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Math Help - two more questions

  1. #1
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    two more questions

    how do you expand and simplify in the easiest way:

    (x-1)(x+3)(x-3)(x-7)

    and

    factorise:
    (1-a^2)(1-b^2) - 4ab

    and X^4 + 4
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  2. #2
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    Re: two more questions

    You multiply it out. What do you mean "the easiest way"? I guess you can do it term-by-term using the Distributive Property:

    $\begin{align*}(x-1)(x+3)(x-3)(x-7) & = x(x+3)(x-3)(x-7) - 1(x+3)(x-3)(x-7) \\ & = x\cdot x(x-3)(x-7) + x\cdot 3(x-3)(x-7) - 1(x+3)(x-3)(x-7) \\ & = \vdots\end{align*}$

    For $(1-a^2)(1-b^2) - 4ab$, multiply out, then factor.

    For $x^4+4$, it is irreducible over the reals. Are you allowed to factor over some larger field, like the complex numbers?
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  3. #3
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    Re: two more questions

    Quote Originally Posted by richardkim4668 View Post
    how do you expand and simplify in the easiest way:

    (x-1)(x+3)(x-3)(x-7)

    and

    factorise:
    (1-a^2)(1-b^2) - 4ab

    and X^4 + 4
    I am afraid that products generally require mechanical working out and that few tricks are available to reduce the work. Factoring is always a matter of "seeing."

    HOWEVER, all these problems involve a difference of powers and so permit reduction of work for your first problem.

    $\displaystyle p^n - q^n \equiv (p - q) * \sum_{i=0}^{n}p^{(n - i)}q^i \implies p^2 - q^2 \equiv (p - q)(p + q).$

    In terms of the first problem, you can use the difference of squares in reverse to reduce the problem from one involving four factors to one involving three factors.

    A clue for the third problem. Do you see that:

    $x^4 + 4 = x^4 - (-4).$ Now you have a difference of fourth powers.

    EDIT: As slipeternal pointed out, $x^4 - (-4)$ is not reducible over the reals, but it is in terms of complex numbers.
    Last edited by JeffM; April 13th 2014 at 07:43 AM. Reason: Address previous post
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  4. #4
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    Re: two more questions

    Hello, richardkim4668!

    The factoring problems are quite tricky.


    Factor: . (1-a^2)(1-b^2) - 4ab

    Expand: . 1 - a^2 - b^2 + a^2b^2 - 4ab

    We have: . a^2b^2 - 2ab + 1 - a^2 - 2ab - b^2

    . . . . =\;(a^2b^2 - 2ab + 1) - (a^2 + 2ab + b^2)

    Factor: . (ab - 1)^2 - (a + b)^2 . .
    Difference of squares!

    Factor: . \big([ab-1] - [a+b]\big)\big([ab-1] +[a+b]\big)

    . . . . =\; (ab-1 - a - b)(ab-1 +a+b)




    Factor: . x^4 + 4

    Add and subtract 4x^2\!:\;\;x^4 \,{\color{red}+\,4x^2} + 4 \,{\color{red}-\, 4x^2}

    We have: . (x^2 + 2)^2 - (2x)^2 . .
    Difference of squares!


    Therefore: . (x^2 + 2 - 2x)(x^2+2 + 2x)
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  5. #5
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    Re: two more questions

    Wow soroban. Very impressed.
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  6. #6
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    Re: two more questions

    Soroban, how did you know to add and subtract 4x??
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  7. #7
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    Re: two more questions

    Hello, sakonpure6!

    Soroban, how did you know to add and subtract 4x^2?

    Many years ago, I saw someone else factor x^4 + 1.

    I said it was impossible; we have the sum of two squares.

    Then they added and subtracted 2x^2.

    . . x^4 + 2x^2 + 1 - 2x^2 \;=\;(x^2+1)^2 - (\sqrt{2}x)^2 .
    Difference of squares!


    And factored: . (x^2 + 1 - \sqrt{2}x)(x^2+1 + \sqrt{2}x)

    . . . . . . . . . =\;(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Since then I can factor the sum of two even powers.
    (We sort of "complete the square" for the middle term.)

    x^8 + 9 \;=\;x^8 \,{\color{red}+\,6x^4} + 9\,{\color{red}-\,6x^4}

    . . . . . . =\;(x^4 + 3)^2 - (\sqrt{6}x^2)^2

    . . . . . . =\;(x^4 - \sqrt{6}x^2 + 3)(x^4 + \sqrt{6}x^2 + 3)
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