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Math Help - Factoring

  1. #1
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    Factoring

    Factor 2x^3-98a^2x+28x^2+98x

    QUESTION:

    Would it make sense to divide EVERY coefficient in the polynomial by the leading coefficient of 2 to reduce the BIG numbers?

    Is there an easier method?
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  2. #2
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    Quote Originally Posted by symmetry View Post
    Factor 2x^3-98a^2x+28x^2+98x

    QUESTION:

    Would it make sense to divide EVERY coefficient in the polynomial by the leading coefficient of 2 to reduce the BIG numbers?

    Is there an easier method?
    Yes, you can divide with 2, so you get:
    2(x^3  - 49a^2 x + 14x^2  + 49x)

    I was just wondering, does a belong in expression? Is that a typo maybe?
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  3. #3
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    Hello, symmetry!

    Factor: . 2x^3-98a^2x+28x^2+98x

    Would it make sense to divide EVERY coefficient in the polynomial
    by the leading coefficient of 2 to reduce the BIG numbers?

    Well, I wouldn't call it "dividing".

    You can factor out a 2 . . . it's a factoring problem, remember?

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  4. #4
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    ok

    Thank you for your tips.

    To OReilly:

    The variable a belongs to the second term.

    If it is a typo, I would not know.

    So, what is the final answer?
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  5. #5
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    Hello again, symmetry!

    If the problem is correctly stated, it's a sneaky one . . .


    Factor: . 2x^3-98a^2x+28x^2+98x

    Factor 2x out of each term: . 2x\left[x^2 - 49a^2 + 14x + 49\right]

    We have: . 2x\left[\underbrace{x^2 + 14x + 49} \:- \:\underbrace{49a^2}\right]

    . . . . . . . . = \;2x\left[(x + 7)^2 \;- \;(7a)^2\right] . . . a difference of squares!

    . . . . . . . . = \;2x\,(x+ 7 - 7a)\,(x + 7 + 7a)

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  6. #6
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    ok

    Soroban,

    I guess the answer to OReilly's question is: Yes, the letter a is part of the question.

    Thanks!
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