Aim: How do we factor complex base and using regrouping

Do Now: Factor the following

1. 5x2y – 10xy2

2. x2 – 5x + 6

3. x(y – 1) + 3(y – 1)

The GCF can be a monomial or a binomial

x(y – 1) + 3(y – 1) The GCF is y – 1

Factor y – 1, then write the remaining in another parenthesis

= (y – 1)(x + 3)

x2(y + 2) – (y + 2)

Factor y + 2= (y + 2)(x2 – 1)

Factor (x2 – 1) again= (y + 2)(x – 1)(x + 1)

Factor: 2(5x + 2)2 – 7(5x + 2)

Factor the GCF (5x + 2)

=(5x + 2)[2(5x + 2) – 7]

=(5x + 2)(10x + 4 – 7)

Simplify the expression in the bracket

=(5x + 2)(10x – 3)

Factoring Expressions With Complex Bases

(a + 2)2 + 3(a + 2) + 2

Let A = (a + 2).

A2 + 3A + 2

= [(a + 2) + 2] [(a + 2) + 1]

Replace (a + 2) with A.

Factor the trinomial.

Replace (a + 2) with A.

Simplify.= (a + 4)(a + 3)

= (A + 2)(A + 1)

155 23 xxx

)15()5( 23 xxxGroup into two binomials

)15()15(2 xxx Factor by GFC if possible

)1)(15( 2 xx Factor by GFC

)1)(1)(15( xxx Factor completely

Factor each trinomial if possible.

1) y(x – 2) + 3(x – 2)

2) x2 – x – xy + y

3) 5(2x – 3)2 + 9(2x – 3)

4) (x – 3)2 – 6(x – 3) + 8

5) x3 + 3x2 – x – 3