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:

$$\dfrac{x^4}{x^2}=x^2$$

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

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

Step 2:

$$\dfrac{2x^2}{x^2}=2$$

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

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

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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