MHB How Do You Simplify (x^2-2x+1)/(x-1) to x-1?

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To simplify the expression (x^2-2x+1)/(x-1) to x-1, it is essential to factor the numerator as (x-1)(x-1) or (x-1)^2. When dividing, the (x-1) terms can be canceled, resulting in x-1, provided that x is not equal to 1 to avoid division by zero. The discussion also touches on the general method for factoring quadratics, emphasizing the importance of identifying coefficients and their relationships. This approach can become more complex with different coefficients in the quadratic expression.
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Could someone explain how I'd simplify (x2-2x+1)/(x-1) to become x-1? Thanks a bunch!
 
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Re: Easy algebra

linapril said:
Could someone explain how I'd simplify (x2-2x+1)/(x-1) to become x-1? Thanks a bunch!

Hi linapril, :)

Are you familiar with how to factor? Here's how I would do this problem.

[math]\frac{x^2-2x+1}{x-1}=\frac{(x-1)(x-1)}{x-1}=x-1[/math]

Jameson
 
Re: Easy algebra

Jameson said:
Hi linapril, :)

Are you familiar with how to factor? Here's how I would do this problem.

[math]\frac{x^2-2x+1}{x-1}=\frac{(x-1)(x-1)}{x-1}=x-1[/math]

Jameson

Along with the proviso that $x\not=1$. Any time you cancel factors such that there is no longer that factor in the denominator, you must include a proviso so that you preserve the domain of the original function.
 
This is the method I have taught students I have tutored on how to factor quadratics.

Consider the general quadratic:

$\displaystyle ax^2+bx+c$

In the case of $\displaystyle x^2-2x+1=(1)x^2+(-2)x+(1)$, we identify:

$a=1,\,b=-2,\,c=1$

To factor, I first look at the product $ac=(1)(1)=1$. We want to find two factors of 1 whose sum is $b=-1$, and so those factors are -1 and -1, since $(-1)(-1)=1$ and $(-1)+(-1)=-2$.

Since $a=1$, we know the factorization will be of the form:

$(x\cdots)(x\cdots)$

To understand why the method I outlined works, we could set:

$x^2-2x+1=(x+d)(x+e)=x^2+(d+e)x+de$

Equating coefficients, we see we need two numbers $d$ and $e$ that simultaneously satisfy:

$d+e=-2$

$de=1$

As we already found, we need $d=e=-1$.

And so we may replace the dots with the two factors we found:

$(x-1)(x-1)=(x-1)^2$

Thus, we may state:

$x^2-2x+1=(x-1)^2$

The method I outlined can get a little more involved if $a$ is not 1, even more involved still if $a$ is a composite number. Consider the quadratic:

$8x^2+34x+35$

We want two factors of $8\cdot35=280$ whose sum is $34$. They are $14$ and $20$. Now we must observe that $20$ is divisible by $4$ and $14$ is divisible by $2$. The product of $2$ and $4$ will give is $a=8$.

So, we will have the form:

$(4x\cdots)(2x\cdots)$

$\displaystyle \frac{20}{4}=5$ and $\displaystyle \frac{14}{2}=7$ so we have:

$8x^2+34x+35=(4x+7)(2x+5)$
 
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