Difference of Squares Recall that, when multiplying conjugate binomials, the product is a difference of squares. E.g., (x - 7)(x + 7) = x 2 - 49 Therefore,

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Presentation transcript:

Difference of Squares Recall that, when multiplying conjugate binomials, the product is a difference of squares. E.g., (x - 7)(x + 7) = x Therefore, when factoring a difference of squares, the factors will be conjugate binomials. Factor: x = (x - 9)(x + 9) 16x = (4x - 11)(4x + 11) 5x = 5(x ) = 5(x - 4)(x + 4) (x) 2 - (9) 2 (4x) 2 - (11) 2

x CANNOT BE FACTORED Factor completely: x = (x 2 - 4) x 8 - 1= (x 4 - 1)(x 4 + 1) x x = (x 2 - 9)(x 2 - 4) = (x - 2)(x + 2) (x 2 + 4) = (x - 1)(x + 1) (x 4 + 1) = (x 2 - 1)(x 2 + 1) (x 2 + 4) (x 2 + 1)(x 4 + 1) = (x - 3)(x + 3)(x - 2)(x + 2) Factoring Completely

= [(x + y) - 4](x + y) (x + 5) = [(x + 5) - 7] (3x + 2) = [(3x + 2) - 9] 25(x + 4) = [5(x + 4) - 7] Factor completely: = (x + y - 4)(x + y + 4) [(x + y) + 4] = (x - 2)(x + 12) [(x + 5) + 7] = (x )(x ) = (3x - 7)(3x + 11) [(3x + 2) + 9] = (3x )(3x ) = (5x + 13)(5x + 27) = (5x )(5x ) [5(x + 4) + 7] Factoring a Difference of Squares with a Complex Base

25 - (x + 3) 2 = [5 - (x + 3)] 16 - (x - y) 2 = [4 - (x - y)] 81 - (3x + 2) 2 = [9 - (3x + 2)] (x + 4) 2 = [7 - 5(x + 4)] = (4 - x + y)(4 + x - y) [4 + (x - y)] = (-x + 2)(8 + x) = (5 - x - 3)(5 + x + 3) [5 + (x + 3)] = (-3x + 7)(3x + 11) = (9 - 3x - 2)(9 + 3x + 2) [9 + (3x + 2)] = (-5x - 13)(5x + 27) = (7 - 5x - 20)(7 + 5x + 20) [7 + 5(x + 4)] Factoring a Difference of Squares with a Complex Base

Factor:

How do you factor the sum or difference of cubes? You’ll need to memorize the factorization of the sum or difference of two cubes: Sum of Two Cubes: a 3 + b 3 = (a + b)(a 2 – ab + b 2 ) Difference of Two Cubes: a 3 – b 3 = (a – b)(a 2 + ab + b 2 ) Example: Factor x The cube roots of the terms are x and 3 = (x + 3)(x 2 – 3x + 9) Example: Factor 128x 3 – 250 Factor out the GCF first… = 2(64x 3 – 125) The cube roots are 4x and –5 = 2(4x – 5)(16x x + 25)