Question

Find the quotient using polynomial long division:

Answer 1

(1) $50y^3=5y\cdot 10y^2$, and

$10y^2\cdot(5y-4)=50y^3-40y^2$

Subtract this from $50y^3+10y^2-35y-7$ to get a remainder of

$(50y^3+10y^2-35y-7)-(50y^3-40y^2)=50y^2-35y-7$

What we've done here is show that

$\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+\dfrac{50y^2-35y-7}{5y-4}$

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, $50y^2=5y\cdot 10y$, and

$10y\cdot(5y-4)=50y^2-40y$

Subtract this from the previous remainder to get a new remainder of

$(50y^2-35y-7)-(50y^2-40y)=5y-7$

which means

$\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+10y+\dfrac{5y-7}{5y-4}$

Finally, $5y=5y\cdot1$, and

$1\cdot(5y-4)=5y-4$

Subtract this from the previous remainder to get

$(5y-7)-(5y-4)=-3$

So we end up with

$\dfrac{50y^3+10y^2-35y-7}{5y-4}=\boxed{10y^2+10y+1}-\dfrac3{5y-4}$

(the quotient is the expression in the box)

(2) Using the same process as before:

$8m^4=2m\cdot4m^3$

$4m^3\cdot(2m+1)=8m^4+4m^3$

$(8m^4-4m^2+m+4)-(8m^4+4m^3)=-4m^3-4m^2+m+4$

$\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-\dfrac{4m^3+4m^2+m+4}{2m+1}$

$-4m^3=2m\cdot(-2m^2)$

$-2m^2\cdot(2m+1)=-4m^3-2m^2$

$(-4m^3-4m^2+m+4)-(-4m^3-2m^2)=-2m^2+m+4$

$\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-\dfrac{2m^2-m-4}{2m+1}$

$-2m^2=2m\cdot(-m)$

$-m\cdot(2m+1)=-2m^2-m$

$(-2m^2+m+4)-(-2m^2-m)=2m+4$

$\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-m+\dfrac{2m+4}{2m+1}$

$2m=2m\cdot1$

$1\cdot(2m+1)=2m+1$

$(2m+4)-(2m+1)=3$

$\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=\boxed{4m^3-2m^2-m+1}+\dfrac3{2m+1}$