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    Part 3, I was able to do part 3 although I called 3x-1, 6x-2 instead but I was able to see that.

    But the second part I simply cannot do. I thought the polynomial on the numerator 9x^3 + 21x^2 + 6x would probably go into 3x+1 as well.

    Doing so gave me 3x^2 + 6x. I factorised this and got (x+2) and (x) as factors, I don't believe these are correct, so how would I do this?
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    (Original post by Dinasaurus)
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    Part 3, I was able to do part 3 although I called 3x-1, 6x-2 instead but I was able to see that.

    But the second part I simply cannot do. I thought the polynomial on the numerator 9x^3 + 21x^2 + 6x would probably go into 3x+1 as well.

    Doing so gave me 3x^2 + 6x. I factorised this and got (x+2) and (x) as factors, I don't believe these are correct, so how would I do this?
    Doing this quickly in my head gives me f(x) = (x+2)(3x-1)(3x+1)

    and

    \displaystyle

 \begin{equation*}9x^3 + 21x^2 + 6x = x(9x^2 +21x+ 6) = 3x(x+2)(3x+1)\end{equation*}

    So:

    \displaystyle

\begin{equation*} \frac{9x^3 + 21x^2 + 6x}{f(x)} = \frac{3x(x+2)(3x+1)}{(x+2)(3x-1)(3x+1)}\end{equation*}

    What cancels with and what are you left with?
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    (Original post by Dinasaurus)
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    Part 3, I was able to do part 3 although I called 3x-1, 6x-2 instead but I was able to see that.

    But the second part I simply cannot do. I thought the polynomial on the numerator 9x^3 + 21x^2 + 6x would probably go into 3x+1 as well.

    Doing so gave me 3x^2 + 6x. I factorised this and got (x+2) and (x) as factors, I don't believe these are correct, so how would I do this?
    You think correcty. 3x+1 is indeed a factor of the numerator however 3x^2+6x \neq x(x+2), you are forgetting a number.

    You need to express the numerator and denominator in terms of their factors and "cancel" appropriately.
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    (Original post by Kvothe the arcane)
    You think correcty. 3x+1 is indeed a factor of the numerator however 3x^2+6x \neq x(x+2), you are forgetting a number.

    You need to express the numerator and denominator in terms of their factors and "cancel" appropriately.
    Yeah I know that part but it's getting the numerator as factors which is the problem for me, I used the grid method to work it out and I can't seem to get another number. I have tried once again with the grid method and I don't go any further than (3x^2+6x)

    If I use the quadratic formula for this, (-6+-sqrt(36))/6

    I only get 0 and -2.
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    (Original post by Dinasaurus)
    Yeah I know that part but it's getting the numerator as factors which is the problem for me, I used the grid method to work it out and I can't seem to get another number. I have tried once again with the grid method and I don't go any further than (3x^2+6x)

    If I use the quadratic formula for this, (-6+-sqrt(36))/6

    I only get 0 and -2.
    Zacken has factorized it nicely for you. Forget this complex problem. If you have 3x^2+6x=Ax(x+B), what are A and B? You've worked out B, A is all that is all you're missing.
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    (Original post by Dinasaurus)
    Yeah I know that part but it's getting the numerator as factors which is the problem for me, I used the grid method to work it out and I can't seem to get another number. I have tried once again with the grid method and I don't go any further than (3x^2+6x)

    If I use the quadratic formula for this, (-6+-sqrt(36))/6

    I only get 0 and -2.
    9x^3 + 21x^2 + 6x = 3x(3x^2 + 7x + 2)

    Using the quadratic formula on the quadratic factor gets you: \displaystyle x = \frac{-7 \pm \sqrt{25}}{6} = \frac{-7 \pm 5}{6}

    So the quadratic 3x^2 + 7x + 2 factorises as (x+2)(3x+1).
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    (Original post by Zacken)
    9x^3 + 21x^2 + 6x = 3x(3x^2 + 7x + 2)

    Using the quadratic formula on the quadratic factor gets you: \displaystyle x = \frac{-7 \pm \sqrt{25}}{6} = \frac{-7 \pm 5}{6}

    So the quadratic 3x^2 + 7x + 2 factorises as (x+2)(3x+1).
    Thank you! If I take out a factor of 3x+1 instead of 3x, I get 3x^2+6x, which gives me the roots -2 and 0 from the quadratic formula? Why does it not work this way?
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    (Original post by Dinasaurus)
    Thank you! If I take out a factor of 3x+1 instead of 3x, I get 3x^2+6x, which gives me the roots -2 and 0 from the quadratic formula? Why does it not work this way?
    It does work. If you take out a factor of 3x+1 you're left with 3x^2 + 6x, this has roots x=-2 and x=0 meaning that 3x^2 + 6x = 3(x-0)(x+2) = 3x(x+2).
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    (Original post by Zacken)
    It does work. If you take out a factor of 3x+1 you're left with 3x^2 + 6x, this has roots x=-2 and x=0 meaning that 3x^2 + 6x = (x-0)(x+2) = x(x+2).

    Would (x)(3x+1)(x+2)/ (3x-1)(3x+1)(x+2)

    which cancels to x/(3x-1) not be correct then?

    Sorry for asking so much, I see that the mark scheme says it is 3x on the top.

    But if you take out a factor of 3x+1, then surely the other two linear factors are x and (x+2)?
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    (Original post by Dinasaurus)
    Would (x)(3x+1)(x+2)/ (3x-1)(3x+1)(x+2)

    which cancels to x/(3x-1) not be correct then?

    Sorry for asking so much, I see that the mark scheme says it is 3x on the top.

    But if you take out a factor of 3x+1, then surely the other two linear factors are x and (x+2)?
    Ah, okay - I see your confusion. The problem is that any quadratic equation of the form k(3x^2 + 6x) =0 will have those same roots, so you can't factorise it knowing just the roots. What you need to be able to do is just see:

    3x^2 + 6x = 3(x^2 + 2x) = 3x(x+2).
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    (Original post by Zacken)
    Ah, okay - I see your confusion. The problem is that any quadratic equation of the form k(3x^2 + 6x) =0 will have those same roots, so you can't factorise it knowing just the roots. What you need to be able to do is just see:

    3x^2 + 6x = 3(x^2 + 2x) = 3x(x+2).
    So if I have no C value, I cannot use the quadratic formula?

    Thank you for this explanation.
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    (Original post by Dinasaurus)
    So if I have no C value, I cannot use the quadratic formula?

    Thank you for this explanation.
    You can use the quadratic formula and it will give you the linear factors. It is your job for you to multiply the entire thing by whatever you need to get it to look like the unexpanded version.

    For example, if I gave you 5x^2 + 15x the quadratic formula will gives you roots as x=-3 and x=0, this allows you to factorise it as A(x-0)(x+3), but you now need to set A=5 to get the proper factorisation. In general, the A will be the coefficient of the x^2 term.
 
 
 
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