Problems from Quotient of polynomials

Development:

We complete and we design the initial table

$x^4$

$0$

$0$

$-2x$

$3$

$x^2-2$

Step 1:

${\displaystyle \frac{x^4}{x^2}}=x^2$

$x^2\cdot (x^2-2)=x^4-2x^2$

$x^4$	$0$	$0$	$-2x$	$3$	$x^2-2$
$-x^4$	$0$	$+2x^2$	$0$	$0$	$x^2$
$0$	$0$	$+2x^2$	$-2x$	$3$

Step 2:

${\displaystyle \frac{2x^2}{x^2}}=2$

$2\cdot (x^2-2)=2x^2-4$

$x^4$	$0$	$0$	$-2x$	$3$	$x^2-2$
$-x^4$	$0$	$+2x^2$	$0$	$0$	$x^2+2$
$0$	$0$	$+2x^2$	$-2x$	$3$
		$-2x^2$	$0$	$+4$
		$0$	$-2x$	$7$

The process ends here because:

degree$(-2x+7)=1<2=$degree$(x^2-2)$

Solution:

Quotient: $x^2+2$

Reminder: $-2x+7$

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Development:

First we do the addition,

	r(x)	q(x)	r(x)+q(x)
degree 0	$1$	$-3$	$-2$
degree 1	$-x$	$-x$	$-2x$
degree 2	$0$	$0$	$0$
degree 3	$0$	$2x^3$	$2x^3$

$r(x)+q(x)=2x^3-2x-2$

Then, the product by the monomials of $p(x)$,

$x\cdot (r(x)+q(x))=x\cdot (2x^3-2x-2)=2x^4-2x^2-2x$

$-x^3\cdot (r(x)+q(x))=-x^3\cdot (2x^3-2x-2)=-x^6+2x^4+2x^3$

Solution:

We join both polynomials and make groups of similar terms:

$(r(x)+q(x))\cdot p(x)=(2x^4-2x^2-2x)+(-x^6+2x^4+2x^3)=$

$=-x^6+4x^4+2x^3-2x^2-2x$

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