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Thread: Factorising

  1. #1
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    Factorising

    How would I factor these:

    1) $\displaystyle x^7+x$

    2) x^12 - y^12

    And I'm also having trouble with this problem:

    1) If $\displaystyle x+y=1$ and $\displaystyle x^3+y^3=19$. What is $\displaystyle x^2+y^2$
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  2. #2
    Senior Member topher0805's Avatar
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    1) divide by x to get: $\displaystyle x(1+6x)$

    2) this is like a really complicated difference of squares. $\displaystyle x^2-y^2 = (x+y)(x-y)$, etc.

    3) use the first equation to get $\displaystyle x = 1 - y$, then plug that into the second equation to figure out $\displaystyle y$. Once you have $\displaystyle y$, plug it into the first equation to find $\displaystyle x$, then plug them both into $\displaystyle x^2 + y^2$ to get your answer.
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  3. #3
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by acevipa View Post
    How would I factor these:

    1) $\displaystyle x^7+x$

    2) x^12 - y^12

    And I'm also having trouble with this problem:

    1) If $\displaystyle x+y=1$ and $\displaystyle x^3+y^3=19$. What is $\displaystyle x^2+y^2$
    Question 1

    $\displaystyle x^7+x=x(x^6+1)=x((x^2)^3+(1)^3)=x(x^2+1)(x^4-x^2+1)$


    Question 2

    $\displaystyle x^{12}-y^{12}=(x^6-y^6)(x^6+y^6)=((x^2)^3+(y^2)^3)(x^3-y^3)(x^3+y^3)$

    $\displaystyle =(x^2+y^2)(x^4-x^2y^2+y^4)(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$

    Last Question

    $\displaystyle (x+y)^3=x^3+y^3+3xy(x+y)$

    Subbing in values:

    $\displaystyle 1^3=19+3xy(1) \Leftrightarrow 3xy=-18 \Leftrightarrow xy=-6$

    Also, $\displaystyle (x+y)^2=x^2+y^2+2xy$

    Subbing in again,

    $\displaystyle (1)^2=x^2+y^2+2(-6)$

    $\displaystyle \Rightarrow x^2+y^2=13$
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