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Learn how to factor special cases such as difference of squares and perfect square trinomials, and use a check list to ensure polynomials are factored correctly.
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CHAPTER 8.4 Special Case Factors!
Recognizing the binomial product Difference of Squares • Recall special case F.O.I.L. where middle terms subtracts out. • e.g. (a+b)(a-b)= a2 - b2 • When factoring Difference of Squares both terms must be perfect squares, the sign must be negative, and when you check by FOIL the middle term subtracts out.
Factor A. x2 – 16B. x2 – 10 C. z6 - 25 • A. (x)2- (4)2 Both terms are perfect squares and sign is ( - ). = (x+4)(x-4) • B. (x)2-(?)2 x is perfect square, but10 is not, therefore NF over Integers. • C. (z3)2-(5)2 Both terms are perfect squares and sign is ( - ). = (z3+5)(z3-5)
Factor A. 25x2 – b2 B. 6x2 – 1 C. n8 - 36 • A. (5x)2- (b)2 Both terms are perfect squares and sign is ( - ). = (5x+b)(5x-b) • B. (?)2-(1)2 6x is not a perfect square, but1 is, therefore NF over Integers. • C. (n4)2-(6)2 Both terms are perfect squares and sign is ( - ). = (n4+6)(n4-6)
Factor z4 - 16 • (z2)2- (4)2 Is the difference of squares and factors to (z2+4)(z2-4) always continue to factor. • (z2+4) (z2-4)second binomial factor is also a difference of squares and factors to(z-2)(z+2). (z2+4) Is simplified! • (z2+4)(z+2)(z-2)
Factor n4 - 81 • (n2)2- (9)2 Is the difference of squares and factors to (n2+9)(n2-9) always continue to factor. • (n2+9) (n2-9)second binomial factor is also a difference of squares and factors to(n+3)(n-3). • (n2+9) Is simplified! • (n2+9)(n+3)(n-3)
Factoring Perfect Square Trinomials • Recall patterns of special case FOIL • (a+b)2 and (a-b)2 • a2+2ab+b2 = (a+b)(a+b) = (a+b)2 • a2 - 2ab+b2 = (a -b)(a -b) = (a-b)2
Things to consider; • Can the first and last terms be rewritten as perfect squares? • Does the middle term equal two times the perfect squares of the first and last terms? • If yes, then you can factor as a perfect square trinomial.
Factor 9x2 – 30x +25 • 1st Rewrite first and last term as perfect squares (3x)2 -30x + (5)2 • 2nd Does 2(a)(b)= middle term? 2(3x)(5)= 30x • Then factor as a perfect square trinomial. • (3x-5)(3x-5) = (3x-5) 2
Factor 16x2 +8x +1 • 1st Rewrite first and last term as perfect squares (4x)2 +8x + (1)2 • 2nd Does 2(a)(b)= middle term? 2(4x)(1)= 8x • Then factor as a perfect square trinomial. • (4x+1)(4x+1) = (4x+1) 2
Factor 4x2 +37x +9 • 1st Rewrite first and last term as perfect squares. (2x)2 +37x + (3)2 • 2nd Does 2(a)(b)= middle term? 2(2x)(3)= 12x = 37x • Then factor is not a perfect square trinomial. Factor by trial and error or grouping.
Factor 4x2 +37x +9 by grouping. • 1st (a)(b) = (4)(9) = 36 • 2nd Consider factors of 36 • 1, 36 • 2, 18 • 3, 12 • 4, 9 • 6, 6 • 3rd Rewrite middle terms 4x2+36x + x +9 • 4th Factor by grouping 4x (x+9) +1(x+9) • (x+9)(4x+1) • Check by F.O.I.L. 4x2+37x+9
Factor x2 +14x +36 • 1st Rewrite first and last term as perfect squares (x)2 +14x + (6)2 • 2nd Does 2(a)(b)= middle term? 2(x)(6)= 12x = 14x • Then factor is not a perfect square trinomial. Factor by trial and error or grouping.
Factor x2 +14x +36 by grouping. • 1st (a)(b) = (1)(36) = 36 • 2nd Consider factors of 36 • 1, 36 • 2, 18 • 3, 12 • 4, 9 • 6, 6 • None of the factors sum to 14, therefore does not factor over integers.
Chapter 8.4 Objective 2 • While factoring, one must always ask are the terms in simplest form? Could the elements be factored further? • This objective will answer that question, and supply a check list for factoring polynomials
Factoring Check List • 1. Is there a common factor? If so, factor the common factor out. • 2. Is the polynomial the difference of two squares? If so, factor. • 3. Is the polynomial a perfect square trinomial? If so, factor. • 4. Is the polynomial a product of two binomials? If so, factor. • 5. Does the polynomial have four terms? If so, try factoring by the grouping method.
Factor 3x2 -48 • 1stGCF 3(x2-16) • Continue to factor as difference of squares. • 3(x+4)(x-4) • Factoring complete, check by FOIL.
Factor x3 -3x2 -4x+12 • 1st There is not a GCF, but contains four terms. Try factor by grouping. x2(x3 -3x2 )-4(-4x+12)= • x2 (x-3)-4(x-3)= • (x-3)(x2 -4)= Continue to factor • (x-3)(x+2)(x-2) • Factoring complete.
Factor 4x2y2 + 12xy2 +9y2 • 1st Factor GCF • y2(4x2 +12x+9)Is remaining trinomial a perfect square trinomial? • First and last term can be written as perfect squares. Two times first times the last perfect squares[2(a)(b)] equals middle term. Therefore, factors as. • y2(2x+3)2
NOW YOU TRY! • 1. x4 -256 • (x2 -16)(x+4)(x-4) • 2. 9x2 -81 • 9(x+3)(x-3) • 3. a2b– 3a2 -16b +48 • (b-3)(a+4)(a-4)