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Math Help - Factoring Help!

  1. #1
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    Factoring Help!

    I'm having trouble doing these problems:

    1. Solve the equation

    What are the 2 solutions? I get -1 and 4 but apparently I'm wrong

    2. The polynomial can be factored into the product of three polynomials, A*B*C. Find A B and C.

    No clue how to do this one.

    3. The polynomial can be factored into the product of two polynomials, A*B. Find A and B.

    Now I got one of the factors to be 9x-8y but i can't find the other.
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  2. #2
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    Krizalid's Avatar
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    Quote Originally Posted by badandy328 View Post
    I'm having trouble doing these problems:

    1. Solve the equation

    What are the 2 solutions? I get -1 and 4 but apparently I'm wrong

    2. The polynomial can be factored into the product of three polynomials, A*B*C. Find A B and C.

    No clue how to do this one.

    3. The polynomial can be factored into the product of two polynomials, A*B. Find A and B.

    Now I got one of the factors to be 9x-8y but i can't find the other.
    For the first one, multiply both sides by 6(x+1)(x+2)

    For the third one, use the following identity a^3-b^3=(a-b)(a^2+ab+b^2)

    I'm too lazy now for the second one
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    For the first one, multiply both sides by 6(x+1)(x+2)

    For the third one, use the following identity a^3-b^3=(a-b)(a^2+ab+b^2)

    I'm too lazy now for the second one
    For the first one though I end up with 0= x^2+3x-16
    I need to get two solutions, and I don't think this could be factored. Am I doing something wrong?
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  4. #4
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    Hello, badandy328!


    2. The polynomial . 81x^3 + 486x^2 - x - 6 can be factored
    into the product of three polynomials, A\cdot B\cdot C
    Find A,\,B\text{ and }C.
    Factor "by grouping"


    \text{We have: }\;81x^2\underbrace{(x+6)}_{\text{common}} - \underbrace{(x + 6)}_{\text{factor}}

    \text{Factor: }\;(x + 6)\underbrace{(81x^2 - 1)}_{\text{diff. of squares}}

    \text{Factor: }\;(x + 6)(9x-1)(9x+1)

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  5. #5
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    thanks! All i need now is number 1. I can't figure it out
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  6. #6
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    Hello, badandy328!

    Did you take Krizalid's advice for the first one?



    1. Solve the equation: . \frac{1}{x+1} - \frac{1}{x+2} \;=\;\frac{1}{6}
    Multiply through by 6(x+1)(x+2)

    . . 6(x+1)(x+2)\cdot\frac{1}{x+1} - 6(x+1)(x+2)\cdot\frac{1}{x+2} \;=\;6(x+1)(x+2)\cdot\frac{1}{6}


    Reduce: . 6(x+2) - 6(x+1) \;=\;(x+1)(x+2)


    Expand: . 6x + 12 - 6x - 6 \;=\;x^2 + 3x + 2


    We have the quadratic: . x^2 + 3x - 4 \;=\;0

    . . which factors: . (x - 1)(x + 4) \;=\;0

    . . and has roots: . x \:=\:1,\,-4

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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, badandy328!

    Did you take Krizalid's advice for the first one?



    Multiply through by 6(x+1)(x+2)

    . . 6(x+1)(x+2)\cdot\frac{1}{x+1} - 6(x+1)(x+2)\cdot\frac{1}{x+2} \;=\;6(x+1)(x+2)\cdot\frac{1}{6}


    Reduce: . 6(x+2) - 6(x+1) \;=\;(x+1)(x+2)


    Expand: . 6x + 12 - 6x - 6 \;=\;x^2 + 3x + 2


    We have the quadratic: . x^2 + 3x - 4 \;=\;0

    . . which factors: . (x - 1)(x + 4) \;=\;0

    . . and has roots: . x \:=\:1,\,-4

    OHHH i forgot to bring over that subtraction sign. I got -1 and 4. Thank you so much.
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