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Math Help - Factorising a quadratic

  1. #1
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    Factorising a quadratic

    Could someone explain to me as simply as possible how to factorise

    2x^2 + x - 3

    I know the answer is (2x + 3) (x - 1) from my textbook but I don't understand the method used.
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  2. #2
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    Quote Originally Posted by Natasha1 View Post
    Could someone explain to me as simply how to simplify

    2x^2 + x -3

    I know the answer is (2x + 3) (x - 1) from my textbook but I don't understand the method used.
    I think the method is called guess and check

    A more procedural method would be solve for the roots r1,r2 of Ax^2 + Bx + C = 0.
    Then the equation factors into A(x - r1)(x - r2).
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  3. #3
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    I saw on the net a method called the AC method but that is also confusing
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  4. #4
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    Quote Originally Posted by Natasha1 View Post
    Could someone explain to me as simply as possible how to factorise

    2x^2 + x - 3

    I know the answer is (2x + 3) (x - 1) from my textbook but I don't understand the method used.
    It comes from an understanding that

    (2x+a)(x+b)=2x(x+b)+a(x+b)=2x^2+2xb+ax+ab=2x^2+(a+  2b)x+ab

    so you are looking for the factors of -3 for which one of them plus twice the other sum to 1.

    They will be 3 and twice -1.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by snowtea View Post
    I think the method is called guess and check

    A more procedural method would be solve for the roots r1,r2 of Ax^2 + Bx + C = 0.
    Then the equation factors into A(x - r1)(x - r2).

    snotew mentioned "guess"...


    First guess that one root is x=1(it is an easy guess) then do long division: {2x^2 + x -3}/{x-1} you will get x+3/2
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  6. #6
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    Nawwww....I say if not apparent, use the quadratic formula...but that's me!
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  7. #7
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    (\pm\alpha x\pm\beta)(\pm\lambda x\pm\phi)

    \beta*\phi=-3\Rightarrow 3*1=3 Now need to figure out which one is negative.

    \alpha*\lambda=2\Rightarrow 2*1=2

    \alpha*\phi+\beta*\lambda=1

    Now you need to decide if 3 is beta and 1 is phi, is alpha 1 or 2 and is lambda 1 or 2?
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  8. #8
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    Quote Originally Posted by Also sprach Zarathustra View Post
    First guess that one root is x=1(it is an easy guess) then do long division: {2x^2 + x -3}/{x-1} you will get x+3/2
    Indeed. Or instead of long division write:


    2x^2+x-3 = (x-1)(ax+b) -- it's clear by inspection that b = 3 and a = 2.
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  9. #9
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    Quote Originally Posted by Natasha1 View Post
    2x^2 + x - 3

    I know the answer is (2x + 3) (x - 1) from my textbook but I don't understand the method used.
    Isn't this just the problem from this thread with just the letter w replaced with x?
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  10. #10
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    that is freaky! They are the same but it's pure coincidence. I promise
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  11. #11
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    Quote Originally Posted by Wilmer View Post
    Nawwww....I say if not apparent, use the quadratic formula...but that's me!
    You remind me of Armand Assante, Wilmer!

    "When in doubt, get the ABC Robot out!"
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  12. #12
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    Compliments will get you everywhere, Archie !
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