Answer :
To find the quotient of
[tex]$$
\frac{x^4+5x^3-3x-15}{x^3-3},
$$[/tex]
we perform polynomial long division.
First, write the dividend with all degree terms, even if the coefficient is zero:
[tex]$$
x^4 + 5x^3 + 0x^2 - 3x - 15.
$$[/tex]
1. Divide the leading terms:
Divide the highest degree term in the dividend, [tex]$x^4$[/tex], by the highest degree term in the divisor, [tex]$x^3$[/tex]:
[tex]$$
\frac{x^4}{x^3} = x.
$$[/tex]
This [tex]$x$[/tex] is the first term of the quotient.
2. Multiply and subtract:
Multiply the entire divisor by [tex]$x$[/tex]:
[tex]$$
x \cdot (x^3-3) = x^4 - 3x.
$$[/tex]
Subtract this product from the dividend:
[tex]$$
\begin{aligned}
&\left(x^4+5x^3+0x^2-3x-15\right) - \left(x^4-3x\right) \\
&= x^4+5x^3+0x^2-3x-15 -x^4+3x \\
&= 5x^3 + 0x^2 + 0x - 15 \quad \text{(since } -3x + 3x = 0\text{)}.
\end{aligned}
$$[/tex]
3. Repeat for the new dividend:
Now, the new dividend is [tex]$5x^3 - 15$[/tex]. Divide the leading term [tex]$5x^3$[/tex] by [tex]$x^3$[/tex]:
[tex]$$
\frac{5x^3}{x^3} = 5.
$$[/tex]
This [tex]$5$[/tex] is the next term of the quotient.
4. Multiply and subtract again:
Multiply the divisor by [tex]$5$[/tex]:
[tex]$$
5 \cdot (x^3-3) = 5x^3-15.
$$[/tex]
Subtract this from the current dividend:
[tex]$$
(5x^3-15) - (5x^3-15) = 0.
$$[/tex]
The remainder is [tex]$0$[/tex], which confirms that the division is exact.
The complete quotient is therefore:
[tex]$$
x + 5.
$$[/tex]
Thus, the quotient of the given division is [tex]$\boxed{x+5}$[/tex].
[tex]$$
\frac{x^4+5x^3-3x-15}{x^3-3},
$$[/tex]
we perform polynomial long division.
First, write the dividend with all degree terms, even if the coefficient is zero:
[tex]$$
x^4 + 5x^3 + 0x^2 - 3x - 15.
$$[/tex]
1. Divide the leading terms:
Divide the highest degree term in the dividend, [tex]$x^4$[/tex], by the highest degree term in the divisor, [tex]$x^3$[/tex]:
[tex]$$
\frac{x^4}{x^3} = x.
$$[/tex]
This [tex]$x$[/tex] is the first term of the quotient.
2. Multiply and subtract:
Multiply the entire divisor by [tex]$x$[/tex]:
[tex]$$
x \cdot (x^3-3) = x^4 - 3x.
$$[/tex]
Subtract this product from the dividend:
[tex]$$
\begin{aligned}
&\left(x^4+5x^3+0x^2-3x-15\right) - \left(x^4-3x\right) \\
&= x^4+5x^3+0x^2-3x-15 -x^4+3x \\
&= 5x^3 + 0x^2 + 0x - 15 \quad \text{(since } -3x + 3x = 0\text{)}.
\end{aligned}
$$[/tex]
3. Repeat for the new dividend:
Now, the new dividend is [tex]$5x^3 - 15$[/tex]. Divide the leading term [tex]$5x^3$[/tex] by [tex]$x^3$[/tex]:
[tex]$$
\frac{5x^3}{x^3} = 5.
$$[/tex]
This [tex]$5$[/tex] is the next term of the quotient.
4. Multiply and subtract again:
Multiply the divisor by [tex]$5$[/tex]:
[tex]$$
5 \cdot (x^3-3) = 5x^3-15.
$$[/tex]
Subtract this from the current dividend:
[tex]$$
(5x^3-15) - (5x^3-15) = 0.
$$[/tex]
The remainder is [tex]$0$[/tex], which confirms that the division is exact.
The complete quotient is therefore:
[tex]$$
x + 5.
$$[/tex]
Thus, the quotient of the given division is [tex]$\boxed{x+5}$[/tex].
