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Math Help - factorize

  1. #1
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    factorize

    Hello guys just a few questions

    1.) Factorise completely 2x^3 - 20x^2 + 46x - 28

    here i took out the common factor to give 2(x^3 - 10x^2 + 23x - 14)

    now when i began trying factors i wasnt sure whether to leave the 2 outside like so: 2((1)^3 - 10(1)^2 + 23(1) - 14))

    or just (1)^3 - 10(1)^2 + 23(1) -14

    __________________________________________________ _________

    now the second part is long division... do i leave the two outside when long dividing or do i have to expand i back into the sum?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by andrew2322 View Post
    now the second part is long division... do i leave the two outside when long dividing or do i have to expand i back into the sum?
    You can leave the 2 outside. If you don't want this common factor to appear during the division, you can even factor x^3-10x^2+23x-14 and, once it's done, say that 2x^3-20x^2+46x-28=2(x^3-10x^2+23x-14)=\ldots

    Good luck
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  3. #3
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    hey

    me confused =( and another thing, just for general knowledge, if u had something like this : 3(2x-1)^3 -4 could u times the 3(2x-1)^3 by the three first, or do u have to cube it first?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by andrew2322 View Post
    me confused =(
    P(x)=2x^3-20x^2+46x-28=2(x^3-10x^2+23x-14) and 1 is a root of this polynomial.

    The division of x^3-10x^2+23x-14 by x-1 will give something like x^3-10x^2+23x-14=(x-1)(ax^2+bx+x) hence P(x)=2(x^3-10x^2+23x-14)=2(x-1)(ax^2+bx+x) : P has been factored too. That's why you only need to factor x^3-10x^2+23x-14.

    Is it clearer ?

    just for general knowledge, if u had something like this : 3(2x-1)^3 -4 could u times the 3(2x-1)^3 by the three first, or do u have to cube it first?
    I don't understand what you mean.
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  5. #5
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    hey

    with the 3(2x-1)^3 bit, can u times the 3 in first or do u have to cube (2x-1) first
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