Question

Not a graded or timed assessment. Need help, thank you. :)

Answer 1

We are given the following division problem

$\frac{m^8-1}{m-1}$

We need to factor out the numerator first

$m^8-1=(m^4)^2-1^2$

Apply the difference of squares formula

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

Again, factor out the term on the right side

$m^4-1=(m^2)^2-1^2$

Apply the difference of squares formula

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

Again, factor out the term on the right side

$m^2-1=(m)^2-1^2$

Apply the difference of squares formula

$(m)^2-1^2=(m+1)(m-1)_{}$

Finally, the expression becomes

$\frac{m^8-1}{m-1}=\frac{(m^4+1)(m^2+1)(m+1)(m-1)}{m-1}$

(m-1) cancels out

$\frac{m^8-1}{m-1}=(m^4+1)(m^2+1)(m+1)$

Therefore, the quotient is

$(m^4+1)(m^2+1)(m+1)$

Or in simplified form

$m^7+m^6+m^5+m^4+m^3+m^2+m+1$