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| 10 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Consider the numbers a=2, b=12, and c=10. Find two numbers p and q such that the following conditions are satisfied.
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 3x2+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.
Factors of 3 | Sum of the Factors |
---|---|
-1 and -3 | -1+(-3)=-4 × |
1 and 3 | 1+3=4 ✓ |
LHS/(x+3)=RHS/(x+3)
Cancel out common factors
Simplify quotient
Rearrange equation
Distribute 3
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.
seeits factors. Consider the following expression.
It is known that a=4 and c=3, so ac=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 ac | 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.
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 3x2−5x+2 square inches and that it has a width of x−1 inches.
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 |
Commutative Property of Addition
Factor out 3x
Rewrite 2 as 2⋅1
Factor out -2
Factor out (x−1)
Start by writing a quadratic expression for the area of the tarp. Then, factor the obtained expression.
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.
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 |
Factor out 2w
Split into factors
Factor out 7
Factor out (w−4)
Use the Zero Product Property
(I): LHS+4=RHS+4
(II): LHS−7=RHS−7
(II): LHS/2=RHS/2
The value of h is zero when the ball hits the ground.
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 |
Rearrange equation
Factor out -5t
Rewrite 6 as 2⋅3
Factor out -2
a+(-b)=a−b
Factor out (t−3)
Use the Zero Product Property
(I): LHS+2=RHS+2
(I): LHS/(-5)=RHS/(-5)
(I): Put minus sign in front of fraction
(I): Calculate quotient
(II): LHS+3=RHS+3
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.
LHS−2204=RHS−2204
Rearrange equation
Split into factors
Factor out 4
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 |
LHS/4=RHS/4
Use the Zero Product Property
(I): LHS+4=RHS+4
(II): LHS−44=RHS−44
Substitute the given numbers a, b, and c into the equations of the conditions.
Factors of 6 | Sum of Factors |
---|---|
1 and 6 | 7 |
2 and 3 | 5 |
We will factor the given trinomial. To do so, we first need to check the greatest common factor (GCF) between the terms of the expression. To identify the GCF, we will split each term into its factors.
After factoring out the GCF, we obtained a trinomial with a leading coefficient of 1 in the parentheses. Now let's consider the coefficient of the linear term. x^2+x-6 ⇕ x^2+ 1x+( -6) Next, we will factor this expression into the form (x+ p)(x+ q), where p and q are factor pairs of -6 whose sum is 1. Let's list all factors of -6 and calculate their sums in a table.
Factors of -6 | Sum of Factors |
---|---|
1 and -6 | -5 |
2 and -3 | -1 |
3 and -2 | 1 |
6 and -1 | 5 |
According to the table, 3 and -2 satisfy the conditions. Therefore, we will substitute p= 3 and q= -2 into the product (x+ p)(x+ q).
Finally, we can complete the factorization. Don't forget the 2 that we factored out at the beginning! 2x^2+2x-12 ⇕ 2(x+ 3)(x -2)
We are given a quadratic trinomial in the form am^2+bm+c, where |a|≠1 and there are no common factors between its terms. To factor this expression, we will rewrite the middle term, bm, as a sum of two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 6m^2+ 17m+ 5 In this case, a= 6, b= 17, and c= 5. To rewrite the expression, we need to follow these three steps.
Factors of ac=30 | Sum of Factors |
---|---|
1 and 30 | 31 |
2 and 15 | 17 |
3 and 10 | 13 |
5 and 6 | 11 |
6m^2 + 17m+5 ⇕ 6m^2 + 2m + 15m+5 Now we can factor this expression by grouping.
We will find the factored form of the given trinomial. To do so, we will begin by factoring out -1 from the quadratic expression to make its leading coefficient positive. -4y^2-8y+21 ⇔ - (4y^2+8y-21 ) We now have a quadratic trinomial in the form ay^2+by+c, where |a| ≠ 1 and the greatest common factor of the terms is 1. To factor this expression, we will rewrite the middle term, by, as the sum of two terms. The coefficients of these two terms will be factors of ac whose sum must be b. - ( 4y^2+8y-21 ) ⇔ - ( 4y^2+ 8y+( -21) ) In this expression, a= 4, b= 8, and c= -21. There are now three steps we need to follow to rewrite this expression.
Factors of ac=-84 | Sum of Factors |
---|---|
-1 and 84 | 83 |
-2 and -42 | 40 |
-3 and 28 | 25 |
-4 and 21 | 17 |
-6 and 14 | 8 |
-7 and 12 | 5 |
- (4y^2 + 8y-21 ) ⇕ - ( 4y^2 - 6y + 14y-21 ) Now, we can factor this obtained expression by grouping.
We want to solve the given equation by factoring. Let's write an outline of our plan.
To get things started, notice that the greatest common factor of the terms is 1. That means we do not need to factor anything out of the original equation. We are working with a trinomial. Let's then identify the equation's coefficients. 2k^2+k-15=0 ⇕ 2k^2+ 1k+( -15)=0 Here, a= 2, b= 1, and c= -15. Let's use this information find the value of a c. 2( -15)=-30 Now, let's find the factors of a c by analyzing -30, the product of 2 by -15. Because -30 is a negative value, we need the factors of a c to have different signs — one positive and one negative. a + times a - equals a - Recall that b= 1, which is positive. That means the absolute value of the positive factor will need to be greater than the absolute value of the negative factor.
Factors of ac=- 30 | Sum of Factors |
---|---|
-1 and 30 | - 1+30=29 |
-2 and 15 | - 2+15=13 |
-3 and 10 | - 3+10=7 |
-5 and 6 | - 5+6=1 |
The factor pair of 6 and ¬5. We can now rewrite bk as the sum of two terms. 2k^2+ 1k-15=0 ⇕ 2k^2 - 5k + 6k-15=0 Next, we will factor the expression on the left-hand side of the equation by grouping.
Finally, we will apply the Zero Product Property to find the solutions to the given equation.
The solutions to the equation are k= 52 and k=-3.