The given division is
[tex]\frac{y^2-16}{y+4}[/tex]
At first, we will put zero in the place of the missing term y, then it should be
[tex]\frac{y^2+0y-16}{y+4}[/tex]
Now, divide y^2 by y
[tex]\frac{y^2}{y}=y[/tex]
Multiply the answer by (y + 4)
[tex]y(y+4)=y^2+4y[/tex]
Subtract it from (y^2 + 0y - 16)
[tex]\begin{gathered} y^2+0y-16-(y^2+4y)= \\ (y^2-y^2)+(0y+-4y)-16= \\ 0-4y-16= \\ -4y-16 \\ \frac{y^2+0y-16}{y+4}=y+\frac{-4y-16}{y+4} \end{gathered}[/tex]
Divide -4y by y
[tex]\frac{-4y}{y}=-4[/tex]
Multiply the answer by (y + 4)
[tex]-4(y+4)=-4y-16[/tex]
Subtract it from the numerator of the fraction
[tex]\begin{gathered} -4y-16-(-4y-16= \\ -4y-16+4y+16= \\ (-4y+4y)+(-16+16)= \\ 0+0=0 \\ \frac{y^2-16}{y+4}=(y-4)+\frac{0}{y+4} \end{gathered}[/tex]
The answer to the division is
[tex](y-4)+\frac{0}{y+4}[/tex]
Q = (y - 4)
R = 0
D = y + 4
