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Mathematics 21 Online
OpenStudy (anonymous):

Part A: Factor x2y2 + 6xy2 + 8y2. Show your work. (3 points) Part B: Factor x2 + 8x + 16. Show your work. (3 points) Part C: Factor x2 - 16. Show your work. (2 points) Part D: Do the polynomials in parts A, B, and C have a common factor? Explain your answer. (2 points)

OpenStudy (anonymous):

@phi

OpenStudy (anonymous):

@sourwing

OpenStudy (anonymous):

@Microrobot

OpenStudy (anonymous):

@leticia_9878

OpenStudy (anonymous):

Ok

OpenStudy (anonymous):

B. (x+4)(x+4)

OpenStudy (anonymous):

umm could you show me step please @Microrobot

OpenStudy (anonymous):

A. (x+4)(x+2) y^2

OpenStudy (anonymous):

I will

OpenStudy (anonymous):

So for B

OpenStudy (anonymous):

x^2 + 8x + 16

OpenStudy (anonymous):

You need to pay attention to the signs, since they are all plus signs you know your answer must have x+ on in both parenthesis

OpenStudy (anonymous):

So you always look at the the third digit the one with out the variable which is 16

OpenStudy (anonymous):

So from there you find factors of 16

OpenStudy (anonymous):

So you get 1*16=16 2*8=16 4*4=16

OpenStudy (anonymous):

As a possible factor, next you look at the second digit in your case is 8x

OpenStudy (anonymous):

Remember the second digit is the one that when you add your two numbers you get the exact answer

OpenStudy (anonymous):

1*16=16 2*8=16 4*4=16 was our factors of 16. 4+4=8

OpenStudy (anonymous):

So you know on the spot that it has to be 4, and 4

OpenStudy (anonymous):

and since the trinomial x^2 + 8x + 16 is completely positive

OpenStudy (anonymous):

so that would all be for part b thanks

OpenStudy (anonymous):

(x+4)(x+4) is the answer

OpenStudy (anonymous):

To test use the FOIL method

OpenStudy (anonymous):

No problem

OpenStudy (anonymous):

For C x^2 - 16 this is going to be (x-4)(x+4)

OpenStudy (anonymous):

any steps please?

OpenStudy (anonymous):

Of course

OpenStudy (anonymous):

So this problem is known as a

OpenStudy (anonymous):

for part c?

OpenStudy (anonymous):

Perfect square

OpenStudy (anonymous):

Yeah

OpenStudy (anonymous):

So for your problem you can tell became it's always going to be the same number. For example x^2-64

OpenStudy (anonymous):

8*8=64

OpenStudy (anonymous):

So the rules with these you open two paranthesis with the number as a positive and a negative

OpenStudy (anonymous):

So it would be (x-8)(x+8)

OpenStudy (anonymous):

for what part?

OpenStudy (anonymous):

ohh part c?

OpenStudy (anonymous):

It was an example for. C. It will be. (x-4)(x+4)

OpenStudy (anonymous):

oh kinda need steops but that's ok. could you help with part D?

OpenStudy (anonymous):

With D

OpenStudy (anonymous):

I'm going with yes, because the common factor had to work in the entire equation. When you get your common factor you are able to test it through the foil method and from there you will return back to the original equation.

OpenStudy (anonymous):

ALSO CAN I ASK FOR STEP FOR PART A.)

OpenStudy (anonymous):

This common factor is a number that is followed by a variable which would be x

OpenStudy (anonymous):

Sure

OpenStudy (anonymous):

Factor x^2y^2 + 6xy^2 + 8y^2.

OpenStudy (anonymous):

You need to find a common factor which is y^2

OpenStudy (anonymous):

So you get from there y^2(x^2+6x+8)

OpenStudy (anonymous):

You get the factor for x^2+6x+8

OpenStudy (anonymous):

Which gets you y^2(x+4)(x+2)

OpenStudy (anonymous):

And that's basically it

OpenStudy (anonymous):

@Microbot, that is amazing...I'm surprised that you only have a 59 as your smart score...You're like a Genius.

OpenStudy (anonymous):

@Micorobot How did you get y^2(x+4)(x+2)

OpenStudy (anonymous):

glovemanJR..why don't you stop cheating on your final exam for FLVS

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