Home work Help: Algebraic Expression

[O'Brian]

Hey guys. These kinds of problems are stressing me out! Please help if possible.

Simplify the Expression


1.)

(2x^ - 3x + 1)(4)(3x+2)^3(3) + (3x+2)^4(4x - 3)


2.)

((x^2) - 4)^1/2(3)(2x + 1)^2(2) + (2x + 1)^3(1/2)((x^2) - 4)^-1/2(3)

Thank you!

 

[soroban]

Hello, O'Brian!

At first, I wonder why these Algebra problems are posted under Calculus.
Then I saw you used the Product Rule, didn't you?

You need to do some Factoring . . .

\(\displaystyle 1)\;(2x^2\,-\,3x\,+\,1)(4)(3x\,+\,2)^3(3)\:+\3x\,+\,2)^4(4x\,-\, 3)\)

We have: \(\displaystyle \,12(3x\,+\,2)^3(2x^2\,-3x\,+\,1)\:+\3x\,+\,2)^4(4x\,-\,3)\)

Take out common factors: \(\displaystyle \,(3x\,+\,2)^3\,\cdot\,\left[12(2x^3\,-\,3x\,+\,1)\:+\3x\,+\,2)(4x\,-\,3)\right]\)

And we have: $\displaystyle \,(3x\,+\,2)^3\,\cdot\,\left[24x^2\,-\,36x\,+\,12\,+\,12x^2\,-\,x\,-\,6\right]$

$\displaystyle \;\;= \;(3x\,+\,2)^3\,(36x^2\,-\,37x\,+\,6)$

\(\displaystyle 2) \;(x^2\,-\, 4)^{\frac{1}{2}}(3)(2x\,+\,1)^2(2)\:+\2x\,+\,1)^3\left(\frac{1}{2}\right)(x^2\,-\,4)^{-\frac{1}{2}}(3)\)

We have: $\displaystyle \,6(x^2\,-\,4)^{\frac{1}{2}}(2x\,+\,1)^2\:+\;\frac{3}{2}(x^2\,-\,4)^{-\frac{1}{2}}(2x\,+\,1)^3$

Factor: $\displaystyle \,\frac{3}{2}(x^2\,-\,4)^{-\frac{1}{2}}(2x\,+\,1)^2\,\cdot\,\left[4(x^2\,-\,4)\,+\,(2x\,+\,1)\right]$

and we have: $\displaystyle \,\frac{3}{2}(x^2\,-\,4)(2x\,+\,1)^{-\frac{1}{2}}(2x\,+\,1)^2\,\cdot\,\left(4x^2\,-\,16x\,+\,2x\,+\,1\right)$

$\displaystyle \;\;\;= \;\frac{3}{2}(x^2\,-\,4)^{-\frac{1}{2}}(2x\,+\,1)\,(4x^2\,-\,14x\,+\,1)$