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Maths questions help

I'm stuck on a part of my Maths homework.

3. Express the following as the product of factors

a. (4-z)^2 (2-z) + p(2-z)
b. (r-d)^3 + 5(r-d)^2
c. (b+c)^5 (a+b) - (b+c)^5
d. m(a-2x) + rp^2(2x-a)

4. Simplify each expression, leaving your answer in its factorised form

a. (p+q)^2 + 2q(p+q)
b. 2(2x-y)^2 - 6x(2x-y)
c. (r+6s)^2 - (r+6s) (r-s)

Any help would be appreciated on how to do these questions.
(edited 7 years ago)
Original post by dRaGoN2509
I'm stuck on a part of my Maths homework.

3. Express the following as the product of factors

a. (4-z)^2 (2-z) + p(2-z)
b. (r-d)^3 + 5(r-d)^2
c. (b+c)^5 (a+b) - (b-c)^5
d. m(a-2x) + rp^2(2x-a)

4. Simplify each expression, leaving your answer in its factorised form

a. (p+q)^2 + 2q(p+q)
b. 2(2x-y)^2 - 6x(2x-y)
c. (r+6s)^2 - (r+6s) (r-s)

Any help would be appreciated on how to do these questions.


What have you tried?
Reply 2
Original post by RDKGames
What have you tried?


I don't get what 'product of factors' means..
Original post by dRaGoN2509
I don't get what 'product of factors' means..


Express them as a product. Of factors.

First one: (4z)2(2z)+p(2z)=(2z)[(4z)2+p)](4-z)^2(2-z) + p(2-z) = (2-z)[(4-z)^2+p)]
Reply 4
Original post by RDKGames
Express them as a product. Of factors.

First one: (4z)2(2z)+p(2z)=(2z)[(4z)2+p)](4-z)^2(2-z) + p(2-z) = (2-z)[(4-z)^2+p)]


Is that the answer? For part b I got (r-d)+5, c I got a+b.
Original post by dRaGoN2509
Is that the answer? For part b I got (r-d)+5, c I got a+b.


Yes, both brackets are factors. And those aren't full answers.

b: (rd)2[(rd)+5](r-d)^2[(r-d)+5] can you see it?

c: How did you get that??? From what you've written in the OP, it doesn't factorise. Unless you meant it to be +c in the third bracket, or -c in the first bracket.
Reply 6
Original post by RDKGames
Yes, both brackets are factors. And those aren't full answers.

b: (rd)2[(rd)+5](r-d)^2[(r-d)+5] can you see it?

c: How did you get that??? From what you've written in the OP, it doesn't factorise. Unless you meant it to be +c in the third bracket, or -c in the first bracket.


Sorry, for c it was +c in the third bracket.
Original post by dRaGoN2509
Sorry, for c it was +c in the third bracket.


Then that goes to (b+c)5[(a+b)1](b+c)^5[(a+b)-1]

Try the other ones and write the answer in this sort of format.
Reply 8
Original post by RDKGames
Then that goes to (b+c)5[(a+b)1](b+c)^5[(a+b)-1]

Try the other ones and write the answer in this sort of format.


Is part d (2x-a) [(rp^2) + m] ?
Original post by dRaGoN2509
Is part d (2x-a) [(rp^2) + m] ?


No. The first bracket isn't the same as the second one. Check again and see what you can do to turn them the same.
Reply 10
Original post by RDKGames
No. The first bracket isn't the same as the second one. Check again and see what you can do to turn them the same.


Do you put a minus in front of it?
Original post by dRaGoN2509
Do you put a minus in front of it?


Bit of an awkward way of reasoning, but essentially yes. You are factoring -1 out of it.
Reply 12
Original post by RDKGames
Bit of an awkward way of reasoning, but essentially yes. You are factoring -1 out of it.


d) (2x-a) [(rp^2) -1m)] ?
(edited 7 years ago)
Original post by dRaGoN2509
d) (2x-a) [(rmp^2)-1)] ?


Nope. Expand it and see if it gives you the same thing. Check it again.
Reply 14
Original post by RDKGames
Nope. Expand it and see if it gives you the same thing. Check it again.


Changed it.
Original post by dRaGoN2509
Changed it.


Yes that's correct. Have a go at the rest of them.
Reply 16
Original post by RDKGames
Yes that's correct. Have a go at the rest of them.


4a) (p+q) [(p+q)^2+2q)]
4b) 2(2x-y) [(2x-y)^2-3x)]
4c) (r+6s) [(r+6s)^2-(r-s)]

Are these correct?
Original post by dRaGoN2509
4a) (p+q) [(p+q)^2+2q)]
4b) 2(2x-y) [(2x-y)^2-3x)]
4c) (r+6s) [(r+6s)^2-(r-s)]

Are these correct?


Just expand them and if they're equal to what you've started with, then yes :smile:
Reply 18
Original post by RDKGames
Just expand them and if they're equal to what you've started with, then yes :smile:


Alright, thanks. :smile:

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