# How do you multiply (x-2)(2x+3)(3-x)?

Jul 11, 2015

Construct the coefficient of each power of $x$ in descending order to find:

$\left(x - 2\right) \left(2 x + 3\right) \left(3 - x\right) = - 2 {x}^{3} + 7 {x}^{2} + 3 x - 18$

#### Explanation:

For each power of $x$ in descending order, pick out the different combinations of terms, using one from each binomial, such that when multiplied together they will give the target power of $x$ and add them together. For brevity, omit the $x$'s as you are multiplying and adding the coefficients...

So:

$\textcolor{b l u e}{{x}^{3}}$ : $\left(1 \cdot 2 \cdot - 1\right) = \textcolor{red}{- 2}$

$\textcolor{b l u e}{{x}^{2}}$ : $\left(1 \cdot 2 \cdot 3\right) + \left(1 \cdot 3 \cdot - 1\right) + \left(- 2 \cdot 2 \cdot - 1\right)$

$= 6 - 3 + 4 = \textcolor{red}{7}$

$\textcolor{b l u e}{x}$ : $\left(1 \cdot 3 \cdot 3\right) + \left(- 2 \cdot 2 \cdot 3\right) + \left(- 2 \cdot 3 \cdot - 1\right)$

$= 9 - 12 + 6 = \textcolor{red}{3}$

$\textcolor{b l u e}{1}$ : $\left(- 2 \cdot 3 \cdot 3\right) = \textcolor{red}{- 18}$

Hence $\left(x - 2\right) \left(2 x + 3\right) \left(3 - x\right) = - 2 {x}^{3} + 7 {x}^{2} + 3 x - 18$

Check: Try $x = 1$ ...

$\left(x - 2\right) \left(2 x + 3\right) \left(3 - x\right) = \left(1 - 2\right) \left(2 + 3\right) \left(3 - 1\right)$

$= - 1 \cdot 5 \cdot 2 = \textcolor{red}{- 10}$

$- 2 {x}^{3} + 7 {x}^{2} + 3 x - 18 = - 2 + 7 + 3 - 18 = \textcolor{red}{- 10}$