Complete Factoring of x^2 - 4x + 4 - 4y^2: Homework Solution & Explanation

[frozonecom]

Homework Statement

Factor the polynomial x^2 - 4x + 4 -4y^2 completely.

Homework Equations

The Attempt at a Solution

Rearranging, I get x^2 - 4y^2 - 4x + 4
Then, I know that it is equal to (x -2y)(x+2y) - 4(x-1)

and that is my final answer. but, my teacher only considered the answer :
(x-2)^2 - 4y^2
then, (x-2+2y)(x-2-2y)

---- I know how he got HIS factored form, but is it only the right answer? Is my answer wrong? Isn't it that the polynomial is completely factored if it can no longer be factored(prime) ?

Can there be one and only one factored form for a polynomial? If so, how do you judge if it is the right factored form of the polynomial?

Please help me. I'm really confused. Thanks in advance!

 

[Curious3141]

Homework Statement

Factor the polynomial x^2 - 4x + 4 -4y^2 completely.

Homework Equations

The Attempt at a Solution

Rearranging, I get x^2 - 4y^2 - 4x + 4
Then, I know that it is equal to (x -2y)(x+2y) - 4(x-1)

and that is my final answer. but, my teacher only considered the answer :
(x-2)^2 - 4y^2
then, (x-2+2y)(x-2-2y)

---- I know how he got HIS factored form, but is it only the right answer? Is my answer wrong? Isn't it that the polynomial is completely factored if it can no longer be factored(prime) ?

Can there be one and only one factored form for a polynomial? If so, how do you judge if it is the right factored form of the polynomial?

Please help me. I'm really confused. Thanks in advance!

Well, to factorise something completely, you need to express it as the product of numbers or expressions. When there are two factors (as in this case), the answer should be of the form $a.b$ or $f(x)g(x)$ in this case, with the degree of $f(x), g(x)$ etc. being as low as possible (linear in this case). Your answer is not in that form, so it's not a valid factorisation. $f(x)g(x) - h(x)$ is not a valid complete factorisation.

The trick here was to recognise the expression as: ${(x-2)}^2 - {(2y)}^2$ and then use $a^2 - b^2 = (a+b)(a-b)$ to factorise completely.

 

[frozonecom]

oh. so it really has to be products! thanks for the clear explanation! :)