# How can I factor the polynomial $x^4-2x^3+x^2-1$?

How can I factor the polynomial $$x^4-2x^3+x^2-1$$?

This is an exercise in algebra. I have the solution showing that $$x^4-2x^3+x^2-1=(x^2-x-1)(x^2-x+1).$$

But the solution does not show any details. Using the distributive property I can check that this is indeed true: \begin{align} &(x^2-x-1)(x^2-x+1)\\ &=x^2(x^2-x+1)-x(x^2-x+1)-(x^2-x+1)\\ &=x^4-x^3+x^2-x^3+x^2-x-x^2+x-1\\ &=x^4-2x^3+x^2-1, \end{align}

but I can't figure out the steps to get there. Can anyone help?

• Write $x^4-2x^3+x^2 = x^2(x^2-2x+1)=x^2(x-1)^2$. – Math Lover Oct 16 '17 at 22:16
• what about the -1 at the end? – user453610 Oct 16 '17 at 22:22
• @Icy107 Have you heard of difference of squares? – Crescendo Oct 16 '17 at 22:25
• As a difference of squares. One of the squares matches the formula $(a-b)^2=a^2-2ab+b^2$, the other square that's subtracted from the first one is a constant. – Oscar Lanzi Oct 16 '17 at 22:27

Let’s examine the first three terms of your quartic$$y=x^4-2x^3+x^2-1$$The coefficients are in the order of $1,-2,1$. Thus\begin{align*}x^4-2x^3+x^2-1 & =x^2(x^2-2x+1)-1\\ & =x^2(x-1)^2-1\\ & =\left[x(x-1)-1\right]\left[x(x-1)+1\right]\\ & =(x^2-x-1)(x^2-x+1)\end{align*}where the second to last step is obtained with the difference of squares factorization.

• Nice answer! $(+1)$ – Mr Pie Oct 16 '17 at 23:14

As @MathLover suggests, use the factorization $$x^4-2x^3+x^2=x^2(x-1)^2=(x(x-1))^2.$$ Now use the fact that it is of the form $y^2-1$ to factor it as the difference of squares.

Remember that for all values $$a$$ and $$b$$, $$(a + b)^2 = (a + b)(a + b)$$ which we then FOIL to work out that $$(a + b)^2 = a^2 + b^2 + 2ab$$

Because the last term on the furthest $$RHS$$ (Right Hand Side of the equation) being $$2ab$$ is a multiple of $$2$$, if you have an equation for example:

$$x^2 + 6x = 7$$

What you do to solve for $$x$$ is divide $$6$$ by $$2$$ and then add the square of the result.

$$x^2 + 6x + \bigg(\frac{6}{2}\bigg)^2 - 9 = 7$$

Here we also subtract $$9$$ since $$\dfrac{6}{2} = 3$$ and $$3^2 = 9$$, because we do not want to change the $$RHS$$. And now, we can factor the $$LHS$$ (Left Hand Side of the equation):

$$\underbrace{x^2 + 6x + 9}_{(x + 3)^2} - 9 = 7 \Rightarrow (x + 3)^2 - 9 = 7 \Rightarrow (x + 3)^2 = 7 + 9 = 16 \Rightarrow x + 3 = \pm \sqrt{16} = \pm 4 \ \ \ \ \ \therefore \ \ x = 1 \quad \text{ or } \quad x = -7$$

You can also solve for $$x$$ using the Quadratic Formula as an alternative method by re-writing the equation as $$x^2 + 6x - 7 = 0$$, but this does not help us know how to factor. The technique that I showed you instead, however, is known as Completing the Square.

So now let's see where your equation fits. We have $$x^4 - 2x^3 + x^2 - 1$$ Since $$(a + b)^2 = a^2 + b^2 + 2ab \Rightarrow (a - b)^2 = a^2 + b^2 - 2ab$$, we can rearrange the terms. $$x^4 - 2x^3 + x^2 - 1 = \underbrace{(x^2)^2 + x^2 - 2\cdot x^2\cdot x}_{(x^2 - x)^2} - 1 = (x^2 - x)^2 - 1$$ Now we can expand the bracket of $$x^2 - x$$ (simply, expand $$x^2 - x$$) so that $$x^2 - x = x(x - 1)$$, thus we have $$[x(x - 1)]^2 - 1$$ And now we can actually use a seperate rule: $$a^2 - b^2 = (a + b)(a - b)$$ since $$1 = 1^2$$. \begin{align} \therefore [x(x - 1)]^2 - 1 &= [x(x - 1)]^2 - 1^2 \\ &= \big[[x(x - 1)] + 1\big]\big[[x(x - 1) - 1\big] \\ &= (x^2 - x + 1)(x^2 - x - 1) \\ &= (x^2 - x - 1)(x^2 - x + 1) \qquad \color{green}{\checkmark} \end{align} Remember that these rules are called Binomial Formulas (or Formulae) derived from the Binomial Theorem (or go here instead if you found the previous link too difficult to understand) and there you go! In fact, since the equation you want to factor has all of its terms being a product of $$x$$ (apart from $$1$$ of course), you can expand the entire equation and re-write it as $$x\cdot c - 1$$ for some value $$c$$. Then use the binomial formulas to factor $$c$$ and go from there! So there are two ways to complete the problem!

After you have completed the problem yourself, try and factor $$a^3 - b^3$$ to improve your knowledge! If you don't know, tap/click the yellow box below for the answer or go here!

$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$

Another way to factor $$a^3 - b^3$$ that is seperate to the link provided is shown below if needed.

$$\frac{a^3 - b^3}{a - b} = \frac{a^3 - ba^2 + ba^2 - b^3}{a - b} = a^2 + \frac{ba^2 - b^3}{a - b} = a^2 + ab + \frac{b^2a - b^3}{a - b} = a^2 + ab + b^2$$

:)

• You should also include the factorization of $a^3-b^3$, but hide it with the >! – Crescendo Oct 16 '17 at 23:52
• @Crescendo haha thanks man!! =D – Mr Pie Oct 17 '17 at 0:33