665x^(2)-7x-3

sqrt(3x^(2)-7x-30)-sqrt(2x^(2)-7x-5)=x-5

Prove that the function f(x)=x^(2)-7x+3 is continous at x=1,x=2,x=3

Add: 5x^(2) - 7x + 3, -8x^(2) + 2x -5 and 7x^(2) - x - 2

Factorise : 3x^(3) + 14x^(2) - 7x - 10

Factorize :x^(3)+3x^(2)+3x-7x^(3)-3x^(2)+3x+7x^(6)-7x^(3)-8

Factorise : 2x^(3) + 7x^(2) + 2x -3

Simplify: 5x - 2x ( x- 1) A. 3x - 2 B. 3x + 2 C. 7x + 2x^(2) D. 7x - 2x^(2)

From the sum of 6x^(4) - 3x^(3) + 7x^(2) - 5x + 1 and -3x^(4) + 5x^(3) - 9x^(2) + 7x - 2 subtract 2x^(4) - 5x^(3) + 2x^(2) - 6x - 8