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    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.
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    (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?
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    (Original post by RDKGames)
    What have you tried?
    I don't get what 'product of factors' means..
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    (Original post by dRaGoN2509)
    I don't get what 'product of factors' means..
    Express them as a product. Of factors.

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

    First one: (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.
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    (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: (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.
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    (Original post by RDKGames)
    Yes, both brackets are factors. And those aren't full answers.

    b: (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.
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    (Original post by dRaGoN2509)
    Sorry, for c it was +c in the third bracket.
    Then that goes to (b+c)^5[(a+b)-1]

    Try the other ones and write the answer in this sort of format.
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    (Original post by RDKGames)
    Then that goes to (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] ?
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    (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.
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    (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?
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    (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.
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    (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)] ?
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    (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.
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    (Original post by RDKGames)
    Nope. Expand it and see if it gives you the same thing. Check it again.
    Changed it.
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    (Original post by dRaGoN2509)
    Changed it.
    Yes that's correct. Have a go at the rest of them.
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    (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?
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    (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
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    (Original post by RDKGames)
    Just expand them and if they're equal to what you've started with, then yes
    Alright, thanks.
 
 
 
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