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Factoring

linapril

New member
Dec 21, 2012
23
Could someone explain how I'd simplify (x2-2x+1)/(x-1) to become x-1? Thanks a bunch!
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
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

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

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