To answer this question, we have to do the long division process for polynomials. We can do the operation as follows:
To do this division process, we have:
1. Divide the first term of the dividend by the first element of the divisor. They are:
[tex]\frac{-4x^3}{4x^2}=-x[/tex]
2. Now, we have to multiply this result by the divisor, and the result will change its sign since we have to subtract that result from the dividend as follows:
[tex]-x\cdot(4x^2_{}-4x-4)=-4x^3+4x^2+4x[/tex]
And since we to subtract this result from the dividend, we end up with:
[tex]-(-4x^3+4x^2+4x)=4x^3-4x^2-4x[/tex]
3. Then we have the following algebraic addition:
[tex]\frac{\begin{cases}-4x^3+24x^2-15x-15 \\ 4x^3-4x^2-4x\end{cases}}{20x^2-19x-15}[/tex]
4. Again, we need to divide the first term of the dividend by the first term of the divisor as follows:
[tex]\frac{20x^2}{4x^2}=5[/tex]
5. And we have to multiply 5 by the divisor, and the result will be subtracted from the dividend:
[tex]5\cdot(4x^2-4x-4)=20x^2-20x-20[/tex]
Since we have to subtract this from the dividend, we have:
[tex]-(20x^2-20x-20)=-20x^2+20x+20[/tex]
6. And we have to add this algebraically to the dividend we got in the previous step:
[tex]\frac{\begin{cases}20x^2-19x-15 \\ -20x^2+20x+20\end{cases}}{x+5}[/tex]
And this is the remainder of the division, x + 5.
As we can see from the division process, we got as:
1. The quotient: -x + 5
[tex]q=-x+5[/tex]
2. The remainder: x + 5.
[tex]R=x+5[/tex]
Since we have that the dividend = divisor * quotient + remainder.
Therefore, the result for this division is:
[tex]-4x^3+24x^2-15x-15=(4x^2-4x-4)\cdot(-x+5)+(x+5)[/tex]
