Factoring X^4 - 2X^3 - 8X^2 + 18X - 9
- Thread starter ca.chick
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- Feb 4, 2004
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I'd start with the Rational Roots Test, and test the potential roots with synthetic division. I'd keep going until I got down to a quadratic.
If you are supposed to use some other method, please reply with specifics. Thank you!
Eliz.
If you are supposed to use some other method, please reply with specifics. Thank you!
Eliz.
You can group this.
x^4 - 2x^3 - 8x^2 + 18x - 9 = (x^4 - 8x^2 - 9) + 18x - 2x^3
The expression in brackets is (x^2 - 9)(x^2 + 1), so it all becomes:
(x^2 - 9)(x^2 + 1) + (18x - 2x^3) = (x^2 - 9)(x^2 + 1) - (2x^3 - 18x)
Finish off by collecting (or "grouping") the common factor. Can you see it?
x^4 - 2x^3 - 8x^2 + 18x - 9 = (x^4 - 8x^2 - 9) + 18x - 2x^3
The expression in brackets is (x^2 - 9)(x^2 + 1), so it all becomes:
(x^2 - 9)(x^2 + 1) + (18x - 2x^3) = (x^2 - 9)(x^2 + 1) - (2x^3 - 18x)
Finish off by collecting (or "grouping") the common factor. Can you see it?
Hello, ca.chick!
Another grouping . . .
We are given: \(\displaystyle \:x^4\,-\,2x^3\,-\,8x^2\,+\,18x\,-\,9\)
. . . . . . . \(\displaystyle =\:x^4\,-\,2x^3\,+\,\overbrace{x^2\,-\,9x^2}\,+\,18x\,-\,9\)
. . . . . . . \(\displaystyle = \:x^2(x^2\,-\,2x\,+\,1)\,-\,9(x^2\,-\,2x\,+\,1)\)
. . . . . . . \(\displaystyle = \x^2\,-\,2x\,+\,1)(x^2\,-\,9)\)
. . . . . . . \(\displaystyle = \x\,-\,1)^2(x\,-\,3)(x\,+\,3)\)
Another grouping . . .
We are given: \(\displaystyle \:x^4\,-\,2x^3\,-\,8x^2\,+\,18x\,-\,9\)
. . . . . . . \(\displaystyle =\:x^4\,-\,2x^3\,+\,\overbrace{x^2\,-\,9x^2}\,+\,18x\,-\,9\)
. . . . . . . \(\displaystyle = \:x^2(x^2\,-\,2x\,+\,1)\,-\,9(x^2\,-\,2x\,+\,1)\)
. . . . . . . \(\displaystyle = \
x^2\,-\,2x\,+\,1)(x^2\,-\,9)\). . . . . . . \(\displaystyle = \
x\,-\,1)^2(x\,-\,3)(x\,+\,3)\)