# Factoring $1 - y-x^2-y^2-yx^2+y^3$

I was struggling to factorize this expression, I need it cleaned to bracket form but I was struggling to understand the process to get the answer? Is there a special formula or technique for factorizing these sorts of expressions?

$$1 - y-x^2-y^2-yx^2+y^3$$

Answer: $$\ ( 1+y)[(1-y)^2-x^2]$$

• There is no easy or obvious way to factor multivariate polynomials. Jan 2 '20 at 18:03
• Yeah, you just need to play around with the expression and hopefully recognize a pattern that resembles something you've seen before. Jan 2 '20 at 18:45

Factoring multivariable polynomials usually needs a bit of creativity. However, when looking for an approach, you can always think of multivariable polynomials as polynomials over a single variable whose coefficients are another polynomials!

With a bit of creativity: \begin{align} 1 - y-x^2-y^2-yx^2+y^3 &= \underbrace{(1-y-y^2 +y ^3)}_{\text{only y dependent}}-\overbrace{(x^2 +yx^2)}^\text{also x dependent}\\ \\ &= (1-y-y^2 +y ^3)-x^2(1+y). \\\\ \end{align}

If we plug $$y = -1$$, we have $$(1+1-1-1)+x^2(1-1) \equiv 0.$$ So we can divide this expression by $$(1+y)$$, and obtain the desired result or factorize even further to get

\begin{align}(1+y)( (y^2-2y+1) - x^2) &= (1+y)((y-1)^2-x^2)\\\\ &= \boxed{(1+y)(y-1-x)(y-1+x).}\end{align}

A more deep argument of why I tried this method: note that this expression is cubic over $$y$$. So if you can factorize this polynomial $$p(x,y)$$ into $$q(x,y)r(x,y)$$, either $$q$$ has degree $$1$$ over $$y$$ or $$q$$ has degree $$2 \implies r$$ has degree $$1$$. So there's a rational function $$g(x)$$ such that $$p(x,g(x)) = 0$$. So we need a factor of the form $$k(x)y - l(x)$$.

Note that you could use this last argument to deduce the same thing for $$x$$. So getting $$x^2 = k(y)/l(y)$$ is also a good guess.

• Why did this get downvoted? Jan 2 '20 at 17:21

Note $$x^2$$ is the only $$x$$-term,

$$1 - y-x^2-y^2-yx^2+y^3$$ $$=1-y-y^2+y^3 -(1+y)x^2$$ $$=(1-y) -y^2(1-y) -(1+y)x^2$$ $$=(1-y)(1-y^2)-(1+y)x^2 =(1+y)(1-y)^2-(1+y)x^2$$ $$= ( 1+y)[(1-y)^2-x^2]$$

Once you gave us the answer, the factoring became easy. You just have to look for the factor $$1+y$$

$$1 - y-x^2-y^2-yx^2+y^3= (1+y^3)-(y+y^2)-x^2(1+y)$$

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

$$=(1+y)[(1-y)^2-x^2]$$

Here is how I factored it:

$$1-y-x^2-y^2-yx^2+y^3=$$ $$(1-y^2)-(y-y^3)-(x^2+yx^2)=$$ $$(1-y^2)-y(1-y^2)-x^2(1+y)=$$ $$(1-y^2)(1-y)-x^2(1+y)=$$ $$(1-y)(1+y)(1-y)-x^2(1+y)=$$ $$(1+y)(1-y)^2-(1+y)x^2=$$ $$\boxed{(1+y)[(1-y)^2-x^2]}$$

You can factorize it further:

$$(1+y)[(1-y)^2-x^2]=$$ $$(1+y)[(1-y-x)(1-y+x)]=$$ $$\boxed{(1+y)(1-y-x)(1-y+x)}$$

We can expand $$(1+y)(1-y-x)(1-y+x)$$ to make sure we got it:

$$\require{cancel} (1+y)(1-y-x)(1-y+x)=$$ $$(1\cancel{-y}-x\cancel{+y}-y^2-xy)(1-y+x)=$$ $$(1-x-y^2-xy)(1-y+x)=$$ $$(1-x-y^2-xy)-(y-xy-y^3-xy^2)+(x-x^2-xy^2-x^2y)=$$ $$1\cancel{-x}-y^2\cancel{-xy}-y\cancel{+xy}+y^3\cancel{+xy^2}\cancel{+x}-x^2-\cancel{xy^2}-x^2y=$$ $$1-y^2-y+y^3-x^2-x^2y=$$ $$\boxed{1-y-x^2-y^2-yx^2+y^3}$$