Math 10 Week # 10

For Week 10 of Math 10 this semester, my class is continuing on expanding and factoring of polynomials. This week we continued with factoring for the GCF of polynomials, factoring simple trinomials, and factoring the difference of squares. For my week 10 blog post, I will be showing you how to factor the difference of squares.

So before we begin, let me explain how exactly we factor the difference of squares. There are a few basic rules that we have when we know that we need to use this process for an equation. Firstly, we need to see if all the terms in the polynomial can divide by a GCF. If not, than we know that we can proceed. Next, all of the numbers and variables have to be squares or be able to be square rooted evenly or this process wont work. Thirdly, the polynomial needs to be a binomial, if not, this process wont work. And lastly, the leading term needs to be subtracting the second term because we are finding the difference between the terms or subtracting one from another.

For our first example, we have 9x^2 – 36. So to start off, we need to see if there are any common factors or a GCF between these two terms that we can divide them by. As you can see, we have a GCF in 9 so we divide both terms by 9 and we write it to the left of our factored terms below in brackets. Next, we can see that we are not done yet and that we are able to factor the difference of squares. In the brackets on line 2, we see that we have x^2 – 4 and both terms have perfect squares. Because of this, we square root them and we write down their square roots below. Now on our answer, you can see we have (x+2) and (x-2) and this is because our constant of 4 in line two is a negative number or subtracting x^2 and we can only get negatives by multiplying one negative with one positive. Finally, we end up with 9(x+2)(x-2).

For our second example, we have a little more difficult question in x^4 – 16. Once again we start of by seeing if the two terms have a GCF that they can both be divided by and in this case, they do not. Following this, we move on to the factoring for the difference of squares. On line 1, we can see that the polynomial is a binomial, both terms are perfect squares, and it is a subtracting equation so we can continue with this process. So once we square root x^4 – 16, we end up with (x^2 – 4)(x^2 + 4). However, we are not done yet as we can once again repeat this process with (x^2 – 4) on the left. We do not do the same on the right polynomial in brackets because it is not a subtraction or difference question anymore so we are done with that. Once this step is complete, we end up with (x – 2)(x + 2)(x^2 + 4).

 

 

 

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