Part A: Factor 3x2y2 - xy2 - 2y2. Show your work. (3 points) Part B: Factor x2 + 6x + 9. Show your work. (3 points) Part C: Factor x2 - 25. 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) HELPPP WITH ALLLL OF THIS AND SHOWINGG THE WORK PLEASEEEEEEEE TOO HELPP MEEE GET CREDIT :) :)
@Guyzrc00l
Could you show where the numbers are exponents? So we didn't do any mistakes.
@Guyzrc00l OKAY.
well for the 1st question remove the common factor \[y^2(3x^2 - x - 2)\] now you need to factor the quadratic....
For Part A: (sorry campbell_st) 3x^(2)y^(2)-xy^(2)-2y^(2) Factor out the GCF of y^(2) from the expression 3x^(2)y^(2). y^(2)(3x^(2))-xy^(2)-2y^(2) Factor out the GCF of y^(2) from the expression -xy^(2). y^(2)(3x^(2))+y^(2)(-x)-2y^(2) Factor out the GCF of y^(2) from the expression -2y^(2). y^(2)(3x^(2))+y^(2)(-x)+y^(2)(-2) Factor out the GCF of y^(2) from 3x^(2)y^(2)-xy^(2)-2y^(2). y^(2)(3x^(2)-x-2)
@Guyzrc00l A: 3x^2y^2-xy^2-2y^2 B: x^2+6x+9 C: x2-25
Thanks
@Guyzrc00l welcome
but the problem is that in PART A... @Guyzrc00l hasn't factored the quadratic equation for you...
For part A: Checked with a powerful calculator also. 3x^2y^2 - xy^2 - 2y^2 y^2(3x^2) - xy^2 - 2y^2 y^2(3x^2) + y^2(-x)+y^2(-2) y^2 (3x^2 - x - 2)
Same result @campbell_st
@Guyzrc00l TO make sure the answer is right??
but if you read my post... I said that the quadratic can be factored.... without a powerful calculator.... PART A \[y^2(3x^2 - x -2) = y^2 (3x +2)(x -1)\]
The question just asked to factor it. Simple.
yes.... andthe question is about factoring quadratics.... which means you would need to fully factor each expression....
Well, ok fine. You are right too. The result will be then.... (3x + 2) (x - 1) y^2
For Part B: x^2 + 6x + 9 (x + 3)(x + 3) (x + 3)^(1+1) (x + 3)^2
I'm assuming that x2-25 is actually "x^2-25" For Part C: x^2 - 25 The binomial can be factored using the difference of squares formula, because both terms are perfect squares. The difference of squares formula is a^(2)-b^(2)=(a-b)(a+b). (x - 5)(x + 5)
@Guyzrc00l for (3x+2) (x-1) y^2 that is for part a??
Yes
That is the last step
@Guyzrc00l Nope, Part C dose not have a exponent
@campbell_st A little help here? If you don't mind.
@Guyzrc00l Nope, it does not I looked it's a regular two that's where i am thrown for a loop
oh, make you could take a snippet of it. Is it on a computer?
the last looks like (x -5)(x+5) the difference of 2 squares...
@Guyzrc00l Yes, i am on a computer I don't know how to take a picture of it and put it on here??
@pbrianna100 said that there isn't an exponent after x. And that it isn't a normal 2
Are you using windows or mac?
@Guyzrc00l I have a windows 7 on my computer
click Start and type Snipping Tool
@Guyzrc00l Okay, I did what to do next?
Select the area with the problem and save it on your computer
You'll have to open the program first of course. "Snipping Tool"
@Guyzrc00l I did that, I just don't know how to upload it to here now
Oh, there is a attach File on the bottom of the text box where you type. :)
@Guyzrc00l Okay I uploaded it i think you can see it now lol
@Guyzrc00l has aleady posted the correct solution... I'd say a typo.. bu the teacher
I really do think it is a mistake. I use FLVS curriculum too and I noticed a few mistakes also.
@Guyzrc00l So you really think it is a exponent for part C??
What do you think for part D @campbell_st ?
Yes
That's how it usually works.
they all have a term in x^2.... and they could all be factored
Ok, Campbell_st answered for part D.
@Guyzrc00l Do we know what part D is??
@Guyzrc00l Yes, I think they have a common what about you??
Yes, I believe that all of them have a common factor which is x^2.
@Guyzrc00l Okay good I thought it was me. :) do you mind still helping me still :)??
Well, I don't mind but I have to go do my own work now. I'm really sorry. Hope some around from here could help you.
Really sorry.
@Guyzrc00l Okay. Thanks anyway you really did help me
You're welcome. Bye
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