`x^2-4x-5=(x-5)(x+1)`

Factorize the following polynmials: (1) x^(2)-4x-5 =x^(2)-5x+x-5 =x(x-5)+1(x-5) =(x-5)(x+1)

Simplify: (x^2-3x+2)(5x-2)-\ (3x^2+4x-5)\ (2x-1)

Simplify: (x^(2)-3x+2)(5x-2)-(3x^(2)+4x-5)(2x-1)

Evaluate int(5x^4+4x^5)/((x^5+x+1)^2)dx

Evaluate int(5x^4+4x^5)/((x^5+x+1)^2)dx

Check whether the following are quadratic equations : (1) (x-3)^(2)=x(2x-5) (2) (2x-3)(8x+1) = (4x+5)(4x-5) (3) (5x+3)(x-2)=(4x+3)(2x-1) (4) (2x+5)^(3)=8(x-1)^(3) (5) x^(2)+7x-8=x(x+5) (6) x^(3)+9x^(2)-7x+2=(x+3)^(3)

log_(x^(2))((4x-5)/(|x-2|))>=-(1)/(2)

Sum of squares of the roots of the equation slogs 5^(log_(5)(x^(2)-4x+5))=x-1 is equal to

((x+1)(4x-3)-4x^(2)+5)/(4x+1)=2

int((9x^(2)-4x+5))/((3x^(3)-2x^(2)+5x+1))dx