If r || s and M<5 = 128, what’s is M<1?? A. 128 B. 59 C. 62 D. 64
1 answer:
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Notation
I imagine that the expression you are asked to work with is:
When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.
PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.
This gives us:
To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.
PART B
Start with the answer in part A. Namely,
. For now let's focus only on what is in the parenthesis. We have four terms so let's take them two at a time. I am separating the expression in two using square brackets. ![[( x^{3}+5x)]-[3 x^{2} -15]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D)
Let's next factor what is in each bracket:
![[( x^{3}+5x)]-[3 x^{2} -15] = [x( x^{2} +5)]-[3( x^{2} +5)]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D%20%3D%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D)
Notice that both brackets have the same expression in them so now we factor that out:![[x( x^{2} +5)]-[3( x^{2} +5)] = (x-3)( x^{2} +5)](https://tex.z-dn.net/?f=%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D%20%3D%20%28x-3%29%28%20x%5E%7B2%7D%20%2B5%29)
Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get:
I imagine that the expression you are asked to work with is:
When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.
PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.
This gives us:
To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.
PART B
Start with the answer in part A. Namely,
Let's next factor what is in each bracket:
Notice that both brackets have the same expression in them so now we factor that out:
Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get: