What are the steps to factor $x^2 - 1$ into $(x+1)(x-1)$?

Question

Does $(x+1)(x-1) = x^2+1x-1x-1$? If so where are the $+1x$ and the $-1x$ when it is being factored from $x^2-1$ into $(x+1)(x-1)$?

What exactly are we dividing $x^2-1$ by to get $(x+1)(x-1)$ and how did you know what to divide it by?

Answer 1

Method 0:

$+1x - 1x = 0$

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

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

$x^2 + 0 -1 =$

$x^2 -1 $.

Method 1:

$x^2 - 1 = $

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

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

$(x\cdot x + x\cdot 1) + (-1)\cdot(x+1) =$

$x\cdot(x + 1) + (-1)\cdot (x+1) =$

$\color{blue}x\cdot\color{red}{(x + 1)} + \color{blue}{(-1)}\color{red}{(x + 1)} = $

$(\color{blue}x + \color{blue}{(-1)})\color{red}{(x+1)} = $

$(\color{blue}x-\color{blue}1)\color{red}{(x+1)}$

Method 2:

$(x^2 - 1)\div (x+1) = ?????$

We need to find a firs term, $a$ so that when we multiply $a$ times $x+1$ and get $ax+a$ that $ax = x^2$. What can be $a$? Well $ax=x^2$ and so (if we assume $x$ isn't always $0$) then $a= x$.

So $x(x+1) = x^2 + x$.

But $x^2 -1 \ne x^2 + x$ we must find a "remainder".

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

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

And $(x^2 - 1)\div (x+1) = x + \frac {-x-1}{x+1}$

Now we must divide $x+1$ into $-x-1$.

We must find term $b$ so that when we multiply $b$ by $x+1$ and get $bx + b$ the first term $bx = -x$. What can that $b$ be. Clearly it is $b=-1$.

So $-1(x+1) = -x-1$.

And $-x-1$ DOES equal $-1(x+1)$ so we have no remainder.

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

And $(x^2 - 1)\div (x+1) = x + (-1) = x-1$

So $(x^2 - 1) \div (x+1) = x-1$ so that mean $(x^2-1)= (x-1)(x+1)$.

Method 3:

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

$= x^2 +ax +bx + ab =$

$x^2 + (a+b)x + ab$

And that is supposed to be $x^2 -1 = x^2 + 0*x + (-1)$.

So we need $a,b$ so that $a*b = -1$ and $a+b = 0$.

Okay so $ab = -1$ so $a = -\frac 1b$

And $-\frac 1b + b = 0$ so

$b = \frac 1b$ so

$b^2 = 1$.

So $b = \pm 1$.

And $a +b = a\pm 1 = 0$ so $a =\mp 1$.

So one of them is $1$ and the other is $-1$.

So $x^2 -1 = (x+a)(x+b) = (x+1)(x-1)$.

Method 4:

If $ax^2 + bx + c = 0$ has two solutions $x= m$ and $x = n$

Then $ax^2 + bx + c = a(x-m)(x+n)$.

So what are the two solutions to $x^2 -1 = 0$

They are:

$x^2 - 1 =0$ so

$x^2 = 1$ so

$x = \pm 1$

So $m = 1$ and $n=-1$ are the two solutions and

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

........

But in ALL of these methods the $+1x -1x =0$ and the "cancel out".

You can cancel them to $0$ or pop them out of nowhere from $0$.

The this is you just have to think of it.

Answer 2

When you factor a polynomial, its roots must be roots of the factors. (If $p(x)=f(x)\cdot g(x)$, $p(r)=0\iff f(r)=0\lor g(r)=0$.)

As

$$x^2-1=0\iff x=\pm1$$ there must be a factor that vanishes with $x=1$ and another with $x=-1$.

Hence

$$x^2-1\propto (x-1)(x+1).$$

Answer 3

Let $y:=x-1$. Then

$$x^2-1=(y+1)^2-1=y^2+2y$$ which obviously factors as $$(y+2)y=(x+1)(x-1).$$

Answer 4

We have the equation $f(x)=x^2-1=(x-x_1)\cdot (x-x_2)$, where $x_1,x_2$ are the roots of $f(x)$. Thus we have to find the solution of $x^2-1=0$. This is not difficult. Adding $1$ on both sides of the equation.

$x^2=1$

square root

$|x|=1$

Thus $x_1=\color{red}1$ and $x_2=\color{blue}{-1}$. Therefore

$$f(x)=x^2-1=(x-\color{red}1)\cdot (x-(\color{blue}{-1}))=(x-1)\cdot (x+1)$$

I was thinking about square roots, but I knew the square root of -1 was a no no. The x with no subscript, x subscript 1, and x subscript 2 are still confusing me. Could you at least use a and b

You don´t calculate the square root of -1 here. You calculate the square root of $1$. And the equation $x^2=1$ has two solutions.

Sure you can use $a$ and $b$, where $a$ and $b$ are the roots of $f(x)$:

$f(x)=x^2-1=(x-a)\cdot (x-b)=0$. This is the representation of $f(x)$ with linear factors (!). Here we can use the zero product property which states: If at least one of the factors is $0$, then the product is $0$.

Then you solve $f(x)=x^2-1=0$ like above. After you´ve got the solution you can write the function with linear factors.

"Adding 1 on both sides of the equation." I don't understand why this needs to happen

If you solve an equation, it is always a good idea to separate the terms with the variable $x(,x^2,x^3,...)$ and the terms with out the variable $x$.