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| | 10 Theory slides |
| | 10 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.
A quadratic trinomial in the form x^2+bx+c can be factored as (x+p)(x+q) if there exist p and q such that p+q=b and pq=c.
x^2+bx+c = (x+p)(x+q)
Substitute expressions
Distribute x
Factor out x
Factor out q
Factor out x+p
Commutative Property of Multiplication
For some p and q, the aim is to write the given trinomial as follows. x^2+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. x^2+ 7x+ 12 Here, b= 7 and c= 12, so both b and c are positive.
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.
For the given trinomial, two integers with a sum of 7 and a product of 12 were found. x^2+7x+12 ⇒ lp=3 q=4 Therefore, the trinomial can be written as the product of the binomials x+3 and x+4. x^2+7x+12 = (x+3)(x+4)
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+13x+42 and its length as x+7. Help Tearrik answer the following questions.
A & = l * w & ⇓ x^2 + 13x+42 & = (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+ 13x+ 42 Here, b= 13 and c= 42, so both b and c are 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+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.
w= 9, l= 10
Add terms
Multiply
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+3x-10 is twice the area of the triangle and that the length of BC is greater than the length of AB.
x^2+3x-10 = 2 * AB * BC/2 To find AB and BC, the quadratic trinomial needs to be written as a product of two binomials. x^2+3x-10= (x+p)(x+q) Now identify b and c to get an idea about the signs of p and q. x^2+3x-10 ⇓ x^2+ 3x+( - 10) For this expression, b= 3 and c= - 10, so b is positive but c is negative.
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+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-2 & x=10 8 x+5 & 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.
P & = (x-2) + (x+5) + (x+7) & = 3x + 10
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.
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)
Start by identifying the values of b and c for the given quadratic trinomial.
First, b and c will be identified. x^2-9x-90 ⇓ x^2+( - 9)x+( - 90) Here, b= - 9 and c= - 90, meaning that both b and c are 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^2-9x-90 = (x-15)(x+6)
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.
Davontay needs to write a quadratic trinomial in the form x^2+bx+c. It is also given that b is negative and c=60. x^2+bx+c ⇓ x^2 +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. x^2+bx+60= (x+p)(x+q) Now, two facts can be inferred from this trinomial.
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^2-16x+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. x^2-16x+60 = (x-6)(x-10)
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.
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 12 250 square centimeters.
Width: 70 centimeters
Distribute w
LHS-12 250=RHS-12 250
Rearrange equation
| Factors of - 12 250 | 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 |
Use the Zero Product Property
(I): LHS+70=RHS+70
(II): LHS-175=RHS-175
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.
4 and 7
q= 11-p
Distribute p
LHS-28=RHS-28
LHS * (- 1)=RHS* (- 1)
Commutative Property of Addition
| 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 -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.
A map of the USA is displayed on a whiteboard. The area of the map, in square centimeters, is represented by x^2+3x-54. Its length is represented by (x+9).
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.
A lawyer's office and garden are on a rectangular piece of land. The area of the land, in square meters, is represented by x^2+7x-30.
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.
The graph of y=x^2-5x-14 is shown.
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.
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)
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.