# Factoring

• Jan 20th 2007, 07:28 AM
symmetry
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?
• Jan 20th 2007, 08:04 AM
OReilly
Quote:

Originally Posted by symmetry
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?
• Jan 20th 2007, 08:46 AM
Soroban
Hello, symmetry!

Quote:

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?

• Jan 20th 2007, 10:02 AM
symmetry
ok

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, 10:52 AM
Soroban
Hello again, symmetry!

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

Quote:

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!

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

• Jan 20th 2007, 01:51 PM
symmetry
ok
Soroban,

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

Thanks!