2x^3 - 11x^2 + 10x + 3
my professor told me to use group factoring but I don't see the proper way to do this
any help is greatly appreciated
"By grouping" certainly wouldn't have been
my first choice! But since we're given that this is the method you're supposed to use, we'll have to try doing it that way.
We can try to create the grouping necessary to factor, by splitting terms. Since the first term is 2x^3 and the last is 3, then it stands to reason that the first term of whatever we'll be factoring out will have a first term with factors of 2 as its coefficient and a second term with factors of 1 as a coefficient. What if we use "2" and "3"?
. . . . .\(\displaystyle 2x^3\, +\, 3x^2\, -\, 14x^2\, ...\)
Why the "3x^2" and the "-14x^2"? Because I'm hoping that 2's and 3's go together (so a 3x^2 goes with a 2x^3), and +3x^2 and -14x^2 equal the original 11x^2. Continuing, the -14x^2 = (-7)(2x^2) will need a (-7)(3x) = -21x, so:
. . . . .\(\displaystyle 2x^3\, +\, 3x^2\, -\, 14x^2\, -\, 21x\, +\, 31x\, ...\)
But "31x" isn't a multiple of 2, so this isn't working. What if we use "2" and "-3"?
. . . . .\(\displaystyle 2x^3\, -\, 3x^2\, -\, 8x^2\, +\, 12x\, -\, 2x\, +\, 3\)
Ooo... that actually worked! So the grouping is:
. . . . .\(\displaystyle (2x^3\, -\, 3x^2)\, -\, (8x^2\, -\, 12x)\, -\, (2x\, -\, 3)\)
...so the factoring is:
. . . . .\(\displaystyle x^2\,(2x\, -\, 3)\, -\, 4x\,(2x\, -\, 3)\, +\, 1\,(2x\, -\, 3)\, =\, ...\)
...and so forth.
Confession: I first did a graph of the corresponding function, y = 2x^3 - 11x^2 + 10x + 3, in my graphing calculator, and found a rational zero (which I could have found using the Rational Roots Test and synthetic division), and then "found" the factorization. I would
strongly suggest that you learn the
real ways of finding zeroes of polynomials (
here), so you can "fudge" the answers on the next test!
