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

[linapril]

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

 

[Jameson]

Re: Easy algebra

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.

\(\displaystyle \frac{x^2-2x+1}{x-1}=\frac{(x-1)(x-1)}{x-1}=x-1\)

Jameson

 

[Ackbach]

Re: Easy algebra

Hi linapril, :)

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

\(\displaystyle \frac{x^2-2x+1}{x-1}=\frac{(x-1)(x-1)}{x-1}=x-1\)

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.

 

[MarkFL]

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:

$8x^2+34x+35=(4x+7)(2x+5)$

 

[CellCoree]

To simplify this expression, we can use the factoring method. We first need to factor the numerator, which is a quadratic expression. We can use the "ac" method, where we find two numbers whose product is equal to the product of the first and last terms of the expression, and whose sum is equal to the coefficient of the middle term. In this case, the first and last terms are x^2 and 1, and the middle term is -2x.

So, we need to find two numbers whose product is x^2 and whose sum is -2x. These numbers are -1 and -1. Therefore, we can rewrite the numerator as (x-1)(x-1).

Now, we can rewrite the original expression as [(x-1)(x-1)]/(x-1).

Using the property of division, we can rewrite this as (x-1) * [(x-1)/(x-1)].

The term (x-1) in the numerator and denominator cancels out, leaving us with (x-1) as the simplified expression.

In summary, we can simplify (x^2-2x+1)/(x-1) to x-1 by factoring the numerator and using the property of division.