${\left(3 x-2\right)}^2 \left(x^3\+2 x\right)$

${\left(3x-2\right)}^2\left(x^3\+2x\right)$

$\stackrel{\text{expand}}{\=}$

$9x^5-12x^4\+22x^3-24x^2\+8x$

$\stackrel{\text{factor}}{\=}$

${\left(3x-2\right)}^{2}x\left(x^2\+2\right)$

Loading Student:-Basics

Obtain an annotated stepwise expansion of the given expression

${\left(3 x-2\right)}^2 \left(x^3\+2 x\right)$$\stackrel{\text{expand steps}}{\=}$$\begin{array}{c}{\left(3\cdot x-2\right)}^2\cdot \left(x^3+2\cdot x\right)\\ \=\left(3x-2\right)\cdot \left(3x-2\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Rewrite exponentiation as multiplication}\right)\\ \=\left(3x\cdot \left(3x-2\right)-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Distributive multiply}\right)\\ \=\left(3x\cdot \left(3x\right)+3x\cdot \left(\mathrm{-2}\right)-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Distributive multiply}\right)\\ \=\left(9\cdot x\cdot x+3x\cdot \left(\mathrm{-2}\right)-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=\left(9\cdot x^2+3x\cdot \left(\mathrm{-2}\right)-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply terms to exponential form}\right)\\ \=\left(9x^2-6\cdot x-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=\left(9x^2-6x+\left(\left(\mathrm{-2}\right)\cdot \left(3x\right)-2\cdot \left(\mathrm{-2}\right)\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Distributive multiply}\right)\\ \=\left(9x^2-6x+\left(\left(\mathrm{-6}\right)\cdot x-2\cdot \left(\mathrm{-2}\right)\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=\left(9x^2-6x+\left(-6x+4\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=\left(9x^2-6x+\left(4-6x\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Reorder terms}\right)\\ \=\left(9x^2-12x+4\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Add terms}\right)\\ \=\left(9x^2-12x+4\right)\cdot x^3+\left(9x^2-12x+4\right)\cdot \left(2x\right)& \left(\text{Distributive multiply}\right)\\ \=x^3\cdot \left(9x^2\right)+x^3\cdot \left(-12x\right)+x^3\cdot 4+\left(9x^2-12x+4\right)\cdot \left(2x\right)& \left(\text{Distributive multiply}\right)\\ \=9\cdot x^5+x^3\cdot \left(-12x\right)+x^3\cdot 4+\left(9x^2-12x+4\right)\cdot \left(2x\right)& \left(\text{Add exponents with common base}\right)\\ \=9x^5-12\cdot x^4+x^3\cdot 4+\left(9x^2-12x+4\right)\cdot \left(2x\right)& \left(\text{Add exponents with common base}\right)\\ \=9x^5-12x^4+4x^3+\left(2x\cdot \left(9x^2\right)+2x\cdot \left(-12x\right)+2x\cdot 4\right)& \left(\text{Distributive multiply}\right)\\ \=9x^5-12x^4+4x^3+\left(18\cdot x\cdot x^2+2x\cdot \left(-12x\right)+2x\cdot 4\right)& \left(\text{Multiply constants}\right)\\ \=9x^5-12x^4+4x^3+\left(18\cdot x^3+2x\cdot \left(-12x\right)+2x\cdot 4\right)& \left(\text{Add exponents with common base}\right)\\ \=9x^5-12x^4+4x^3+\left(18x^3-24\cdot x\cdot x+2x\cdot 4\right)& \left(\text{Multiply constants}\right)\\ \=9x^5-12x^4+4x^3+\left(18x^3-24\cdot x^2+2x\cdot 4\right)& \left(\text{Multiply terms to exponential form}\right)\\ \=9x^5-12x^4+4x^3+\left(18x^3-24x^2+8\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=9x^5-12x^4+22x^3-24x^2+8x& \left(\text{Add terms}\right)\end{array}$

Factor the expanded form

$9x^5-12x^4+22x^3-24x^2+8x$$\stackrel{\text{factor}}{\=}$$x\left(x^2+2\right){\left(3x-2\right)}^2$

Table A-10.1(a)   Syntax-free annotated stepwise expansion and factoring

$\mathrm{Student}:-\mathrm{Basics}:-\mathrm{ExpandSteps}\left({\left(3 x-2\right)}^2 \left(x^3\+2 x\right)\right)$

$\begin{array}{c}{\left(3\cdot x-2\right)}^2\cdot \left(x^3+2\cdot x\right)\\ \=\left(3x-2\right)\cdot \left(3x-2\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Rewrite exponentiation as multiplication}\right)\\ \=\left(3x\cdot \left(3x-2\right)-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Distributive multiply}\right)\\ \=\left(3x\cdot \left(3x\right)+3x\cdot \left(\mathrm{-2}\right)-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Distributive multiply}\right)\\ \=\left(9\cdot x\cdot x+3x\cdot \left(\mathrm{-2}\right)-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=\left(9\cdot x^2+3x\cdot \left(\mathrm{-2}\right)-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply terms to exponential form}\right)\\ \=\left(9x^2-6\cdot x-2\cdot \left(3x-2\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=\left(9x^2-6x+\left(\left(\mathrm{-2}\right)\cdot \left(3x\right)-2\cdot \left(\mathrm{-2}\right)\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Distributive multiply}\right)\\ \=\left(9x^2-6x+\left(\left(\mathrm{-6}\right)\cdot x-2\cdot \left(\mathrm{-2}\right)\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=\left(9x^2-6x+\left(-6x+4\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=\left(9x^2-6x+\left(4-6x\right)\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Reorder terms}\right)\\ \=\left(9x^2-12x+4\right)\cdot \left(x^3+2\cdot x\right)& \left(\text{Add terms}\right)\\ \=\left(9x^2-12x+4\right)\cdot x^3+\left(9x^2-12x+4\right)\cdot \left(2x\right)& \left(\text{Distributive multiply}\right)\\ \=x^3\cdot \left(9x^2\right)+x^3\cdot \left(-12x\right)+x^3\cdot 4+\left(9x^2-12x+4\right)\cdot \left(2x\right)& \left(\text{Distributive multiply}\right)\\ \=9\cdot x^5+x^3\cdot \left(-12x\right)+x^3\cdot 4+\left(9x^2-12x+4\right)\cdot \left(2x\right)& \left(\text{Add exponents with common base}\right)\\ \=9x^5-12\cdot x^4+x^3\cdot 4+\left(9x^2-12x+4\right)\cdot \left(2x\right)& \left(\text{Add exponents with common base}\right)\\ \=9x^5-12x^4+4x^3+\left(2x\cdot \left(9x^2\right)+2x\cdot \left(-12x\right)+2x\cdot 4\right)& \left(\text{Distributive multiply}\right)\\ \=9x^5-12x^4+4x^3+\left(18\cdot x\cdot x^2+2x\cdot \left(-12x\right)+2x\cdot 4\right)& \left(\text{Multiply constants}\right)\\ \=9x^5-12x^4+4x^3+\left(18\cdot x^3+2x\cdot \left(-12x\right)+2x\cdot 4\right)& \left(\text{Add exponents with common base}\right)\\ \=9x^5-12x^4+4x^3+\left(18x^3-24\cdot x\cdot x+2x\cdot 4\right)& \left(\text{Multiply constants}\right)\\ \=9x^5-12x^4+4x^3+\left(18x^3-24\cdot x^2+2x\cdot 4\right)& \left(\text{Multiply terms to exponential form}\right)\\ \=9x^5-12x^4+4x^3+\left(18x^3-24x^2+8\cdot x\right)& \left(\text{Multiply constants}\right)\\ \=9x^5-12x^4+22x^3-24x^2+8x& \left(\text{Add terms}\right)\end{array}$

Table A-4.1(b)   Application of the Student Basics package command ExpandSteps