# How do you multiply (2x²-3x+2)(3x³-x²+√2x)?

Aug 30, 2015

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

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

$= 6 {x}^{5} - 11 {x}^{4} + \left(9 + 2 \sqrt{2}\right) {x}^{3} - \left(2 + 3 \sqrt{2}\right) {x}^{2} + 2 \sqrt{2} x$

#### Explanation:

For each power of $x$ in descending order, pick out pairs of terms - one from each trinomial - whose product results in that power of $x$ and add them together:

$\textcolor{g r e e n}{{x}^{5}}$: $2 {x}^{2} \cdot 3 {x}^{3} = \textcolor{b l u e}{6 {x}^{5}}$

$\textcolor{g r e e n}{{x}^{4}}$: $\left(2 {x}^{2} \cdot - {x}^{2}\right) + \left(- 3 x \cdot 3 {x}^{3}\right) = - 2 {x}^{4} - 9 {x}^{4} = \textcolor{b l u e}{- 11 {x}^{4}}$

$\textcolor{g r e e n}{{x}^{3}}$: $\left(2 {x}^{2} \cdot \sqrt{2} x\right) + \left(- 3 x \cdot - {x}^{2}\right) + \left(2 \cdot 3 {x}^{3}\right)$

$= 2 \sqrt{2} {x}^{3} + 3 {x}^{3} + 6 {x}^{3} = \textcolor{b l u e}{\left(9 + 2 \sqrt{2}\right) {x}^{3}}$

$\textcolor{g r e e n}{{x}^{2}}$: $\left(- 3 x \cdot \sqrt{2} x\right) + \left(2 \cdot - {x}^{2}\right) = - 3 \sqrt{2} {x}^{2} - 2 {x}^{2}$

$= \textcolor{b l u e}{- \left(2 + 3 \sqrt{2}\right) {x}^{2}}$

$\textcolor{g r e e}{x}$: $2 \cdot \sqrt{2} x = \textcolor{b l u e}{2 \sqrt{2} x}$

$\textcolor{g r e e n}{1}$: $\textcolor{b l u e}{0}$

So:

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

$= 6 {x}^{5} - 11 {x}^{4} + \left(9 + 2 \sqrt{2}\right) {x}^{3} - \left(2 + 3 \sqrt{2}\right) {x}^{2} + 2 \sqrt{2} x$

Actually in practice, I would just work with the coefficients:

$\textcolor{g r e e n}{{x}^{5}}$: $2 \cdot 3 = \textcolor{b l u e}{6}$

$\textcolor{g r e e n}{{x}^{4}}$: $\left(2 \cdot - 1\right) + \left(- 3 \cdot 3\right) = - 2 - 9 = \textcolor{b l u e}{- 11}$

$\textcolor{g r e e n}{{x}^{3}}$: $\left(2 \cdot \sqrt{2}\right) + \left(- 3 \cdot - 1\right) + \left(2 \cdot 3\right) = 2 \sqrt{2} + 3 + 6 = \textcolor{b l u e}{9 + 2 \sqrt{2}}$

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

$\textcolor{g r e e n}{x}$: $2 \cdot \sqrt{2} = \textcolor{b l u e}{2 \sqrt{2}}$

$\textcolor{g r e e n}{1}$: $\textcolor{b l u e}{0}$

...and I would just write out the sum as I went along:

$6 {x}^{5} - 11 {x}^{4} + \left(9 + 2 \sqrt{2}\right) {x}^{3} - \left(2 + 3 \sqrt{2}\right) {x}^{2} + 2 \sqrt{2} x$

Aug 30, 2015

$6 {x}^{5} - 11 {x}^{4} + \left(9 + 2 \sqrt{2}\right) {x}^{3} - \left(2 + 3 \sqrt{2}\right) {x}^{2} + 2 \sqrt{2} x$
{: (color(orange)(" X")," | ", color(orange)(2x^2), color(orange)(-3x), color(orange)(+2)), (" -"," | ", "-", "-", "-"), (color(orange)(3x^3)," | ", 6x^5,color(red)( -9x^4), color(blue)(+6x^3)), (color(orange)(-x^2)," | ",color(red)( -2x^4),color(blue)(+3x^3), color(green)(-2x^2)), (color(orange)(+sqrt(2)x)," | ",color(blue)(2sqrt(2)x^3), color(green)(-3sqrt(2)x^2), +2sqrt(2)x) :}
Combine terms with identical exponents of $x$ to get the Answer above.