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Concept

Example 1 Sum and Difference of Cubes A. Factor the polynomial x 3 – 400. If the polynomial cannot be factored, write prime. Answer:The first term is a perfect cube, but the second term is not. It is a prime polynomial.

Example 1 Sum and Difference of Cubes B. Factor the polynomial 24x 5 + 3x 2 y 3. If the polynomial cannot be factored, write prime. 24x 5 + 3x 2 y 3 =3x 2 (8x 3 + y 3 )Factor out the GCF. 8x 3 and y 3 are both perfect cubes, so we can factor the sum of the two cubes. (8x 3 + y 3 ) =(2x) 3 + (y) 3 (2x) 3 = 8x 3 ; (y) 3 = y 3 =(2x + y)[(2x) 2 – (2x)(y) + (y) 2 ] Sum of two cubes

Example 1 Sum and Difference of Cubes =(2x + y)[4x 2 – 2xy + y 2 ] Simplify. 24x 5 + 3x 2 y 3 =3x 2 (2x + y)[4x 2 – 2xy + y 2 ] Replace the GCF. Answer: 3x 2 (2x + y)(4x 2 – 2xy + y 2 )

A.A B.B C.C D.D Example 1 A. Factor the polynomial 54x x 2 y 3. If the polynomial cannot be factored, write prime. A. B. C. D.prime

A. B. C. D.prime A.A B.B C.C D.D Example 1 B. Factor the polynomial 64x y 5. If the polynomial cannot be factored, write prime.

Concept

Example 2 Factoring by Grouping A. Factor the polynomial x 3 + 5x 2 – 2x – 10. If the polynomial cannot be factored, write prime. x 3 + 5x 2 – 2x – 10Original expression = (x 3 + 5x 2 ) + (–2x – 10)Group to find a GCF. = x 2 (x + 5) – 2(x + 5)Factor the GCF. = (x + 5)(x 2 – 2)Distributive Property Answer: (x + 5)(x 2 – 2)

Example 2 Factoring by Grouping B. Factor the polynomial a 2 + 3ay + 2ay 2 + 6y 3. If the polynomial cannot be factored, write prime. a 2 + 3ay + 2ay 2 + 6y 3 Original expression = (a 2 + 3ay) + (2ay 2 + 6y 3 )Group to find a GCF. = a(a + 3y) + 2y 2 (a + 3y)Factor the GCF. = (a + 3y)(a + 2y 2 )Distributive Property Answer: (a + 3y)(a + 2y 2 )

A.A B.B C.C D.D Example 2 A.(d + 2)(d 2 + 2) B.(d – 2)(d 2 – 4) C.(d + 2)(d 2 + 4) D.prime A. Factor the polynomial d 3 + 2d 2 + 4d + 8. If the polynomial cannot be factored, write prime.

Example 3 Combine Cubes and Squares B. Factor the polynomial 64x 6 – y 6. If the polynomial cannot be factored, write prime. This polynomial could be considered the difference of two squares or the difference of two cubes. The difference of two squares should always be done before the difference of two cubes for easier factoring. Difference of two squares

Example 3 Combine Cubes and Squares Sum and difference of two cubes

Concept

Example 5 Quadratic Form A. Write 2x 6 – x in quadratic form, if possible. 2x 6 – x = 2(x 3 ) 2 – (x 3 ) + 9 Answer: 2(x 3 ) 2 – (x 3 ) + 9

Example 5 Quadratic Form B. Write x 4 – 2x 3 – 1 in quadratic form, if possible. Answer: This cannot be written in quadratic form since x 4 ≠ (x 3 ) 2.

A.A B.B C.C D.D Example 5 A.3(2x 5 ) 2 – (2x 5 ) – 3 B.6x 5 (x 5 ) – x 5 – 3 C.6(x 5 ) 2 – 2(x 5 ) – 3 D.This cannot be written in quadratic form. A. Write 6x 10 – 2x 5 – 3 in quadratic form, if possible.

A.A B.B C.C D.D Example 5 A.(x 8 ) 2 – 3(x 3 ) – 11 B.(x 4 ) 2 – 3(x 3 ) – 11 C.(x 4 ) 2 – 3(x 2 ) – 11 D.This cannot be written in quadratic form. B. Write x 8 – 3x 3 – 11 in quadratic form, if possible.

Example 6 Solve Equations in Quadratic Form Solve x 4 – 29x = 0. Original equation Factor. Zero Product Property Replace u with x 2.

Example 6 Solve Equations in Quadratic Form Take the square root. Answer: The solutions of the equation are 5, –5, 2, and –2.

A.A B.B C.C D.D Example 6 A.2, 3 B.–2, –3 C.–2, 2, –3, 3 D.no solution Solve x 6 – 35x = 0.

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