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?
Hello, symmetry!
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?
Hello again, symmetry!
If the problem is correctly stated, it's a sneaky one . . .
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) $