3/10/10

 Steps: Ex1) 5x +10 Step 1: Examine ? Common Number Notice 5 is a multiple of both Step 2: Remove 5(x + 2)

Practice:  Yes or No – There is a common multiple that can be removed? 1.4x – x – 7 3.x 2 + 2x 4.x Yes 2, 2(x - 5) No, 3x - 7 Yes x, x(x + 2) No, (x 2 + 4)

So far we’ve used distributive property for simple binomials, but you can also use it for polynomials. Example 2) 4ab + 8b + 3a + 6 First group the monomials to make groups with common multiples: (4ab + 8b) + (3a + 6) Remove the common monomial: 4b(a + 2) + 3(a + 2) Associate Property (4b + 3)(a + 2) 4b(a + 2) + 3(a + 2)

So far we’ve used distributive property for simple binomials, but you can also use it for polynomials. Example 3) c – 2cd + 8d - 4 First group the monomials to make groups with common multiples: (c – 2cd) + (8d - 4) Remove the common monomial: c(1 - 2d) - 4(-2d + 1) Associate Property (c – 4)(1 – 2d) c(1 – 2d) - 4(-2x + 1)

Practice:  Factor the following: 1.3p – 2p2 – 18p p 2 q – 9pq pq (p + 9)(-2p + 3) 3pg(p – 3q + 12)

 Now that we know: c – 2cd + 8d - 4 factors to (c – 4)(1 – 2d) we can solve. c – 2cd + 8d – 4 = 0 or (c – 4)(1 – 2d) = 0 (c – 4) = 0 and (1-2d) = 0 c = 4 -2d = -1 d = 1/2 If we wanted to check our answer you can sub c=4 and d=1/2 back into the problem

 Lets put it all together x 2 – 24x = 0 x(x – 24) = 0 x = 0 and x-24 = 0 x = 24 So x = 0,24 If we wanted to check our answer you can sub c=4 and d=1/2 back into the problem

 Lets put it all together 2 5y 2 – 15y +4y = 12 5y 2 – 15y + 4y -12 = 0 Make it equal to 0 (5y 2 – 15y) + (4y-12) = 0 5y(y – 3) + 4(y – 3) = 0 (5y + 4)(y – 3) = 0 5y + 4 = 0 y – 3 = 0 5y = -4 y = 3 y = -4/5 y = -4/5, 3 Reminder: To check your work you can always input your answer back into the original problem.

 Homework