# How from $x^2-1$ you go to $(x-1)(x+1)$ Can you show all the steps? [closed]

How from $x^2-1$ you go to $(x-1)(x+1)$? Can you show me all the steps?

• Just go from $(x-1)(x+1)$ to $x^2 - 1$ and then work backwards
– RGS
Feb 2, 2017 at 12:40
• Try foiling the result. Feb 2, 2017 at 12:40
• $x^2-1=((x^2-x)+x)-1=(x^2-x)+(x-1)=x(x-1)+1\cdot(x-1)=(x+1)(x-1)=(x-1)(x+1)$ Feb 2, 2017 at 12:42
• Why the downvotes? Feb 2, 2017 at 12:46
• @VishalGupta the question is low quality, low effort, no research, etc. Feb 2, 2017 at 12:51 $x^2 -1=$ the green enclosed area. i.e all small squares - 1 the bottom right.
Now think this way - Extend the $x-1$ rows one column right. this column has length $x-1$. It is equal to the area in the bottom $x-1$ cells. So the area of the green enclosed area of previous picture is $(x+1)(x-1)$.
So, $x^2 - 1 = (x+1)(x-1)$

• +1: This is an excellent approach. However, your picture seems to be suggesting that $x$ is a specific positive integer, when it needn't be anything of the sort. One can readily adapt it to be more general by erasing most of the grid lines, though it still requires $x$ to be greater than $1$ for the picture to make sense. Feb 2, 2017 at 13:05
• @CameronBuie I was expecting this question will be deleted :P So I hurried up.. and make all of them straight line :V Feb 2, 2017 at 13:09
• more natural imo is to go to $2x(x-1)-(x-1)^2=(x-1)(2x-(x-1))$
– JMP
Feb 2, 2017 at 13:23
• @JonMarkPerry I tried to prove without any algebra. Your ones need little bit algebra at last :P Feb 2, 2017 at 13:29
• The answer I was looking for was what @wolfram provided above (algebraic manipulation). But your answer Rezwan Arefin was a pleasant surprise. I never though about the visual/geometrical representation of this equation. Which makes me wonder what is the visual representation of x^2 - 2x + 1. As you might have guessed I am learning algebra on my own and I need help to deepen my understanding, not just memorise. I don't understand why they put this subject on hold. Is it trivial for some people? Is not for someone like me. It was a great learning experience. Thanks to everyone how contributed. Feb 3, 2017 at 14:20

Put $a=x+1$. Then $x=a-1$, hence

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

You go the other way: $$(x-1)(x+1) = x(x+1) - 1(x+1) = x^2 + x- x - 1 = x^2 - 1$$ Once you know that it works in one direction, you're allowed to use the identity in the other direction. That's all there is to it.

You can also try to find the roots of the polynomial $p(x) = x^2-1$. It easy to see that $1$ is root of $p$ ($p(1) = 0$). This means that $p$ should be divisible by $(x-1)$. If you apply the division algorithm to polynomial you will see that $$x^{2}-1 = p(x) = q(x)(x-1)$$ where $q(x) = (x+1)$.

• Valid answer, but someone who knows the division algorithm would not ask this question. Feb 2, 2017 at 12:45
• @VishalGupta I see your point! I have the impression that I've learnt multiplication and division of polynomials at the same time (in high school), and used to prefer these kind of arguments. Show I delete it ? Feb 2, 2017 at 12:50
• Its not wrong, so you do not need to delete it. You might get some upvotes (maybe some downvotes too :P) and once you get it, you cannot delete it. Its up to you. Feb 2, 2017 at 12:52
• Nice answer! If it would not have been already there I would have written this answer. Feb 2, 2017 at 13:19

As $x^2-1$ is a polynomial of degree $2$, it has $2$ zeroes, so we can write:

$$x^2-1=(ax+b)(cx+d)=acx^2+(bc+ad)x+bd$$

This tells us we need:

As polynomials over $\mathbb{R}$ are an UFD up to multiplication by units, we can claim any values for $a$ and $c$ with $ac=1$.
Let's try $a=2, c=\frac12$:
So $\frac b2+2d=0$, or $b+4d=0$.
As $b=-\frac1d$ we have $-\frac1d+4d=0$, and so $-1+4d^2=0$, and so $d=\pm \frac12$.
So a solution is $(2x+2)(\frac x2-\frac12)=(x+1)(x-1)$.