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Math Help - factoring trinomials

  1. #1
    Newbie n_duncan2010's Avatar
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    factoring trinomials

    Factor each trinomial , if not possible , write irreducible

    Dont understand any of it, can anyone teach me how to do this???

    1. 3x ^2 - 5x + 2

    2. 5x ^2 + 17x + 6
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  2. #2
    Junior Member
    Joined
    Apr 2007
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    hope it helps! very long

    okay
    if trinomial
    it must be in the form (lx+a)(mx+b)
    (x-a)(x-b) =lmx^2+(lb+am)x +ab
    to obtain the left hand side from the right
    we can do by (i) inspection or (ii) completing the sq
    (i) we see that lm is the coefficient of x^2, (lb+am) of x and ab of x^0
    therefore we find all the prime factors of lmab, then we can find 2 factors d, e such that de=lmab and d+e =lb+am
    in number 1)3x ^2 - 5x + 2
    lmab=6=2*3 lb+am=-5 therefore one d or e must be negative, but de =6 therefore both d and e must be negative
    d e d+e
    -6 -1 -7
    -3 -2 -5 this is the one we want

    so we do this
    3x ^2 - 5x + 2 =3x ^2 - 3x -2x + 2
    =(3x-2)(x-1)

    we can do the step of finding d and e in our head with practise

    (ii)3x ^2 - 5x + 2 = 3(x ^2 - 5/3x + 2/3)
    we use the fact that (x+a)^2= x^2 +2ax +a^2
    and here 2a= (5/3) a =5/6
    3x ^2 - 5x + 2 = 3(x ^2 - 5/3x + 2/3)
    = 3[(x-5/6)^2 -25/36+2/3]
    =3[((x-5/6)^2 -(1/6)^2]
    we use a^2 - b^2 = (a+b)(a-b)
    3[((x-5/6)^2 -(1/6)^2]=3(x-1)(x-2/3)
    =(3x-2)(x-1) same result
    now u can try the 2nd one, usu we use the 1st method, but the second one always works but is very troublesome!
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  3. #3
    Newbie n_duncan2010's Avatar
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    thank youuu!!

    Thank You sooo!!!! much
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