Question

Let a(x)=-4x^5-3a(x)=−4x5−3a, left parenthesis, x, right parenthesis, equals, minus, 4, x, start superscript, 5, end superscript, minus, 3, and b(x)=x^3b(x)=x3b, left parenthesis, x, right parenthesis, equals, x, cubed.

When dividing aaa by bbb, we can find the unique quotient polynomial qqq and remainder polynomial rrr that satisfy the following equation:

\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}b(x)a(x)​=q(x)+b(x)r(x)​start fraction, a, left parenthesis, x, right parenthesis, divided by, b, left parenthesis, x, right parenthesis, end fraction, equals, q, left parenthesis, x, right parenthesis, plus, start fraction, r, left parenthesis, x, right parenthesis, divided by, b, left parenthesis, x, right parenthesis, end fraction,

where the degree of r(x)r(x)r, left parenthesis, x, right parenthesis is less than the degree of b(x)b(x)b, left parenthesis, x, right parenthesis.

What is the quotient, q(x)q(x)q, left parenthesis, x, right parenthesis?

q(x)=q(x)=q, left parenthesis, x, right parenthesis, equals 


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Answer 1

Given data:

$a(x)=-4x^{5}-3,\:b(x)=x^{3}$

$a(x)$ upon division by $b(x)$ leaves unique quotient $q(x)$ and remainder $r(x)$

The following relation is also satisfied:

$\dfrac{a(x)}{b(x)}=q(x)+\dfrac{r(x)}{b(x)}$

To find:

What is the quotient $q(x)$?

Step-by-step explanation:

Given, $\dfrac{a(x)}{b(x)}=q(x)+\dfrac{r(x)}{b(x)}$

In the left hand side of the above relation, we put $a(x)=-4x^{5}-3$ and $b(x)=x^{3}$

$\Rightarrow \dfrac{-4x^{5}-3}{x^{3}}$

$\Rightarrow \dfrac{(-4x^{2})x^{3}-3}{x^{3}}$

$\Rightarrow \dfrac{(-4x^{2})x^{3}}{x^{3}}+\dfrac{-3}{x^{3}}$

$\Rightarrow (-4x^{2})+\dfrac{-3}{x^{3}}$

This is equivalent to the right hand side of the given relation

$\Rightarrow (-4x^{2})+\dfrac{-3}{x^{3}}=q(x)+\dfrac{r(x)}{b(x)}$

$\Rightarrow q(x)+\dfrac{r(x)}{b(x)}=(-4x^{2})+\dfrac{-3}{b(x)}$ since $b(x)=x^{3}$

Comparing among both sides, we get

$\quad \boxed{q(x)=-4x^{2}}$

Answer: Quotient, $q(x)=-4x^{2}$