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Mathematics 17 Online
OpenStudy (anonymous):

3. Find (-4x)^2 4. Multiply and simplify (x-3)(x+2) 5.Factor out the largest possible common factor 3xyz+8xy^2 6. Factor as much as possible: x^2+10x+25

OpenStudy (anonymous):

3 would be 16x^2 correct?

OpenStudy (anonymous):

4 would be x^2-1

OpenStudy (anonymous):

3) 16x^2 4) x^2-x-6 5) xy(3z+8y) 6) (x+5)^2 or (x+5)*(x+5)

OpenStudy (anonymous):

#4 is not as simple as says x*x + (-3+2) = x^-1 are you familiar with FOIL?

OpenStudy (anonymous):

Yeah sort of

jimthompson5910 (jim_thompson5910):

robz8 don't just give out answers it's against the code of conduct

OpenStudy (anonymous):

sorry

OpenStudy (anonymous):

I dont' understand how you do 5. I came up with a different answer for that

OpenStudy (anonymous):

I see where I messed up with 4 by using foil.

jimthompson5910 (jim_thompson5910):

to factor x^2+10x+25, you look for two numbers that a) multiply to 25 (last term) AND b) add to 10 (middle coefficient)

jimthompson5910 (jim_thompson5910):

the two numbers are 5 and 5 5+5 = 10 5*5 = 25 so this means x^2+10x+25 factors to (x+5)(x+5)

OpenStudy (anonymous):

so you have 3xyz+8xy^2 both those terms have "xy" in common about them, so you can 'factor' them out

jimthompson5910 (jim_thompson5910):

since you have a copy of the same factor, you can condense (x+5)(x+5) into (x+5)^2

jimthompson5910 (jim_thompson5910):

oh sry, saw the wrong one...

OpenStudy (anonymous):

I got 2x(yz+4y^2

jimthompson5910 (jim_thompson5910):

2 is a factor of 8, but not of 3

jimthompson5910 (jim_thompson5910):

so 3xyz+8xy^2 doesn't factor to 2x(yz+4y^2)

OpenStudy (anonymous):

I'm sorry its not supposed to be 3 its supposed to be 2*

jimthompson5910 (jim_thompson5910):

ah ok

jimthompson5910 (jim_thompson5910):

2xyz+8xy^2 2x(yz+4y^2) that's a good start, but you can factor more

jimthompson5910 (jim_thompson5910):

you can pull out y as well

jimthompson5910 (jim_thompson5910):

what do you get when you do that

OpenStudy (anonymous):

2x(z+4y)

jimthompson5910 (jim_thompson5910):

when you pull out the y, it goes outside the parenthesis

jimthompson5910 (jim_thompson5910):

so you should have this 2xy(z+4y)

OpenStudy (anonymous):

Ohhh ok

jimthompson5910 (jim_thompson5910):

as a check, you can distribute back through and you should get the original expression as a result

OpenStudy (anonymous):

THanks that helped alot

jimthompson5910 (jim_thompson5910):

you're welcome

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