We have to prove: (2x-1)^2-(x-3)^2 = (3x-4)(x+2)

Take the left hand side

(2x-1)^2-(x-3)^2

Use the relation x^2 - y^2 = (x - y)(x + y)

=> (2x - 1 - x + 3)(2x - 1 + x - 3)

=> (x + 2)(3x - 4)

which is the right hand side.

**This proves that (2x-1)^2-(x-3)^2 = (3x-4)(x+2)**

We'll raise to square 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)

If we'll subtract (2) from (1), we'll get:

4x^2 - 4x + 1 - x^2 + 6x - 9

We'll combine like terms:

3x^2 + 2x - 8

We'll cancel the quadratic to determine it's x intercepts:

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

We'll write the quadratic 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)

** **