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4.4 Solve Quadratic Equations of the form b y Factoring. EXAMPLE 1. Factor ax 2 + bx + c where c > 0. Factor 5 x 2 – 17 x + 6. SOLUTION. - PowerPoint PPT Presentation
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4.4 Solve Quadratic Equations of the form
by Factoring
EXAMPLE 1 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.
EXAMPLE 1 Factor ax2 + bx + c where c > 0
ANSWER
The correct factorization is 5x2 –17x + 6 = (5x – 2)(x – 3).
EXAMPLE 2 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).
GUIDED PRACTICE for Examples 1 and 2GUIDED PRACTICE
Factor the expression. If the expression cannot be factored, say so.
1. 7x2 – 20x – 3
(7x + 1)(x – 3)
2. 5z2 + 16z + 3
3. 2w2 + w + 3
cannot be factored
ANSWER (5z + 1)(z + 3).
ANSWER
ANSWER
GUIDED PRACTICE for Examples 1 and 2GUIDED PRACTICE
5. 4u2 + 12u + 5
(2u + 1)(2u + 5)
6. 4x2 – 9x + 2
(4x – 1)(x – 2)
4. 3x2 + 5x – 12
(3x – 4)(x + 3)ANSWER
ANSWER
ANSWER
EXAMPLE 3 Factor with special patterns
Factor the expression.a. 9x2 – 64
= (3x + 8)(3x – 8)Difference of two squares
b. 4y2 + 20y + 25
= (2y + 5)2
Perfect square trinomial
c. 36w2 – 12w + 1= (6w – 1)2
Perfect square trinomial
= (3x)2 – 82
= (2y)2 + 2(2y)(5) + 52
= (6w)2 – 2(6w)(1) + (1)2
GUIDED PRACTICEGUIDED PRACTICE for Example 3
Factor the expression.7. 16x2 – 1
(4x + 1)(4x – 1)
8. 9y2 + 12y + 4
(3y + 2)2
9. 4r2 – 28r + 49(2r – 7)2
10. 25s2 – 80s + 64
(5s – 8)2ANSWER
ANSWER
ANSWER
ANSWER
GUIDED PRACTICEGUIDED PRACTICE for Example 3
11. 49z2 + 4z + 9
(7z + 3)2
12. 36n2 – 9 = (3y)2
(6n – 3)(6n +3)
ANSWER
ANSWER
EXAMPLE 4 Factor out monomials first
Factor the expression.a. 5x2 – 45
= 5(x + 3)(x – 3)
b. 6q2 – 14q + 8
= 2(3q – 4)(q – 1)
c. –5z2 + 20z
d. 12p2 – 21p + 3
= 5(x2 – 9)
= 2(3q2 – 7q + 4)
= –5z(z – 4)
= 3(4p2 – 7p + 1)
GUIDED PRACTICEGUIDED PRACTICE for Example 4
Factor the expression.13. 3s2 – 24
14. 8t2 + 38t – 10
2(4t – 1) (t + 5)
3(s2 – 8)
15. 6x2 + 24x + 15 3(2x2 + 8x + 5)
16. 12x2 – 28x – 24
4(3x + 2)(x – 3)
17. –16n2 + 12n –4n(4n – 3)ANSWER
ANSWER
ANSWER
ANSWER
ANSWER
GUIDED PRACTICEGUIDED PRACTICE for Example 4
18. 6z2 + 33z + 36
3(2z + 3)(z + 4)
ANSWER
EXAMPLE 5 Solve quadratic equations
Solve (a) 3x2 + 10x – 8 = 0 and (b) 5p2 – 16p + 15 = 4p – 5.
a. 3x2 + 10x – 8 = 0 (3x – 2)(x + 4) = 0
3x – 2 = 0 or x + 4 = 0
Write original equation.Factor.Zero product propertySolve for x.or x = –4x = 2
3
EXAMPLE 5 Solve quadratic equations
b. 5p2 – 16p + 15 = 4p – 5. Write original equation.5p2 – 20p + 20 = 0
p2 – 4p + 4 = 0(p – 2)2 = 0
p – 2 = 0p = 2
Write in standard form.Divide each side by 5.Factor.Zero product propertySolve for p.
GUIDED PRACTICEGUIDED PRACTICE for Examples 5, 6 and 7
Solve the equation.19. 6x2 – 3x – 63 = 0
or –3 3 12
20. 12x2 + 7x + 2 = x +8
no solution
21. 7x2 + 70x + 175 = 0
–5
ANSWER
ANSWER
ANSWER
