 # EXAMPLE 1 Factor ax 2 + bx + c where c > 0 Factor 5x 2 – 17x + 6. SOLUTION You want 5x 2 – 17x + 6 = (kx + m)(lx + n) where k and l are factors of 5 and.

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EXAMPLE 1 Factor ax 2 + bx + c where c > 0 Factor 5x 2 – 17x + 6. SOLUTION You want 5x 2 – 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 ax 2 + bx + c where c > 0 ANSWER The correct factorization is 5x 2 –17x + 6 = (5x – 2)(x – 3).

EXAMPLE 2 Factor ax 2 + bx + c where c < 0 Factor 3x 2 + 20x – 7. SOLUTION You want 3x 2 + 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 3x 2 + 20x – 7 = (3x – 1)(x + 7).

GUIDED PRACTICE for Examples 1 and 2 GUIDED PRACTICE Factor the expression. If the expression cannot be factored, say so. 1. 7x 2 – 20x – 3 (7x + 1)(x – 3) 2. 5z 2 + 16z + 3 3. 2w 2 + w + 3 cannot be factored ANSWER (5z + 1)(z + 3). ANSWER

GUIDED PRACTICE for Examples 1 and 2 GUIDED PRACTICE 5. 4u 2 + 12u + 5 (2u + 1)(2u + 5) 6. 4x 2 – 9x + 2 (4x – 1)(x – 2) 4. 3x 2 + 5x – 12 (3x – 4)(x + 3) ANSWER

EXAMPLE 3 Factor with special patterns Factor the expression. a. 9x 2 – 64 = (3x + 8)(3x – 8) Difference of two squares b. 4y 2 + 20y + 25 = (2y + 5) 2 Perfect square trinomial c. 36w 2 – 12w + 1 = (6w – 1) 2 Perfect square trinomial = (3x) 2 – 8 2 = (2y) 2 + 2(2y)(5) + 5 2 = (6w) 2 – 2(6w)(1) + (1) 2

GUIDED PRACTICE for Example 3 Factor the expression. 7. 16x 2 – 1 (4x + 1)(4x – 1) 8. 9y 2 + 12y + 4 (3y + 2) 2 9. 4r 2 – 28r + 49 (2r – 7) 2 10. 25s 2 – 80s + 64 (5s – 8) 2 ANSWER

GUIDED PRACTICE for Example 3 11. 49z 2 + 4z + 9 (7z + 3) 2 12. 36n 2 – 9 = (3y) 2 (6n – 3)(6n +3) ANSWER

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

GUIDED PRACTICE for Example 4 Factor the expression. 13. 3s 2 – 24 14. 8t 2 + 38t – 10 2(4t – 1) (t + 5) 3(s 2 – 8) 15. 6x 2 + 24x + 15 3(2x 2 + 8x + 5) 16. 12x 2 – 28x – 24 4(3x + 2)(x – 3) 17. –16n 2 + 12n –4n(4n – 3) ANSWER

GUIDED PRACTICE for Example 4 18. 6z 2 + 33z + 36 3(2z + 3)(z + 4) ANSWER

EXAMPLE 5 Solve quadratic equations Solve (a) 3x 2 + 10x – 8 = 0 and (b) 5p 2 – 16p + 15 = 4p – 5. a. 3x 2 + 10x – 8 = 0 (3x – 2)(x + 4) = 0 3x – 2 = 0 or x + 4 = 0 Write original equation. Factor. Zero product property Solve for x. or x = –4 x = 2323

EXAMPLE 5 Solve quadratic equations b. 5p 2 – 16p + 15 = 4p – 5. Write original equation. 5p 2 – 20p + 20 = 0 p 2 – 4p + 4 = 0 (p – 2) 2 = 0 p – 2 = 0 p = 2 Write in standard form. Divide each side by 5. Factor. Zero product property Solve for p.

GUIDED PRACTICE for Examples 5, 6 and 7 Solve the equation. 19. 6x 2 – 3x – 63 = 0 or –3 3 1212 20. 12x 2 + 7x + 2 = x +8 no solution 21. 7x 2 + 70x + 175 = 0 –5 ANSWER

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