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Thread: Factoring

  1. #1
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    Factoring

    Can anyone factor this please. i'm bad on factoring polynomials.

    6x^3+17x^2+x-10


    thanks...
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  2. #2
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    Quote Originally Posted by Anemori View Post
    Can anyone factor this please. i'm bad on factoring polynomials.

    6x^3+17x^2+x-10


    thanks...
    $\displaystyle 6x^3+17x^2=10-x$

    $\displaystyle x^2(6x+17)=10-x$

    $\displaystyle x^2=\frac{10-x}{6x+17}$

    Positive integers are unhelpful here, but if $\displaystyle x=-1
    $

    $\displaystyle 1=\frac{11}{11}$

    hence $\displaystyle x=-1$ is a solution, so $\displaystyle [x-(-1)]=(x+1)$ is a factor.

    $\displaystyle (x+1)(ax^2+bx+c)=6x^3+17x^2+x-10$

    a=6, c=-10

    $\displaystyle ax^3+bx^2+cx+ax^2+bx+c=ax^3+(a+b)x^2+(b+c)x+c=6x^3 +17x^2+x-10$

    Hence, a+b=17, b=17-a=17-6=11

    or b+c=1, b-10=1, b=10+1=11

    Using this technique, "a" and "c" are found immediately,
    "b" can be found from either of the other two co-efficients using "a" or "c".

    $\displaystyle 6x^3+17x^2+x-10=(x+1)\left(6x^2+11x-10\right)$

    The quadratic is easier to factorise.

    We get

    $\displaystyle 6x^3+17x^2+11x-10=(x+1)(3x-2)(2x+5)$
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    $\displaystyle 6x^3+17x^2=10-x$

    $\displaystyle x^2(6x+17)=10-x$

    $\displaystyle x^2=\frac{10-x}{6x+17}$

    Positive integers are unhelpful here, but if $\displaystyle x=-1
    $

    $\displaystyle 1=\frac{11}{11}$

    hence $\displaystyle x=-1$ is a solution, so $\displaystyle [x-(-1)]=(x+1)$ is a factor.

    $\displaystyle (x+1)(ax^2+bx+c)=6x^3+17x^2+x-10$

    a=6, c=-10

    $\displaystyle ax^3+bx^2+cx+ax^2+bx+c=ax^3+(a+b)x^2+(b+c)x+c=6x^3 +17x^2+x-10$

    Hence, a+b=17, b=17-a=17-6=11

    or b+c=1, b-10=1, b=10+1=11

    Using this technique, "a" and "c" are found immediately,
    "b" can be found from either of the other two co-efficients using "a" or "c".

    $\displaystyle 6x^3+17x^2+x-10=(x+1)\left(6x^2+11x-10\right)$

    The quadratic is easier to factorise.

    We get

    $\displaystyle 6x^3+17x^2+11x-10=(x+1)(3x-2)(2x+5)$
    oh ok thanks....
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  4. #4
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    Quote Originally Posted by Anemori View Post
    Can anyone factor this please. i'm bad on factoring polynomials.

    6x^3+17x^2+x-10


    thanks...
    By inspection x = -1 is a zero of 6x^3+17x^2+x-10 and so x + 1 is a factor. Now divide x + 1 into 6x^3+17x^2+x-10 to get the quadratic factor and then factorise (if possible) the quadratic.
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