Algebra

Jul 30th, 2015
Anonymous
Category:
Engineering
Price: $15 USD

Question description

I will post an example of # 3 and instructions for the ac method at the bottom

  • For problems 1 and 2 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 problem 3 make sure you use the “ac method”, Show the steps of this method in your work in a similar manner of the example.

  • 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


  1.  b^3 + 49b               (Factoring completely)

  2.  a^2 + 7ab 10b^2    (Factoring with Two Variables)

  3.  3a^2 -14a +15         (Factor each trinomial using the ac method. Instructions for this method are                                    below)


The ac Method

The first step in factoring ax^2 + bx + c with a =1 is to find two numbers with a product of c and a sum of b. If a cancels out 1, then the first step is to find two numbers with a product of ac and a sum of b. This method is called the ac method. The strategy for factoring by the ac method follows. Note that this strategy works whether or not the leading coefficient is 1.

How to factor the trinomial ax^2 + bx + c:

  •  Find two numbers that have a product equal to ac and a sum equal to b.
  •  Replace bx by the sum of two terms whose coefficients are the two numbers found in (1).
  •  Factor the resulting four-term polynomial by grouping.


Here is an example of #3


5b^2 – 13b + 6                a = 5 and c = 6, so ac = 5(6) = 30.

                                       The factor pairs of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6

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

5b^2 – 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 


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