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Math Help - factorize

  1. #1
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    factorize

    can you guys please help me with the following questions??!!!

    1. Factorise:
    a. 8x^2 - 2x - 15
    b. 2y^3 + y^2 - 2y - 1
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  2. #2
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    By inspection,

    8x^2 - 2x - 15 = (2x - 3)(4x + 5)

    You could have used the quadratic formula here to find the 2 roots and thus factorise the equation.

    \text{Let} \,\,f(y) = 2y^3 + y^2 - 2y - 1

    Using the remainder theorem, we can see f(1) = 2 + 1 - 2 - 1 = 0
    And therefore y = 1 is a root, therefore (y - 1) is a factor of the polynomial.

    Factorising: 2y^3 + y^2 - 2y - 1 = (y - 1)(2y^2 + 3y + 1) = (y - 1)(2y + 1)(y + 1)
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  3. #3
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    Quote Originally Posted by nzmathman
    By inspection,



    You could have used the quadratic formula here to find the 2 roots and thus factorise the equation.
    thanks A LOT! for the solution of the first equation....but for the second one...can you please show the long division..cuz even i understood HOW to do it....but i go wrong in the procedure....ALWAYS!!!! so i kinda needed help with that....
    thanks in adv...
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  4. #4
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    Hmmm I know how to do the long division but cannot present it here nicely for you to see...I am not that advanced in the math coding system. I'm just able to factorise such expressions straight away...
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  5. #5
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    Talking

    hmmm...sure .. no probz..ill try the long division again and if i still have a doubt..i'll tell u...
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sudha_14 View Post
    can you guys please help me with the following questions??!!!

    1. Factorise:
    a. 8x^2 - 2x - 15
    b. 2y^3 + y^2 - 2y - 1
    There is no need for using polynomial division

    \begin{aligned}2y^3+y^3-2y-1&=(2y^3+y^2)-(2y+1)\\<br />
&=y^2(2y+1)-(2y+1)\\<br />
&=(y^2-1)(2y+1)\\<br />
&=(y-1)(y+1)(2y+1)\end{aligned}
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  7. #7
    MHF Contributor red_dog's Avatar
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    b) Another approach, without the long division:
    2y^3+y^2-2y-1=y^2(2y+1)-(2y+1)=(2y+1)(y^2-1)=(2y+1)(y-1)(y+1)
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  8. #8
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    thanks A LOT guys....i had completely forgotten about that method...
    anyways...thanks again!!!!! alot!!
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