1. HW: Pg. 267 #47-67o, 69, 70

5. 5.2 Solving Quadratic Equations by Finding Square Roots

6. 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).

7. 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.

8. GUIDED PRACTICE
for Example 1
Factor the expression. If the expression cannot be factored, say so.
1. x2 – 3x – 18
2. n2 – 3n + 9
3. r2 + 2r – 63
ANSWER
ANSWER
ANSWER
(r + 9)(r –7)
(x – 6)(x + 3)
cannot be factored

9. Special Factoring Patterns
PATTERN NAME PATTERN EXAMPLE
Difference of Two Squares
(FL pattern)
Perfect Square Trinomial
(smiley face pattern)

10. 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

11. GUIDED PRACTICE
for Example 2
Factor the expression.
4. x2 – 9
7. w2 – 18w + 81
ANSWER
(x – 3)(x + 3)
ANSWER
(w – 9)2
5. q2 – 100
ANSWER
(q – 10)(q + 10)
6. y2 + 16y + 64
ANSWER
(y + 8)2

12. Multiply using FOIL. Write in standard form. Factor.
EXAMPLE 3
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.
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.
Reject the negative value, –1200. The field’s length and width should each be increased by 200 meters. The new dimensions are 800 meters by 600 meters.

13. EXAMPLE 4
Factor ax2 + bx + c where c > 0
Factor 5x2 – 17x + 6.
SOLUTION
You want 5x2 – 17x + 6 = (kx + m)(lx + n) where k and l are factors of 5 and m and n are factors of 6. You can assume that k and l are positive and k ≥ l. Because mn > 0, m and n have the same sign. So, m and n must both be negative because the coefficient of x, –17, is negative.
ANSWER
The correct factorization is 5x2 –17x + 6 = (5x – 2)(x – 3).

14. EXAMPLE 5
Factor ax2 + bx + c where c < 0
Factor 3x2 + 20x – 7.
SOLUTION
You want 3x2 + 20x – 7 = (kx + m)(lx + n) where k and l are factors of 3 and m and n are factors of –7. Because mn < 0, m and n have opposite signs.
ANSWER
The correct factorization is 3x2 + 20x – 7 = (3x – 1)(x + 7).

15. GUIDED PRACTICE
GUIDED PRACTICE
for Examples 4 and 5
Factor the expression. If the expression cannot be factored, say so.
x2 + 5x – 12
x2 – 20x – 3
ANSWER
(7x + 1)(x – 3)
ANSWER
(3x – 4)(x + 3)
u2 + 12u + 5
z2 + 16z + 3
ANSWER
(5z + 1)(z + 3).
ANSWER
(2u + 1)(2u + 5)
w2 + w + 3
x2 – 9x + 2
ANSWER
cannot be factored
ANSWER
(4x – 1)(x – 2)

16. Difference of two squares
EXAMPLE 6
Factor with special patterns
Factor the expression.
a. 9x2 – 64
= (3x)2 – 82
Difference of two squares
= (3x + 8)(3x – 8)
b. 4y2 + 20y + 25
= (2y)2 + 2(2y)(5) + 52
Perfect square trinomial
= (2y + 5)2
c w2 – 12w + 1
= (6w)2 – 2(6w)(1) + (1)2
Perfect square trinomial
= (6w – 1)2

17. GUIDED PRACTICE
GUIDED PRACTICE
for Example 6
Factor the expression.
x2 – 1
z2 + 4z + 9
ANSWER
(4x + 1)(4x – 1)
ANSWER
(7z + 3)2
y2 + 12y + 4
n2 – 9 = (3y)2
ANSWER
(3y + 2)2
ANSWER
(6n – 3)(6n +3)
r2 – 28r + 49
ANSWER
(2r – 7)2
s2 – 80s + 64
ANSWER
(5s – 8)2

18. EXAMPLE 7
Factor out monomials first
Factor the expression.
a. 5x2 – 45
= 5(x2 – 9)
= 5(x + 3)(x – 3)
b. 6q2 – 14q + 8
= 2(3q2 – 7q + 4)
= 2(3q – 4)(q – 1)
c. –5z2 + 20z
= –5z(z – 4)
d. 12p2 – 21p + 3
= 3(4p2 – 7p + 1)

19. GUIDED PRACTICE
GUIDED PRACTICE
for Example 7
Factor the expression.
s2 – 24
z2 + 33z + 36
ANSWER
3(s2 – 8)
ANSWER
3(2z + 3)(z + 4)
t2 + 38t – 10
ANSWER
2(4t – 1) (t + 5)
x2 + 24x + 15
ANSWER
3(2x2 + 8x + 5)
x2 – 28x – 24
ANSWER
4(3x + 2)(x – 3)
–16n2 + 12n
ANSWER
–4n(4n – 3)

20. Write original equation.
EXAMPLE 8
Solve quadratic equations
Solve (a) 3x2 + 10x – 8 = 0 and (b) 5p2 – 16p + 15 = 4p – 5.
a. 3x2 + 10x – 8 = 0
Write original equation.
(3x – 2)(x + 4) = 0
Factor.
3x – 2 = 0
or x + 4 = 0
Zero product property
or x = –4
x = 23
Solve for x.
b. 5p2 – 16p + 15 = 4p – 5.
Write original equation.
5p2 – 20p + 20 = 0
Write in standard form.
p2 – 4p + 4 = 0
Divide each side by 5.
(p – 2)2 = 0
Factor.
p – 2 = 0
Zero product property
p = 2
Solve for p.

21. GUIDED PRACTICE
GUIDED PRACTICE
for Example 8
Solve the equation.
x2 – 3x – 63 = 0
or –3 3 12
ANSWER
x2 + 7x + 2 = x +8
ANSWER
no solution
x2 + 70x = 0
ANSWER –5

22. Pg. 260 #23-91 eoo and prepare for quiz on 5.3 & 5.2 next class
Homework:
Pg. 260 #23-91 eoo
and
prepare for quiz on 5.3 & 5.2 next class