Read the following instructions in order to complete this discussion, and review the example of how to complete the math required for this assignment:
•On pages 345, 346, and 353 of Elementary and Intermediate Algebra, there are many factoring problems. You will find your assignment in the following table.
8 x2 + 8x + 15 2a2 + 1 + 3a
Your initial post should be 150-250 words in length. Respond to at least two of your classmates’ posts by Day 7 in at least a paragraph. Do you agree with how they used the vocabulary? Do their answers make sense?
Carefully review the Grading Rubric for the criteria that will be used to evaluate your discussion.
Response 1 -Edward Williamson
I was assigned the number 36. This means that from pages 345-346 that I will work problems 72 and 100.
72. –No Image -100a The first step to factor this binomial completely is to find the GCF or the greatest common factor.
a is present and can be taken from both terms of this binomial and therefore is the GCF.
a ( No Image – 100) This step shows where the GCF has been taken out of the binomial and then is multiplied by what is left.
The terms in the parenthesis are the difference of two perfect squares. This is true because both No Image and 100 are perfect squares.
The binomial would then be written as a( No Image -102) and then factors to
a (10 + a)(10 – a) which is the final step in factoring this binomial.
*For some reason the first part of this discussion post is copying in a weird way. I hope that although the terms are a little above the problem that it is clear where they belong.*
100. 3x3y2 – 3x2y2 + 3xy2 The first step to factor this polynomial completely is to find the GCF.
The GCF is 3xy2 because this can be pulled from each term in this polynomial.
3xy2(x2 –x +1) This step shows where the GCF was removed and what is left is in the parenthesis.
Because there are no factors of 1 that add to -1 this is a prime polynomial and this is as far as this polynomial can be factored.
84. a2b + 2ab – 15b The first step to factor this polynomial using the ac method is to identify which terms are equal to a and which are equal to c and then multiply those terms together.
a = 1 and c = -15 and so therefore ac = -15
The factor pairs of -15 that would equal a positive 2 are 5 and -3.
The next step is to change the polynomial from having 3 terms to four terms by replacing the 2ab with 5ab and -3ab.
a2b + 5ab -3ab -15b This step shows where the polynomial has four terms and can now be factored by grouping.
ab(a + 5) -3b(a + 5) This step shows that the common binomial factor is (a + 5).
(a + 5)(ab – 3b) Here the problem has been completely factored using the ac method.
Response 2 -Ariel Aker
For this weeks discussion, I was given problems 4 and 84 on page 345 and problem 16 on page 353. The biggest lesson this week was factoring and finding the GCF between two integers.
Problem 4 on page 345
My trinomial for this problem is as states: w² + 6w + 8
The first step is to factor.
( w + 4 )( w + 2 )
This is as far as we can factor this particular problem at this time. In order for us to check our work, we can just use the distributive property with the ( w + 4 ) to the ( w + 2 ) with each of the polynomials to check our work. So first we would distribute the w to the w and the 2.
w(w) + w(2) = w² + 2w
Now we do the same for the 4.
4(w) + 4(2) = 4w + 8
Now we combine the two statements together to achieve one larger equation.
w² + 2w + 4w + 8
When we combine like terms, our answer would then be:
w² + 6w + 8
This is just so we can double check that we did our factoring right to achieve the correct answer.
Problem 84 on page 345.
My binomial for this problem is as states: 5x² + 500
The first step to solving this problem is to factor out the 5.
5 ( x² + 100)
This is as far as we can factor this problem due to the positive sign in front of the 100. Nothing that we can add together and multiply together will cancel each other out and give us a positive 100. Even though the 100 is a perfect square, we can’t factor it further.
One way to double check our work is to use the same method as stated above. We will just distribute the 5 back to the x² and the 100, to achieve our answer.
5x² + 500
This just proves to us that I did my math correctly when I can work backwards and end where I began in my problem.
Problem 16 on page 353.
My trinomial for this problem is as follows: 2h² + 7h + 3.
The first step we take to solving this is to factor.
( 2h + 1 )( h + 3 )
This is the only step we can take at this time for it doesn’t ask for us to solve the equation. This problem is also broken down into all of their prime factors, so we still couldn’t factor it further even if we wanted to. Again, I like to work backwards to make sure that I did all my math correct. So I will start out by distributing the 2h to the h and the 3.
2h(h) + 2h(3) = 2h² + 6h
Next I will do the same for the 1.
1(h) + 1(3) = h + 3
We can then combine the two equations to achieve one large equation.
2h² + 6h + h + 3
Combine like terms.
2h² + 7h + 3
None of my problems could be solved by grouping two binomials together and factoring them separately to achieve my answers.
That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.