Respuesta :
Answer:[tex]\frac{y^3-1}{y+4}=y^2-4y+16+\frac{-65}{y+4}[/tex]Explanation:
Given the expression:
[tex]\frac{y^3+0y^2-1}{y+4}[/tex]Step 1:
Divide the leading term of the dividend by the leading term of the divisor. Multiply the result by the divisor and subtract the final result from the dividend as follows:
[tex]\begin{gathered} \frac{y^3}{y}=y^2 \\ \\ \\ y^2(y+4)=y^3+4y^2 \\ \\ \\ (y^3-1)-(y^3+4y^2)=-4y^2-1 \end{gathered}[/tex]Step 2: Divide the leading term of the obtained remainder by the leading term of the divisor, multiply it by the divisor, and subtract the remainder from the obtained result:
[tex]\begin{gathered} \frac{-4y^2}{y}=-4y \\ \\ \\ -4y(y+4)=-4y^2-16y \\ \\ \\ (-4y^2-1)-(-4y^2-16y)=16y-1 \end{gathered}[/tex]Step 3: Divide the leading term of the obtained remainder by the leading term of the divisor, multiply it by the divisor, and subtract the remainder from the obtained result:
[tex]\begin{gathered} \frac{16y}{y}=16 \\ \\ \\ 16(y+4)=16y+64 \\ \\ \\ (16y-1)-(16y+64)=-65 \end{gathered}[/tex]Since the degree of the remainder is less than the degree of the divisor, we would stop.
Therefore, the answer is:
[tex]\frac{y^3-1}{y+4}=y^2-4y+16+\frac{-65}{y+4}[/tex]