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

In summary, the conversation discusses the process of factoring the polynomial x^2 - 4x + 4 -4y^2 completely. The attempt at a solution involves rearranging the terms and finding a factored form, but the teacher only considers a specific answer. It is clarified that for a polynomial to be completely factored, it must be expressed as the product of numbers or expressions with the lowest possible degree. The solution is then provided using the formula a^2 - b^2 = (a+b)(a-b) to factorise the polynomial completely.
  • #1
frozonecom
63
0

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!
 
Physics news on Phys.org
  • #2
frozonecom said:

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 [itex]a.b[/itex] or [itex]f(x)g(x)[/itex] in this case, with the degree of [itex]f(x), g(x)[/itex] etc. being as low as possible (linear in this case). Your answer is not in that form, so it's not a valid factorisation. [itex]f(x)g(x) - h(x)[/itex] is not a valid complete factorisation.

The trick here was to recognise the expression as: [itex]{(x-2)}^2 - {(2y)}^2[/itex] and then use [itex]a^2 - b^2 = (a+b)(a-b)[/itex] to factorise completely.
 
Last edited:
  • #3
oh. so it really has to be products! thanks for the clear explanation! :)
 

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

What is the meaning of "complete factoring" in this context?

Complete factoring refers to the process of breaking down a polynomial expression into its simplest form by finding its factors. In this case, we are looking for two binomials that when multiplied together, will result in the given polynomial.

What is the general approach to factoring a quadratic polynomial?

The general approach to factoring a quadratic polynomial is to first check if it is in the form of (ax^2 + bx + c). If it is, then we can use the AC method or the quadratic formula to find the factors. If it is not in this form, we can use perfect square trinomial or difference of squares formulas to factor it.

How do we apply the factoring process to the given polynomial x^2 - 4x + 4 - 4y^2?

First, we can rearrange the terms to group the x-terms and the y-terms together: (x^2 - 4x) + (4 - 4y^2). Then, we can factor out the greatest common factor from each group, which is (x-2) and (2-2y^2). This results in (x-2)(x-2) - 4(1-y^2). Finally, we can further simplify this by using the difference of squares formula, which gives us (x-2)^2 - 4(1-y)(1+y). Therefore, the complete factored form is (x-2)^2 - 4(1-y)(1+y).

What is the significance of factoring a polynomial?

Factoring a polynomial can help us find its roots or solutions, which are the values of x that make the polynomial equal to zero. This is useful in solving equations and understanding the behavior of the polynomial. It also allows us to simplify complex expressions and identify common factors, which can make calculations and evaluations easier.

Are there any strategies or tips for factoring more efficiently?

Yes, there are a few strategies that can help make the factoring process more efficient. These include looking for common factors, using special product formulas, checking for patterns, and practicing factoring regularly. It is also helpful to have a good understanding of basic algebraic concepts such as the distributive property and combining like terms.

Similar threads

Obtain a nonzero solution of ##y''-4y'+x^2(y'-4y)=0## by inspection
    • Calculus and Beyond Homework Help
Replies
7
Views
713
Master Factorization with Step-by-Step Guide for x^4 - 3x^2 - 4
    • Precalculus Mathematics Homework Help
Replies
19
Views
1K
Factoring a four term polynomial
    • Precalculus Mathematics Homework Help
Replies
11
Views
2K
Find the factors of an equation
    • Precalculus Mathematics Homework Help
Replies
4
Views
941
Adding and subtracting polynomials
    • Precalculus Mathematics Homework Help
Replies
6
Views
2K
Proof of No Solution for x^2 - 3xy + 2y^2 = 10 Conjecture | Polynomial Homework
    • Precalculus Mathematics Homework Help
Replies
7
Views
2K
I A doubt about the multiplicity of polynomials in two variables
    • General Math
Replies
8
Views
1K
Solving Plane Equation 3x + 2y -z = 4
    • Precalculus Mathematics Homework Help
Replies
8
Views
1K
Proving Inequality $$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 12xy$$
    • Precalculus Mathematics Homework Help
Replies
16
Views
2K
How do you divide polynomials in your head "on sight"?
    • Precalculus Mathematics Homework Help
Replies
27
Views
3K

Hot Threads

Recent Insights

Back
Top