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    Hi there. I'm doing some questions on polynomials and it's just come to my attention that I can't factorise quadratics where there is more than one x^2 value.

    The example I'm working with is (x+2) (3x^2 -8x +5). This is what I've worked out in the question so far, but the point of my question on here is to focus on the quadratic part.

    I know that to factorise it you would normally find two numbers that multiply to give 5 and add up to -8, and I know that those numbers are -3 and -5, but what do I do with the 3 at the beginning?? I tried a few questions like this before and when I checked them they were all wrong. I just never learned this.

    Thank you very much for any and all help (and apologies for any mistakes - the device I am using has a truly terrible keyboard)!
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    (Original post by JustJusty)
    Hi there. I'm doing some questions on polynomials and it's just come to my attention that I can't factorise quadratics where there is more than one x^2 value.

    The example I'm working with is (x+2) (3x^2 -8x +5). This is what I've worked out in the question so far, but the point of my question on here is to focus on the quadratic part.

    I know that to factorise it you would normally find two numbers that multiply to give 5 and add up to -8, and I know that those numbers are -3 and -5, but what do I do with the 3 at the beginning?? I tried a few questions like this before and when I checked them they were all wrong. I just never learned this.

    Thank you very much for any and all help (and apologies for any mistakes - the device I am using has a truly terrible keyboard)!
    When you've found the two numbers, you can rewrite it as (3x^2 - 3x + -5x + 5) (you can see that this is the same thing).

    Now you want to find the common factor of the first two terms and the last two terms. Then you may have used a similar method to factorise that when there is only 1 * x^2.
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    When the x^2 coefficient is 1 you find factors of 5 that add to make -8. When you have a different coefficient you multiply that by your constant and find factors of that which add to -8. E.g. 3*5 = 15 5*3=15

    So you can then write the equation as 3x^2-3x-5x+5=0 and then factorise easily from there.
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    (Original post by SeanFM)
    When you've found the two numbers, you can rewrite it as (3x^2 - 3x + -5x + 5) (you can see that this is the same thing).

    Now you want to find the common factor of the first two terms and the last two terms. Then you may have used a similar method to factorise that when there is only 1 * x^2.
    (Original post by Bruhh)
    When the x^2 coefficient is 1 you find factors of 5 that add to make -8. When you have a different coefficient you multiply that by your constant and find factors of that which add to -8. E.g. 3*5 = 15 5*3=15

    So you can then write the equation as 3x^2-3x-5x+5=0 and then factorise easily from there.
    I really dont understand. I used an algebra solver to see what the answer should be and it came up with (x-1)(3x-5). I pretty much get where the (3x-5) came from, but i have no idea how i would go about working out the other part.
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    Well since you have: 3x^2-3x-5x+5=0 you can now factorise this to get:

    3x(x-1)-5(x-1)=0
    And you can now factor out the (x-1) term to give (3x-5)(x-1)=0

    So x=1 or x=5/3


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    (Original post by JustJusty)
    Hi there. I'm doing some questions on polynomials and it's just come to my attention that I can't factorise quadratics where there is more than one x^2 value.

    The example I'm working with is (x+2) (3x^2 -8x +5). This is what I've worked out in the question so far, but the point of my question on here is to focus on the quadratic part.

    I know that to factorise it you would normally find two numbers that multiply to give 5 and add up to -8, and I know that those numbers are -3 and -5, but what do I do with the 3 at the beginning?? I tried a few questions like this before and when I checked them they were all wrong. I just never learned this.

    Thank you very much for any and all help (and apologies for any mistakes - the device I am using has a truly terrible keyboard)!
    Okay, so for a quadratic in the form:
    ax^2+bx+c

    The first step is find the product ac.
    In your quadratic 3x^2-8x+5, this would be 3 \times 5
    We get 15.

    Next step, we must find two numbers that multiply to give ac. These two numbers added together should also add up to give you b.

    So in your case, we have to find two numbers that multiply to make 15 and add to make b, which is -8 in your case.

    This step will probably take the longest because you'll have to try different number combinations, but in this case, -5 \times -3 gives 15 and -5 + -3 gives -8.

    So in your original quadratic expression, you can change the - 8x to - 3x - 5x.

    i.e. from:
    3x^2 - 8x + 5
    to:
    3x^2 (- 3x - 5x) + 5

    No need for the brackets, just to help you visualise that - 8x is the same as - 3x - 5x, so you can remove the brackets to give you:

    3x^2 - 3x - 5x + 5

    Now split this quadratic into two parts with brackets:

    (3x^2 - 3x) + (- 5x + 5)

    Now factorising like terms out from each bracket:
    Watch out for your signs here, and also if the values in the brackets are not the same as each other, you've gone wrong somewhere.

    3x(x - 1) - 5(x - 1)

    You can see that (x -1) is a common factor of 3x(x -1) and - 5(x -1), so you can factor out the (x - 1)s.

    If you don't understand this step, then let y = (x -1) and substitute to get:

    3xy - 5y

    Then in the same way, factor out the y's to give you:

    y(3x - 5)

    Then because y = (x - 1):

    (x - 1)(3x -5), and you're done!

    The method may seem a bit long, but with practive you'll be able to do it alot quicker, and you won't need to bother with the y = (x - 1) part.

    Sorry for giving a full solution, but it was the only way I could explain.
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    (Original post by The_Big_E)
    Sorry for giving a full solution, but it was the only way I could explain.
    Great explanation, I'm sure the OP will find it helpful - I just wanted to say that this is one of the scenarios where a full solution is acceptable, so there's no need to apologise!
 
 
 
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