1. Lesson 4.3, For use with pages 252-258
Find the product
1. (m – 8) (m – 9)
ANSWER
m2 – 17m + 72
2. (z + 6) (z – 10)
ANSWER
z2 – 4z – 60

2. 3. (y +20) (y – 20) 4. (d +9)2 Lesson 4.3, For use with pages 252-258
Find the product
3. (y +20) (y – 20)
ANSWER
y2 – 400
4. (d +9)2
ANSWER
d2 + 18d +81

3. 6. A car travels at an average speed of (m + 17) miles
Lesson 4.3, For use with pages
Find the product
5. (x – 14)2
ANSWER
x2– 28x +196
6. A car travels at an average speed of (m + 17) miles
per hour for (m + 2) hours. What distance does it
travel?
ANSWER
(m2 + 9m + 14) mi

4. 4.3 Factoring Trinomials

5. Squares 1-20, and 25 Write the squares 1-20, 25 1 2 = 1 2 2 = 4
1 2 = 1
= 4
= 400
= 625

6. Factoring Quadratic Rules
Given
Factored
a 𝑥 2 + bx + c
a 𝑥 bx + c
a 𝑥 bx - c
a 𝑥 bx - c
( ) ( )
( ) ( )
( - Big) ( + Small )
( + Big) ( Small )

7. Factoring Procedures for a Quadratic Equation y = a 𝑥 2 +𝑏𝑥+𝑐
“No Fuse Method”
The trinomial should be set = to y or 0, before beginning to factor.
Examine each term to determine if there is a GCF that will factor from each term (Note: the GCF must factor from all three terms). If there is a GCF that will factor from all three terms, factor each term.
Look at the Given equations and determine the signs for the factored binomial.
If the first term in the trinomial is 𝑥 2 +4x+3, split the 𝑥 2 to x and x and put each x in the first term in each binomial. EX: ( x + ) ( x + ).
Next multiply the outside term a and c together ( 1 x 3 = 3) and determine the factors of 3. The factors of 3 are 1 and 3, determine if you should add or subtract them to obtain the middle term of 4. EX: 1+3 = 4. Put the factors in the last terms of each binomial, the order does not matter because the signs on the factored binomials are both positive (x + 3) ( x + 1). If the factored binomial had been a big and small factor, then the 3 would have been the big factor and the 1 would have been the small factor.
Foil the factored binomial to determine if the quadratic equation was factored properly

8. EXAMPLE 1
Factor trinomials of the form x2 + bx + c
Factor the expression.
a. x2 – 9x + 20
b. x2 + 3x – 12
SOLUTION
a. You want x2 – 9x + 20 = (x + m) (x + n) where mn = 20 and m + n = – 9.
ANSWER
Notice that m = – 4 and n = – 5.
So, x2 – 9x + 20 = (x – 4)(x – 5).

9. EXAMPLE 1
Factor trinomials of the form x2 + bx + c
b. You want x2 + 3x – 12 = (x + m) (x + n) where mn = – 12 and m + n = 3.
ANSWER
Notice that there are no factors m and n such that m + n = 3. So, x2 + 3x – 12 cannot be factored.

10. GUIDED PRACTICE
for Example 1
Factor the expression. If the expression cannot be factored, say so.
1. x2 – 3x – 18
SOLUTION
You want x2 – 3x – 18 = (x + m) (x + n) where mn = – 18 and m + n = – 3.
Factor of – 18 : m, n
1, – 18
–1, 18
– 3, 6
– 2, 9
2, – 9
– 6, 3
Sum of factors: m + n
– 17 17 3 7
– 7
– 3

11. GUIDED PRACTICE
for Example 1
ANSWER
Notice m = – 6 and n = 3 so x2 – 3x – 18 = (x – 6) (x + 3)

12. Factor of 9 : m, n 1, 9 – 1, – 9 3, 3 – 3, – 3 Sum of factors: m + n
GUIDED PRACTICE
for Example 1
2. n2 – 3n + 9
SOLUTION
You want n2 – 3n + 9 = (x + m) (x + n) where mn = 9 and m + n = – 3.
Factor of : m, n
1, 9
– 1, – 9
3, 3
– 3, – 3
Sum of factors: m + n 10
–10 6
– 6

13. GUIDED PRACTICE
for Example 1
ANSWER
Notice that there are no factors m and n such that m + n = – 3 . So, n2 – 3x + 9 cannot be factored.

14. GUIDED PRACTICE
for Example 1
3. r2 + 2r – 63
SOLUTION
You want r2 + 2r – 63 = (x + m) (x + n) where mn = – 63 and m + n = 2.
Factor of – 63 : m, n
– 1, 63
1, – 63
21, – 3
– 21, – 3
9, – 7
Sum of factors: m + n
– 17 11 7 2

15. GUIDED PRACTICE
for Example 1
ANSWER
Notice that m = 9 and n = – 7 . So, r2 + 2r – 63 = (r + 9)(r –7)

16. Difference of two squares
EXAMPLE 2
Factor with special patterns
Factor the expression.
a. x2 – 49
= x2 – 72
Difference of two squares
= (x + 7) (x – 7)
b. d d + 36
= d 2 + 2(d)(6) + 62
Perfect square trinomial
= (d + 6)2
c. z2 – 26z + 169
= z2 – 2(z) (13) + 132
Perfect square trinomial
= (z – 13)2

17. Difference of two squares
GUIDED PRACTICE
for Example 2
Factor the expression.
4. x2 – 9
= x2 – 32
Difference of two squares
= (x – 3) (x + 3)
5. q2 – 100
= q2 – 102
Difference of two squares
= (q – 10) (q + 10)
6. y2 + 16y + 64
= y2 + 2(y)
Perfect square trinomial
= (y + 8)2

18. Perfect square trinomial
GUIDED PRACTICE
for Example 2
7. w2 – 18w + 81
= w2 – 2(w) + 92
Perfect square trinomial
= (w – 9)2

19. Write original equation.
EXAMPLE 3
Standardized Test Practice
SOLUTION
x2 – 5x – 36 = 0
Write original equation.
(x – 9)(x + 4) = 0
Factor.
x – 9 = 0 or
x + 4 = 0
Zero product property
x = 9 or
x = – 4
Solve for x.
ANSWER
The correct answer is C.

20. EXAMPLE 4
Use a quadratic equation as a model
Nature Preserve
A town has a nature preserve with a rectangular field that measures 600 meters by 400 meters. The town wants to double the area of the field by adding land as shown. Find the new dimensions of the field.

21. Multiply using FOIL. Write in standard form. Factor.
EXAMPLE 4
Use a quadratic equation as a model
SOLUTION
480,000 = 240, x + x2
Multiply using FOIL.
0 = x x – 240,000
Write in standard form.
0 = (x – 200) (x )
Factor.
x – 200 = 0
x = 0 or
Zero product property
x = 200 or
x = – 1200
Solve for x.

22. EXAMPLE 4
Use a quadratic equation as a model
ANSWER
Reject the negative value, – The field’s length and width should each be increased by 200 meters. The new dimensions are 800 meters by 600 meters.

23. Write original equation.
GUIDED PRACTICE
for Examples 3 and 4
8. Solve the equation x2 – x – 42 = 0.
SOLUTION
x2 – x – 42 = 0
Write original equation.
(x + 6)(x – 7) = 0
Factor.
x + 6 = or
x – 7 = 0
Zero product property
x = – 6 or
x = 7
Solve for x.

24. Multiply using FOIL. Write in standard form. Factor.
GUIDED PRACTICE
for Examples 3 and 4
9. What If ? In Example 4, suppose the field initially measures 1000 meters by 300 meters. Find the new dimensions of the field.
SOLUTION
New Area
New Length (meters)
New width (meters) =
2(1000)(300) =
( x)
(300 + x) =
x + 300x + x2
Multiply using FOIL.
0 = x x –
Write in standard form.
0 = (x – 200) (x )
Factor.
x – 200 = 0
x = 0 or
Zero product property

25. Solve for x. GUIDED PRACTICE for Examples 3 and 4 x = 200 or
ANSWER
Reject the negative value, – The field’s length and width should each be increased by 200 meters. The new dimensions are 1200 meters by 500 meters.

26. Write original function.
EXAMPLE 5
Find the zeros of quadratic functions.
Find the zeros of the function by rewriting the function in intercept form.
a. y = x2 – x – 12
b. y = x2 + 12x + 36
SOLUTION
a. y = x2 – x – 12
Write original function.
= (x + 3) (x – 4)
Factor.
The zeros of the function are –3 and 4.
Check Graph y = x2 – x – 12. The graph passes through (–3, 0) and (4, 0).

27. Write original function.
EXAMPLE 5
Find the zeros of quadratic functions.
Find the zeros of the function by rewriting the function in intercept form.
a. y = x2 – x – 12
b. y = x2 + 12x + 36
SOLUTION
b. y = x2 + 12x + 36
Write original function.
= (x + 6) (x + 6)
Factor.
The zeros of the function is – 6
Check Graph y = x2 + 12x The graph passes through ( – 6, 0).

28. Write original function.
GUIDED PRACTICE
GUIDED PRACTICE
for Example
for Example 5
Find the zeros of the function by rewriting the function in intercept form.
y = x2 + 5x – 14
SOLUTION
y = x2 + 5x – 14
Write original function.
= (x + 7) (x – 2)
Factor.
The zeros of the function is – 7 and 2
Check Graph y = x2 + 5x – 14. The graph passes through ( – 7, 0) and (2, 0).

29. Write original function.
GUIDED PRACTICE
GUIDED PRACTICE
for Example
for Example 5
y = x2 – 7x – 30
SOLUTION
y = x2 – 7x – 30
Write original function.
= (x + 3) (x – 10)
Factor.
The zeros of the function is – 3 and 10
Check Graph y = x2 – 7x – 30. The graph passes through ( – 3, 0) and (10, 0).

30. Write original function.
GUIDED PRACTICE
GUIDED PRACTICE
for Example
for Example 5
12. f(x) = x2 – 10x + 25
SOLUTION
f(x) = x2 – 10x + 25
Write original function.
= (x – 5) (x – 5)
Factor.
The zeros of the function is 5
Check Graph f(x) = x2 – 10x The graph passes through ( 5, 0).