Question 681763
height of 2x + 3 centimeters, and a hypotenuse of length 4x - 1 

Pythagoras theorem
hypotenuse^2= base^2+height^2

base^2= hypotenuse^2-height^2

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

=16x^2-8x+1-(4x^2+12x+9)

=16x^2-8x+1-4x^2-12x-9

=12x^2-20x-8
=4(3x^2-5x-2)
=4(3x^2-6x+x-2)
=4(3x(x-2)+1(x-2))
=4(3x+1)(x-2)

base = {{{2sqrt((3x+1)(x-2))}}}cm