Show that (2x-1)^2-(x-3)^2 can be simplified to (3x-4)(x+2)?

We'll expand the squares using the
formula:

(a+b)^2 = a^2 + 2ab +
b^2

(a-b)^2 = a^2 - 2ab +
b^2

(2x-1)^2 = (2x)^2 - 2*(2x)*1 +
1^2

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

(x-3)^2 = x^2 - 2*x*3 +
3^2

(x-3)^2 = x^2 - 6x + 9
(2)

We'll subtract (2) from
(1):

4x^2 - 4x + 1 -  x^2 + 6x -
9

We'll combine like
terms:

3x^2 + 2x - 8

We'll
determine the roots of the quadratic:

3x^2 + 2x - 8 =
0

x1 = [-2+sqrt(4+96)]/6

x1 =
(-2+10)/6

x1 = 8/6

x1 =
4/3

x2 = (-2-10)/6

x2 =
-2

The quadratic could be written as a product of linear
factors:

3x^2 + 2x - 8 = 3(x - 4/3)(x +
2)

We'll multiply by 3 the first
factor:

3(x - 4/3)(x + 2) = (3x - 4*3/3)(x +
2)

3x^2 + 2x - 8 =
(3x-4)(x+2)

(2x-1)^2 - (x-3)^2 = (3x-4)(x+2)
q.e.d.