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Factoring Polynomials

Factoring Polynomials. Section 0.4. Factor by finding greatest common factors (GCF). EXAMPLES. Factor the expression by finding what each of the terms have in common. These are called greatest common factors and you should always look for them. 4x + 16. n 2 – 3 n. r 2 + 2 r – 63.

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Factoring Polynomials

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  1. Factoring Polynomials Section 0.4

  2. Factor by finding greatest common factors (GCF) EXAMPLES Factor the expression by finding what each of the terms have in common. These are called greatest common factors and you should always look for them. 4x + 16 n2 – 3n r2 + 2r – 63 ANSWER ANSWER ANSWER 4(x + 4) n(n– 3) (r + 9)(r –7)

  3. Factor trinomials of the form x2+ bx + c as a product of two binomials ANSWER Notice thatm = –4andn = –5. So, x2 – 9x + 20 = (x – 4)(x – 5). EXAMPLE Factor the expression. a. x2 – 9x + 20 b. x2 + 3x – 12 SOLUTION a.You wantx2 – 9x + 20 = (x + m)(x + n)where mn = 20andm + n = –9.

  4. Factor trinomials of the form x2+ bx + c as a product of two binomials ANSWER Notice that there are no factors mand nsuch that m + n = 3. So, x2 + 3x – 12 cannot be factored. We call that trinomial prime or irreducible. EXAMPLE b.You wantx2 + 3x – 12 = (x + m)(x + n)where mn = – 12andm + n = 3.

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

  6. EXAMPLE Factor with special patterns Factor the expression. a. x2 – 49 = x2 – 72 Difference of squares = (x + 7)(x – 7) b. d 2 + 12d + 36 = d 2 + 2(d)(6) + 62 Perfect square trinomial = (d + 6)2 c. z2 – 26z + 169 Perfect square trinomial = z2 – 2(z) (13) + 132 = (z – 13)2

  7. GUIDED PRACTICE Factor the expression, notice special patterns. x2 – 9 ANSWER (x – 3)(x + 3) q2 – 100 ANSWER (q – 10)(q + 10) y2 + 16y + 64 ANSWER (y + 8)2

  8. EXAMPLE More factoring with special patterns Factor the expression. a. 9x2 – 64 = (3x)2 – 82 Difference of squares = (3x + 8)(3x – 8) b. 4y2 + 20y + 25 = (2y)2 + 2(2y)(5) + 52 Perfect square trinomial = (2y + 5)2 c. 36w2 – 12w + 1 = (6w)2 – 2(6w)(1) + (1)2 Perfect square trinomial = (6w – 1)2

  9. GUIDED PRACTICE GUIDED PRACTICE Factor the expression, watch for special patterns. 16x2 – 1 (4x + 1)(4x – 1) ANSWER 9y2 + 12y + 4 (3y + 2)2 ANSWER 4r2 – 28r + 49 (2r – 7)2 ANSWER 25s2 – 80s + 64 ANSWER (5s – 8)2

  10. GUIDED PRACTICE GUIDED PRACTICE 49z2 + 4z + 9 ANSWER (7z + 3)2 36n2 – 9 ANSWER (6n – 3)(6n +3)

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

  12. 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. Write an equation that will represent the new field. EXAMPLE Use a quadratic equation as a model Nature Preserve

  13. EXAMPLE Use a quadratic equation as a model SOLUTION 480,000 = 240,000 + 1000x + x2 Multiply using FOIL. 0 = x2 + 1000x – 240,000 Write in standard form.

  14. EXAMPLE Factor ax2 + bx + c where c > 0 Factor 5x2 – 17x + 6. SOLUTION You want 5x2 – 17x + 6 = (kx + m)(lx + n) where kand lare factors of 5 andmand nare factors of 6. You can assume that kand lare positive and k ≥ l. Because mn> 0, mand nhave the same sign. So, mand nmust both be negative because the coefficient of x, –17, is negative.

  15. ANSWER The correct factorization is5x2 –17x + 6 = (5x – 2)(x – 3). EXAMPLE Factor ax2 + bx + c where c > 0

  16. ANSWER The correct factorization is3x2 + 20x – 7= (3x – 1)(x + 7). EXAMPLE Factor ax2 +bx + c where c < 0 Factor 3x2 + 20x – 7. SOLUTION You want3x2 + 20x – 7 = (kx + m)(lx + n)wherekandlarefactors of3andmandnare factors of–7. Becausemn < 0, mandn have opposite signs.

  17. GUIDED PRACTICE GUIDED PRACTICE Factor the expression. If the expression cannot be factored, say so. 7x2 – 20x – 3 ANSWER (7x + 1)(x – 3) 5z2 + 16z + 3 ANSWER (5z + 1)(z + 3). 2w2 + w + 3 ANSWER cannot be factored

  18. GUIDED PRACTICE GUIDED PRACTICE Factor the expression. If the expression cannot be factored, say so. 3x2 + 5x – 12 ANSWER (3x – 4)(x + 3) 4u2 + 12u + 5 ANSWER (2u + 1)(2u + 5) 4x2 – 9x + 2 ANSWER (4x – 1)(x –2)

  19. EXAMPLE Factor out monomials first Factor the expression. = 5(x2 – 9) a. 5x2 – 45 = 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)

  20. GUIDED PRACTICE GUIDED PRACTICE Factor the expression. 3s2 – 24 ANSWER 3(s2 – 8) 8t2 + 38t – 10 ANSWER 2(4t – 1) (t + 5) 6x2 + 24x + 15 ANSWER 3(2x2 + 8x + 5) 12x2 – 28x – 24 ANSWER 4(3x + 2)(x – 3) –16n2 + 12n ANSWER –4n(4n – 3)

  21. GUIDED PRACTICE GUIDED PRACTICE 6z2 + 33z + 36 ANSWER 3(2z + 3)(z + 4)

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