(1) [tex]50y^3=5y\cdot 10y^2[/tex], and
[tex]10y^2\cdot(5y-4)=50y^3-40y^2[/tex]
Subtract this from [tex]50y^3+10y^2-35y-7[/tex] to get a remainder of
[tex](50y^3+10y^2-35y-7)-(50y^3-40y^2)=50y^2-35y-7[/tex]
What we've done here is show that
[tex]\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+\dfrac{50y^2-35y-7}{5y-4}[/tex]
We can keep going as long as the degree of the remainder's numerator is at least the same as the degree of its denominator.
Next, [tex]50y^2=5y\cdot 10y[/tex], and
[tex]10y\cdot(5y-4)=50y^2-40y[/tex]
Subtract this from the previous remainder to get a new remainder of
[tex](50y^2-35y-7)-(50y^2-40y)=5y-7[/tex]
which means
[tex]\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+10y+\dfrac{5y-7}{5y-4}[/tex]
Finally, [tex]5y=5y\cdot1[/tex], and
[tex]1\cdot(5y-4)=5y-4[/tex]
Subtract this from the previous remainder to get
[tex](5y-7)-(5y-4)=-3[/tex]
So we end up with
[tex]\dfrac{50y^3+10y^2-35y-7}{5y-4}=\boxed{10y^2+10y+1}-\dfrac3{5y-4}[/tex]
(the quotient is the expression in the box)
(2) Using the same process as before:
[tex]8m^4=2m\cdot4m^3[/tex]
[tex]4m^3\cdot(2m+1)=8m^4+4m^3[/tex]
[tex](8m^4-4m^2+m+4)-(8m^4+4m^3)=-4m^3-4m^2+m+4[/tex]
[tex]\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-\dfrac{4m^3+4m^2+m+4}{2m+1}[/tex]
[tex]-4m^3=2m\cdot(-2m^2)[/tex]
[tex]-2m^2\cdot(2m+1)=-4m^3-2m^2[/tex]
[tex](-4m^3-4m^2+m+4)-(-4m^3-2m^2)=-2m^2+m+4[/tex]
[tex]\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-\dfrac{2m^2-m-4}{2m+1}[/tex]
[tex]-2m^2=2m\cdot(-m)[/tex]
[tex]-m\cdot(2m+1)=-2m^2-m[/tex]
[tex](-2m^2+m+4)-(-2m^2-m)=2m+4[/tex]
[tex]\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-m+\dfrac{2m+4}{2m+1}[/tex]
[tex]2m=2m\cdot1[/tex]
[tex]1\cdot(2m+1)=2m+1[/tex]
[tex](2m+4)-(2m+1)=3[/tex]
[tex]\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=\boxed{4m^3-2m^2-m+1}+\dfrac3{2m+1}[/tex]
