On pages 345-6 and 353 of Elementary and Intermediate Algebra, there are many factoring problems. You will find your assignment in the following table.

Since there are several different types of factoring problems assigned from pages 345-346, four types will be demonstrated here to offer a selection, even though individual students will only be working two from these pages.

#73. x3 – 2×2 – 9x + 18 Four terms means start with grouping

x2(x – 2) – 9(x – 2) The common factor for each group is (x – 2)

(x – 2)(x2 – 9) Notice the difference of squares in second group

(x – 2)(x – 3)(x -+ 3) Now it is completely factored.

#81. 6w2 – 12w – 18 Every term has a GCF of 6

6(w2 – 2w – 3) Common factor is removed, now have a trinomial

Need two numbers that add to -2 but multiply to -3

Try with -3 and +1

6(w – 3)(w + 1) This works, check by multiplying it back together

#97. 8vw2 + 32vw + 32v Every term has a GCF of 8v

8v(w2 + 4w + 4) The trinomial is in the form of a perfect square

8v(w + 2)(w + 2) Showing the squared binomial

8v(w + 2)2 Writing the square appropriately

#103. -3y3 + 6y2 – 3y Every term has a GCF of -3y

-3y(y2 – 2y + 1) Another perfect square trinomial

-3y(y – 1)(y – 1) Showing the squared binomial

-3y(y – 1)2 Writing the square appropriately

Here are two examples of problems similar to those assigned from page 353.

5b2 – 13b + 6 a = 5 and c = 6, so ac = 5(6) = 30. The factor pairs of 30

are 1, 30 2, 15 3, 10 5,6

-3(-10)=30 while -3+(-10)= -13 so replace -13b by -3b and -10b

5b2 – 3b – 10b + 6 Now factor by grouping.

b(5b – 3) – 2(5b – 3) The common binomial factor is (5b – 3). (5b – 3)( b – 2) Check by multiplying it back together.

3×2 + x – 14 a = 3 and c = -14, so ac =3(-14)= -42. The factor pairs of – 42 are

1, -42 -1, 42 3, -14 -3, 14

2, -21 -2, 21 6, -7 -6, 7

We see that -6(7) = -42 while -6 + 7 = 1 so replace x with -6x + 7x.

3×2 – 6x + 7x – 14 Factor by grouping.

3x(x – 2) + 7(x – 2) The common binomial factor is (x – 2).

(x – 2)(3x + 7) Check by multiplying it back together

1. Problem 106 : Factor by group Four terms means start with grouping ac + xc + = The common factor for each group is (ac + xc) + ( +) Notice the difference of squares in second group

2. Take out GCF on both parenthesis and it + c(a+x) +(a+x) and if you take out (a+x) it would equal (a+x) (c + Now it is completely factored

On pages 345-6 and 353 of Elementary and Intermediate Algebra, there are many factoring problems. You will find your assignment in the following table.

· For the problems on page 345-6, factor the polynomials using whatever strategy seems appropriate. State what methods you will use and then demonstrate the methods on your problems, explaining the process as you go. Discuss any particular challenges those particular polynomials posed for the factoring.

· For the problem on page 353 make sure you use the “ac method” regardless of what the book’s directions say. Show the steps of this method in your work in a similar manner as how the book shows it in examples.

· Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.):

· Factor

· GCF

· Prime factors

· Perfect square

· Grouping