Learn Step-By-Step Factorization with New Equations

[Rigbee]

Hi, I am new to factorization. Would someone please solve these two equations and explain step by step what was done. thanks1. (x-y)^2 - (x-z)^2 =
2. (5x+2)^2 - (3x-4)^2 =

 

[evinda]

Hi, I am new to factorization. Would someone please solve these two equations and explain step by step what was done. thanks1. (x-y)^2 - (x-z)^2 =
2. (5x+2)^2 - (3x-4)^2 =

For the first one, we have: $(x-y)^2 - (x-z)^2=x^2-2xy+y^2-x^2+2xz-z^2=-2xy+y^2+2xz-z^2\\ =2x(z-y)+(y-z)(y+z)=(z-y)(2x-y-z)$

Can you try the second one?

 

[MarkFL]

Hi, I am new to factorization. Would someone please solve these two equations and explain step by step what was done. thanks1. (x-y)^2 - (x-z)^2 =
2. (5x+2)^2 - (3x-4)^2 =

Hello and welcome to MHB, Rigbee!

Whenever you see an expression of the form:

\(\displaystyle a^2-b^2\)

This is called the difference of squares, and may be factored as follows:

[box=blue]

Difference of Squares​

---

\(\displaystyle a^2-b^2=(a+b)(a-b)\tag{1}\)[/box]

So, for your first problem, you then identify:

\(\displaystyle a=x-y,\,b=x-z\)

And then plug them into the difference of squares formula (1):

\(\displaystyle (x-y)^2-(x-z)^2=((x-y)+(x-z))((x-y)-(x-z))\)

Now, remove the inner parentheses on the right (distributing as needed):

\(\displaystyle (x-y)^2-(x-z)^2=(x-y+x-z)(x-y-x+z)\)

Combine like terms:

\(\displaystyle (x-y)^2-(x-z)^2=(2x-y-z)(-y+z)\)

Now arrange with binomial factor in front, and with no leading negatives.

\(\displaystyle (x-y)^2-(x-z)^2=(z-y)(2x-y-z)\)

This gives you the same result as evinda, but does not require you to expand the squared binomials and your result is already factored. Either method is good, it just depends on your preference.

Incidentally, what we have done is called factoring, not solving. In order to solve, we need expressions on both sides of the equal sign so that we have an equation, and we need to know for which variable we are solving.

So, as suggested, try the second one, using either method...post your work and we will be glad to check it.

 

[Rigbee]

Hello and welcome to MHB, Rigbee!

Whenever you see an expression of the form:

\(\displaystyle a^2-b^2\)

This is called the difference of squares, and may be factored as follows:

[box=blue]

Difference of Squares​

---

\(\displaystyle a^2-b^2=(a+b)(a-b)\tag{1}\)[/box]

So, for your first problem, you then identify:

\(\displaystyle a=x-y,\,b=x-z\)

And then plug them into the difference of squares formula (1):

\(\displaystyle (x-y)^2-(x-z)^2=((x-y)+(x-z))((x-y)-(x-z))\)

Now, remove the inner parentheses on the right (distributing as needed):

\(\displaystyle (x-y)^2-(x-z)^2=(x-y+x-z)(x-y-x+z)\)

Combine like terms:

\(\displaystyle (x-y)^2-(x-z)^2=(2x-y-z)(-y+z)\)

Now arrange with binomial factor in front, and with no leading negatives.

\(\displaystyle (x-y)^2-(x-z)^2=(z-y)(2x-y-z)\)

This gives you the same result as evinda, but does not require you to expand the squared binomials and your result is already factored. Either method is good, it just depends on your preference.

Incidentally, what we have done is called factoring, not solving. In order to solve, we need expressions on both sides of the equal sign so that we have an equation, and we need to know for which variable we are solving.

So, as suggested, try the second one, using either method...post your work and we will be glad to check it.

Thank you MarkFL and Evinda,

I'm only in grade 5 so I apologize if my questions are simplistic. I am just starting to learn factoring. I was told the answer is -(2x-y-z)(y-z) leading with the negative. Why is your answer different? If they are both correct, what did they do different to get this result? I'm trying to understand the steps necessary.

 

[Nono713]

Thank you MarkFL and Evinda,

I'm only in grade 5 so I apologize if my questions are simplistic. I am just starting to learn factoring. I was told the answer is -(2x-y-z)(y-z) leading with the negative. Why is your answer different? If they are both correct, what did they do different to get this result? I'm trying to understand the steps necessary.

The answers are the same, note that Mark's answer has $(z - y)$ while yours has $(y - z)$. And as you hopefully know, $(z - y) = (-1)(y - z) = -(y - z)$ which is where the negative sign comes from :)

 

[Rigbee]

I tried your method with this equation:

(x-3y)^2 - (y-x)^2

step one [(x-3y) + (y-x)] [(x-3y) - (y-x)]

step two (x-3y-y+x) (x-3y+y-x)

step three -2y(2x-4y)

final step -2y(x-2y) I know this is incorrect, I just need some help identifying what I am missing. Thanks

 

[topsquark]

I tried your method with this equation:

(x-3y)^2 - (y-x)^2

step one [(x-3y) + (y-x)] [(x-3y) - (y-x)]

step two (x-3y-y+x) (x-3y+y-x)

step three -2y(2x-4y)

final step -2y(x-2y) I know this is incorrect, I just need some help identifying what I am missing. Thanks

Your answer in step 3 is correct as far as it goes. But the reason it might be considered incorrect is that you can still factor a 2 out of the second factor. ie $$-4y(x - 2y)$$.

-Dan