HELP: Algebra - Trinomials Factorising

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  1. nerd434's Avatar
    • Banned
    • Location: Edinburgh
    • Posts: 899
    HELP: Algebra - Trinomials Factorising
    I've been working through the textbook for Intermediate 2 Maths (Scotland), I was getting everything correct until I came to a specific section.

    I seem to have gotten some of them right however when I look at the answers sections I don't know where they have gotten numbers outside the brackets from, there is no mention of it.

    I'll list the questions (only six), what I wrote and what the answer section says:

    a) 12x² + 23 + 10 = (3x+2)(4x+5) - That was correct.

    b) 12y² + 36y + 15 = (2y+1)(6y+15) - Incorrect, the answer section says - 3(2y+1)(2y+5)

    c) 9m² - 6m - 8 = (3m-4)(3m+2) - That was correct.

    d) 4k² - 12 + 9 = (2k-3)(2k-3) - That was correct.

    e) 10a² + 38a - 8 = (10a-2)(a+4) - Incorrect, the answer section says - 2(a+4)(5a-1)

    f) 6y² - 27y + 12 = (6y-3)(y-4) - Incorrect, the answer sections says - 3(y-4)(2y-1)


    Okay so the ones in which I'm getting wrong, seem to be ones that involve a number outside the brackets. Can someone please tell me how to distinguish whether or not a question requires a number outside the bracket or not? Has it got something to do with the difference of squares? That's the only thing I can think of..

    And also, when I multiply out my answers that I got wrong, I get the correct answer? For example:

    e) (10a-2)(a+4) = 10a² + 40a - 2a - 8 = 10a² - 38a - 2a - Yet the book says it's wrong?

    Can someone PLEASE help? I don't know what I'm doing wrong and can someone explain how the numbers outside the brackets work in trinomials..
  2. sii217's Avatar
    • Respected Member
    • Posts: 195
    Re: HELP: Algebra - Trinomials Factorising
    You've just not fully factorised the ones that you've got incorrect.

    For example your question b: the (6y+15) factor can still be simplified. 3 is a common factor of both 6 and 15, giving you 3(2y+5).

    It doesn't matter where you put the 3 in the answer as all the factors are going to be multiplied. So 3 x (2y+1) x (2y+5) is the same as (2y+1) x 3 x (2y+5).

    It just looks neater putting the number at the beginning that's all.
  3. nerd434's Avatar
    • Banned
    • Location: Edinburgh
    • Posts: 899
    Re: HELP: Algebra - Trinomials Factorising
    (Original post by sii217)
    You've just not fully factorised the ones that you've got incorrect.

    For example your question b: the (6y+15) factor can still be simplified. 3 is a common factor of both 6 and 15, giving you 3(2y+5).

    It doesn't matter where you put the 3 in the answer as all the factors are going to be multiplied. So 3 x (2y+1) x (2y+5) is the same as (2y+1) x 3 x (2y+5).

    It just looks neater putting the number at the beginning that's all.
    Would my answer be wrong? Is it just a matter of simplifying the answer?

    Can you take me through the process you would do to simply the brackets? As in common factors etc..
  4. nerd434's Avatar
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    • Location: Edinburgh
    • Posts: 899
    Re: HELP: Algebra - Trinomials Factorising
    Wait, I think I've got it. I can only use this method if they all have a common factor though am I right?
  5. BigChimp's Avatar
    • Full Member
    • Posts: 93
    Re: HELP: Algebra - Trinomials Factorising
    (Original post by nerd434)
    I've been working through the textbook for Intermediate 2 Maths (Scotland), I was getting everything correct until I came to a specific section.

    I seem to have gotten some of them right however when I look at the answers sections I don't know where they have gotten numbers outside the brackets from, there is no mention of it.







    I'll list the questions (only six), what I wrote and what the answer section says:

    a) 12x² + 23 + 10 = (3x+2)(4x+5) - That was correct.

    b) 12y² + 36y + 15 = (2y+1)(6y+15) - Incorrect, the answer section says - 3(2y+1)(2y+5)

    c) 9m² - 6m - 8 = (3m-4)(3m+2) - That was correct.

    d) 4k² - 12 + 9 = (2k-3)(2k-3) - That was correct.

    e) 10a² + 38a - 8 = (10a-2)(a+4) - Incorrect, the answer section says - 2(a+4)(5a-1)

    f) 6y² - 27y + 12 = (6y-3)(y-4) - Incorrect, the answer sections says - 3(y-4)(2y-1)


    Okay so the ones in which I'm getting wrong, seem to be ones that involve a number outside the brackets. Can someone please tell me how to distinguish whether or not a question requires a number outside the bracket or not? Has it got something to do with the difference of squares? That's the only thing I can think of..

    And also, when I multiply out my answers that I got wrong, I get the correct answer? For example:

    e) (10a-2)(a+4) = 10a² + 40a - 2a - 8 = 10a² - 38a - 2a - Yet the book says it's wrong?

    Can someone PLEASE help? I don't know what I'm doing wrong and can someone explain how the numbers outside the brackets work in trinomials..

    For the last one wouldn't the answer be 10a^2 +38a -8 ?

    Because : 40a - 2a = +38a And -8 is a like term so you just leave it their on it's own?
  6. steve2005's Avatar
    • TSR Demigod
    • Location: LONDON
    Re: HELP: Algebra - Trinomials Factorising
    (Original post by nerd434)
    Can someone PLEASE help? I don't know what I'm doing wrong and can someone explain how the numbers outside the brackets work in trinomials..
    Take any common factors out first.
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