Mathematics
Algebra Discussions on Polynomials - 250 words needed- problems attached

Question Description

I’m stuck on a Algebra question and need an explanation.

Week 5 Discussion factoring problems.pdf 

MAT221.W5.DiscussionExample.pdf 

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.

I have attached an example as well as a copy of the factoring problems, which are

Circled in red.

  • 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

Your initial post should be 150-250 words in length.


Unformatted Attachment Preview

INSTRUCTOR GUIDANCE EXAMPLE: Week Five Discussion Factoring Since there are several different types of factoring problems assigned from pages 345346, four types will be demonstrated here to offer a selection, even though individual students will only be working two from these pages. #73. x3 – 2x2 – 9x + 18 x2(x – 2) – 9(x – 2) (x – 2)(x2 – 9) (x – 2)(x – 3)(x -+ 3) Four terms means start with grouping The common factor for each group is (x – 2) Notice the difference of squares in second group Now it is completely factored. #81. 6w2 – 12w – 18 6(w2 – 2w – 3) Every term has a GCF of 6 Common factor is removed, now have a trinomial Need two numbers that add to -2 but multiply to -3 Try with -3 and +1 This works, check by multiplying it back together 6(w – 3)(w + 1) #97. 8vw2 + 32vw + 32v 8v(w2 + 4w + 4) 8v(w + 2)(w + 2) 8v(w + 2)2 Every term has a GCF of 8v The trinomial is in the form of a perfect square Showing the squared binomial Writing the square appropriately #103. -3y3 + 6y2 – 3y -3y(y2 – 2y + 1) -3y(y – 1)(y – 1) -3y(y – 1)2 Every term has a GCF of -3y Another perfect square trinomial Showing the squared binomial 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 5b2 – 3b – 10b + 6 b(5b – 3) – 2(5b – 3) (5b – 3)( b – 2) Now factor by grouping. The common binomial factor is (5b – 3). Check by multiplying it back together. -3(-10)=30 while -3+(-10)= -13 so replace -13b by -3b and -10b 3x2 + x – 14 3x2 – 6x + 7x – 14 3x(x – 2) + 7(x – 2) (x – 2)(3x + 7) 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. Factor by grouping. The common binomial factor is (x – 2). Check by multiplying it back together. ...
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