Let's divide:-
`(m^2 + 4m - 21) by (m-3)`

Let's divide:- (m^4 - 2m^3 - 7m^2 + 8m + 12) by (m^2 - m - 6)

Factorise:- (m^2 - 4m - 12)

Let's write the value of (m^2 + n^2) (m^3 + n^3) by calculation if m + n = 5 and mn = 6.

Let's Substract: (m^2 -2) From (-3m^2 + 2m +2)

Find the value of m for which the area of the triangle having vertices at (-1,m) , (m-2,1) and (m-2,m) is 12(1)/(2) square units .

Let's resolve into factors the following agebraic expressions : - m^3 - n^3 -m(m^2-n^2) + n(m-n)^2

Let's factorise (8m^3 + 12m^2n + 6mn^2 + n^3)

Let's resolve into factor:- 2m^2 + 7m + 6

Using formula let's show that. (3m + 4n)^2 = (3m - 4n)^2 + 48mn

Using formula let's show that. (l + m)^2 = (l - m)^2 + 4lm