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

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To show that (2x-1)^2-(x-3)^2 can be simplified to (3x-4)(x+2).

(2x-1)^2 = (2x)^2-2*2x+1, as (a-b)^2 = a^2-2ab+b^2.

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

(x-3)^2 = x^2-2x*3 + (-3)^2 .

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

(1)-(2) gives:

(2x-1)^2-(x-3)^2 = (4x^2-x^2)+( -4x+6x) +(1-9).

(2x-1)^2-(x-3)^2 = 3x^2+2x-8.

(2x-1)^2-(x-3)^2= 3x^2+6x-4x-8.

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

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

(2x-1)^2-(x-3)^2 = (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.**