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If $P(x)=2x^4-x^2+3$ and $Q(x)=5x^3+4x^2-3x$, then

$P(x)Q(x)=$ $x^7+$ $x^6-$ $x^5-$ $x^4+$ $x^3+$ $x^2-$ $x$

$P(x)$		$2$	$0$	$-1$	$0$	$3$	$\longleftarrow$ coefficients of $2x^4-x^2+3$
$Q(x)$	$\times$		$5$	$4$	$-3$	$0$	$\longleftarrow$ coefficients of $5x^3+4x^2-3x$
		$0$	$0$	$0$	$0$	$0$	$\longleftarrow$ top row $\times\;$$0$
	$-6$	$0$	$3$	$0$	$-9$	$\cdot$	$\longleftarrow$ placeholder then top row $\times\;$$-3$
					$\cdot$	$\cdot$	$\longleftarrow$ 2 placeholders then top row $\times\;$$4$
				$\cdot$	$\cdot$	$\cdot$	$\longleftarrow$ 3 placeholders then top row $\times\;$$5$
			$18$		$-9$	$0$	$\longleftarrow$ sum of the columns

Expand and simplify:
*Write your answer with descending powers of $x$

$\left(-2x^4+3x^3-4x+5\right)\left(2x^3-x^2-1\right)=$

$\left(2x^2-x+2\right)^2 \left(-x^4+x-3\right)=$

$10$	$8$	$-11$	$-4$	$18$	$12$	$-9$	$0$	are the coefficients of
$x^7$	$x^6$	$x^5$	$x^4$	$x^3$	$x^2$	$x^1$	$x^0$	so