrite
Rewrite 2 as 2 * 1
3x(x-1)-2x+2* 1
FactorOut
Factor out - 2
3x(x-1)-2(x-1)
FactorOut
Factor out (x-1)
(3x-2)(x-1)
Finally, the area can be written as the product of the binomials x-1 and 3x-2. 3x^2-5x+2=(x-1)(3x-2) Since x-1 represents the width of the pamphlet, the binomial 3x-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 ⇔ 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.
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.
The perimeter of the pamphlet is 34 inches.
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Example
Factoring ax^2+bx+c When ac 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.
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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 2w-1.
Since the base is a rectangle, its area is equal to the product of its width and length.
Now, the obtained quadratic expression will be factored. First, its coefficients will be identified. 2w^2-w-28=0 ⇕ 2w^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 ac=- 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. 2w^2 - w-28=0 ⇓ 2w^2 - 8w+7w-28=0 Then, the obtained expression will be factored by grouping.
2w^2-8w+7w-28=0
2w * w-2w* 4+7w-28=0
FactorOut
Factor out 2w
2w(w-4)+7w-28=0
SplitIntoFactors
Split into factors
2w(w-4)+7w-7* 4=0
FactorOut
Factor out 7
2w(w-4)+7(w-4)=0
FactorOut
Factor out (w-4)
(2w+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)(2w+7)=0
▼
Solve using the Zero Product Property
ZeroProdProp
Use the Zero Product Property
lcw-4=0 & (I) 2w+7=0 & (II)
AddEqn
(I): LHS+4=RHS+4
lw=4 2w+7=0
SubEqn
(II): LHS-7=RHS-7
lw=4 2w=-7
DivEqn
(II): .LHS /2.=.RHS /2.
lw=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.
The base of the tarp has a length and a width of 7 and 4 feet, respectively. rc Width:& 4feet Length:& 7 feet
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Example
Factoring ax^2+bx+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 45^(∘) angle.
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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)=-5t^2+13t+6 ⇓ 0=-5t^2+13t+6
Now, the trinomial will be factored. First, its coefficients will be identified.
-5t^2+ 13t+ 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 ac=- 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=-5t^2+ 13t+6 ⇓ 0=-5t^2+ 15t-2t+6 Now, the obtained expression will be factored by grouping.
0=-5t^2+15t-2t+6
RearrangeEqn
Rearrange equation
-5t^2+15t-2t+6=0
-5t * t +5t* 3-2t+6=0
FactorOut
Factor out -5t
-5t(t-3)-2t+6=0
Rewrite
Rewrite 6 as 2 * 3
-5t(t-3)-2t+2 * 3=0
FactorOut
Factor out -2
-5t(t-3)+(-2)(t-3)=0
AddNeg
a+(- b)=a-b
-5t(t-3)-2(t-3)=0
FactorOut
Factor out (t-3)
(-5t-2)(t-3)=0
The factoring process is finally done. Now, the Zero Product Property can be used to find the t-values.
(-5t-2)(t-3)=0
▼
Solve using the Zero Product Property
ZeroProdProp
Use the Zero Product Property
lc-5t-2=0 & (I) t-3=0 & (II)
AddEqn
(I): LHS+2=RHS+2
l-5t=2 t-3=0
DivEqn
(I): .LHS /(- 5).=.RHS /(- 5).
lt= 2- 5 t-3=0
MoveNegDenomToFrac
(I): Put minus sign in front of fraction
lt=- 25 t-3=0
CalcQuot
(I): Calculate quotient
lt=- 0.4 t-3=0
AddEqn
(II): LHS+3=RHS+3
lt=- 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.
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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.
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b The total area for the campground and the jogging track altogether is 2204 square meters. What is the width of the jogging track?
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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:& x + 30 + x ⇔ 2x+30 Length:& x + 50 + x ⇔ 2x+50 Since the given figure is a rectangle, its area is the product of its width and length. A=w l ⇓ A=(2x+30)(2x+50) The product will be rewritten as a trinomial.
A=(2x+30)(2x+50)
A=4x^2+160x+1500
The trinomial that represents the total area is 4x^2+160x+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= 4x^2+160x+1500 The equation will be rewritten to make the right-hand side equal to zero.
2204=4x^2+160x+1500
SubEqn
LHS-2204=RHS-2204
0=4x^2+160x-704
RearrangeEqn
Rearrange equation
4x^2+160x-704=0
Next, the Greatest Common Factor (GCF) will be factored out.
4x^2+160x-704=0
SplitIntoFactors
Split into factors
4 * x^2 + 4 * 40x- 4 * 176=0
FactorOut
Factor out 4
4(x^2+40x-176)=0
Now that the GCF has been factored out, the quadratic trinomial in parentheses will be factored. The expression x^2+40x-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. 4x^2+160x-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
lcx-4=0 & (I) x+44=0 & (II)
AddEqn
(I): LHS+4=RHS+4
lx=4 x+44=0
SubEqn
(II): LHS-44=RHS-44
lx=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.
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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. cc Condition1 & Condition2 [1em] pq=c/a & p+q=b/a If p is less than q, find these two values.
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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.
cc
Condition1 & Condition2 [1em] p* q=12/2 & p+ q=10/2 ⇓ & ⇓ p* q=6 & 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. ax^2+ bx+ c ⇓ 2x^2+ 10x+ 12 First, the Greatest Common Factor of all therms, 2, must be factored out. 2x^2+ 10x+ 12 ⇕ 2(x^2+5x+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. 2x^2+ 10x+ 12 ⇕ 2(x+ 2)(x+ 3)
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Factoring Any Quadratic Expression
Exercise 2.1
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Note that the shape of the mirror is a rectangle. This means we can substitute the given expressions for the area and the width into the formula for the area of a rectangle to write an equation. A= l w ⇓ 4x^2+5x-6= l( x+2) We can rewrite the trinomial on the left-hand side of the equation so that the width is one of the factors and the other will be the expression for the length. Then we will factor by grouping. Let's determine the values of a, b, and c of the trinomial. 4x^2+5x-6 ⇕ 4x^2+ 5x+( - 6) Here, a= 4, b= 5, and c= -6. With this in mind, we can make the following conclusions.
- The Greatest Common Factor (GCF) of a, b, and c is 1. Therefore, we will not factor a GCF out.
- Since a c=-24, which is less than zero, the factors of a c must have different signs — one positive and one negative.
- Since b= 5, which is positive, the absolute value of the positive factor of a c will be greater than the absolute value of the negative factor.
Using this information, let's now list the factor pairs of -24 and look for the sum of 5.
| Factors of ac=-24 | Sum of Factors |
|---|---|
| 24 and -1 | 24+(- 1)=23 |
| 12 and -2 | 12+(- 2)=10 |
| 8 and -3 | 8+(- 3)=5 |
| 6 and -4 | 6+(- 4)=2 |
Now that we know which factor pair to use, we can rewrite bx as the sum of two terms by using the factors 8 and -3. 4x^2 + 5x-6 ⇕ 4x^2 - 3x + 8x-6 Next, we will factor the new expression by grouping its terms.
Now we have factored the given trinomial. Note that one of the factors, x+2, is the given expression for the width of the mirror. This means that the expression 4x-3 represents the length! Using this information, we can finish writing the equation for the area of the mirror. 4x^2+5x-6= l( x+2) ⇕ 4x^2+5x-6=( 4x-3)( x+2)
One way to find the area of the mirror when x= 2 feet is to substitute 2 into the given expression for the area 4x^2+5x-6 and then simplify. Let's do it.
The area of the mirror is 20 square feet when x= 2 feet. Notice that we can also find the same area by substituting x= 2 into the product of the factors x+2 and 4x-3.
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We are told that the length l of the signboard is two feet less than three times its width w. Using this information, we can write an expression for l. l= 3w-2 The shape of the signboard is a rectangle. This means its area is given by the product of its length l and its width w. A= l w Additionally, we are given that the area of the signboard is 21. Therefore, we can substitute 21 for A and 3w-2 for l in the formula for the area to get an equation for the area of the signboard. A= l w Substitute 21=( 3w-2) w We will now rewrite the this equation in the form aw^2+bw+c=0. This will help us find the value of w.
The trinomial on the left-hand side of the equation can now be factored. To do so, let's first identify its coefficients. 3w^2-2w-21=0 ⇕ 3w^2+( -2)w+( - 21) =0 Here, a= 3, b= -2, and c= -21. With this in mind, we can make the following conclusions.
- The Greatest Common Factor (GCF) of a, b, and c is 1. Therefore, we will not factor a GCF out.
- Since a c=-63, which is less than zero, its factors must have different signs — one positive and one negative.
- Because b= -2, which is negative, the absolute value of the negative factor will be greater than the absolute value of the positive factor.
We will list the factor pairs of -63 and look a pair for a sum of -2.
| Factors of ac=-63 | Sum of Factors |
|---|---|
| -63 and 1 | (- 63)+1=-62 |
| -21 and 3 | (- 21)+3=-18 |
| -9 and 7 | (- 9)+7=-2 |
Now that we know which factor pair to use, we can rewrite bw as the sum of two terms by using the factors -9 and 7. 3w^2 - 2w-21=0 ⇕ 3w^2 - 9w + 7w-21=0 Next, we will factor the expression by grouping its terms.
Notice that the product of these two factor expressions is zero. This means that we can use the Zero Product Property to identify the solutions to the equation.
We will use the positive solution since a negative width does not make sense. Therefore, the width of the signboard is 3 feet.
Consider the equation for l that we wrote in Part A. l=3w-2 We also found that the width of the signboard is 3 feet. We will substitute this value for w in the equation and simplify to find the value of l.
The length of the signboard is 7 feet.
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A diver's height above the water h is modeled by h=-10t^2+5t+15, where t is the time in seconds after beginning their dive. How long will it take the diver to reach the water?
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Let's start this exercise by assuming that the water is at ground level h=0. Then we can factor the given equation to solve for the time it takes to hit the water. h=-10t^2+5t+15 ⇓ 0 =-10t^2+5t+15 The terms of the quadratic equation have as a common factor, -5. Therefore, we will rewrite this equation by factoring that out.
The obtained trinomial on the left-hand side of the equation needs to be factored to identify the solutions of the equation. To do so, let's first identify the coefficients of its terms. 2t^2-t-3=0 ⇕ 2t^2+( -1)t+( - 3) =0 Here, a= 2, b= -1, and c= -3. With this in mind, we can make the following conclusions.
- Since a c=-6, which is less than zero, its factors must have different signs — one positive and one negative.
- Since b= -1, which is negative, the absolute value of the negative factor will be greater than the absolute value of the positive factor.
Using this information, let's list the factor pairs of -6 and look for the pair with a sum of -1.
| Factors of ac=-6 | Sum of Factors |
|---|---|
| -6 and 1 | (- 6)+1=-5 |
| -3 and 2 | (- 3)+2=-1 |
Now that we know which factor pair to use, we can rewrite bt as the sum of two terms by using the factors -3 and 2. 2t^2 - 1t-3=0 ⇕ 2t^2 + 2t - 3t -3=0 Next, we will factor the expression by grouping its terms.
Notice that the product of the factor expressions is zero. This means that we can use the Zero Product Property to set each factor equal to 0 and solve for t.
The negative solution is not relevant in this case because time cannot be negative. Therefore, the diver will reach the water 1.5 seconds after the they start the dive.
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Factoring Any Quadratic Expression
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