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EXAMPLE 3

EXAMPLE 3. Factor by grouping. Factor the polynomial x 3 – 3 x 2 – 16 x + 48 completely. = x 2 ( x – 3 ) – 16 ( x – 3 ). x 3 – 3 x 2 – 16 x + 48. Factor by grouping. = ( x 2 – 16 )( x – 3 ). Distributive property. = ( x + 4 )( x – 4 )( x – 3 ). Difference of two

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EXAMPLE 3

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  1. EXAMPLE 3 Factor by grouping Factor the polynomial x3 – 3x2 – 16x + 48 completely. = x2(x – 3) –16(x – 3) x3– 3x2– 16x + 48 Factor by grouping. = (x2– 16)(x – 3) Distributive property = (x+ 4)(x – 4)(x – 3) Difference of two squares

  2. EXAMPLE 4 Factor polynomials in quadratic form Factor completely:(a) 16x4 – 81and(b) 2p8 + 10p5 + 12p2. a. 16x4 – 81 = (4x2)2 – 92 Write as difference of two squares. = (4x2 + 9)(4x2 – 9) Difference of two squares = (4x2 + 9)(2x + 3)(2x – 3) Difference of two squares Factor common monomial. b. 2p8 + 10p5 + 12p2 = 2p2(p6 + 5p3 + 6) Factor trinomial in quadratic form. =2p2(p3 + 3)(p3 + 2)

  3. for Examples 3 and 4 GUIDED PRACTICE Factor the polynomial completely. 5. x3 + 7x2 – 9x – 63 SOLUTION = x2(x + 7) –9(x + 3) x3+ 7x2– 9x – 63 Factor by grouping. = (x2– 9)(x + 7) Distributive property = (x+ 3)(x – 3) (x + 7) Difference of two squares

  4. for Examples 3 and 4 GUIDED PRACTICE 6. 16g4 – 625 SOLUTION a. 16g4 – 625 = (4g2)2 – 252 Write as difference of two squares. = (4g2 + 25)(4g2 – 25) Difference of two squares = (4g2 + 25)(2g + 5)(2g – 5) Difference of two squares

  5. for Examples 3 and 4 GUIDED PRACTICE 7. 4t6 – 20t4 + 24t2 SOLUTION Factor common monomial. 4t6 – 20t4 + 24t2 = 4t2(t4 –5t2 – 6t) = 4t2(t2 – 3)(t2 – 2 ) Factor trinomial in quadratic form.

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