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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- Jan 20th 2007, 06:28 AMsymmetryFactoring
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? - Jan 20th 2007, 07:04 AMOReilly
- Jan 20th 2007, 07:46 AMSoroban
Hello, symmetry!

Quote:

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

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 $\displaystyle 2$ . . . it's a**factoring**problem, remember?

- Jan 20th 2007, 09:02 AMsymmetryok
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? - Jan 20th 2007, 09:52 AMSoroban
Hello again, symmetry!

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

Quote:

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

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

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

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

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

- Jan 20th 2007, 12:51 PMsymmetryok
Soroban,

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

Thanks!