[MOVED] Factoring expression: (3x - 2)^-4 (x + 3) + ....
- Thread starter kC_Lee
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Re: Factoring expression
Hello, kC_Lee!
This is not a Calculus problem . . .
\(\displaystyle \;\;\)but I bet it came from the Product Rule.
We have: \(\displaystyle \,ab^{-4}\,+\,a^2b^{-3}\) . . . which factors: \(\displaystyle \,ab^{-4}(1\,+\,ab)\)
Hence: \(\displaystyle \,(x\,+\,3)(3x\,-\,2)^{-4}\:+\
x\,+\,3)^2 (3x\,-\,2)^{-3} \;= \;(x\,+\,3)(3x\,-\,2)^{-4}\left[1\,+\,(x\,+\,3)(3x\,-\,2)\right]\)
. . \(\displaystyle = \;(x\,+\,3)(3x\,-\,2)^{-4}\left[1\,+\,3x^2\,+\,7x\,-\,6\right] \;= \;(x\,+\,3)(3x\,-\,2)(3x^2\,+\,7x\,-\,5)\)
Hello, kC_Lee!
This is not a Calculus problem . . .
\(\displaystyle \;\;\)but I bet it came from the Product Rule.
How would you factor: \(\displaystyle \,(x\,+\,3)(3x\,-\,2)^{-4}\:+\
We have: \(\displaystyle \,ab^{-4}\,+\,a^2b^{-3}\) . . . which factors: \(\displaystyle \,ab^{-4}(1\,+\,ab)\)
Hence: \(\displaystyle \,(x\,+\,3)(3x\,-\,2)^{-4}\:+\
x\,+\,3)^2 (3x\,-\,2)^{-3} \;= \;(x\,+\,3)(3x\,-\,2)^{-4}\left[1\,+\,(x\,+\,3)(3x\,-\,2)\right]\). . \(\displaystyle = \;(x\,+\,3)(3x\,-\,2)^{-4}\left[1\,+\,3x^2\,+\,7x\,-\,6\right] \;= \;(x\,+\,3)(3x\,-\,2)(3x^2\,+\,7x\,-\,5)\)