# Factor $a^4-3a^2-2ab+1-b^2$

In order to factor $a^4-3a^2-2ab+1-b^2$, I find that $a=1, b=-1$ makes the value of the expression 0. Thus, I assume $b=-a$.

I rewrite the expression on the assumption as: $$a^4-3a^2-2ab+1-b^2$$ $$=a^4-3a^2+2a^2+1-a^2$$ $$=a^4-2a^2+1$$ $$=(a^2-1)^2$$

Then I insert $a+b$, which is another form of the assumption above, into one of the factors. Since $a+b=0$, this insertion should cause no change to whatever relationships in the original expression. I arbitrarily set its coefficient as $1$. A factor, therefore, would be $(a^2+a+b-1)$.

I then divide the original expression by the factor and find that it indeed can be divided without a remainder and I also find the other factor. Thus, $$a^4-3a^2-2ab+1-b^2$$ $$=(a^2+a+b-1)(a^2-a-b-1)$$ Q.E.D.

The problem is that I don't know what I am doing. I tried to use the factor theorem for a bivariate (if this is the proper word for it) expression but I might be fabricating a fortuitous reasoning from the answer. Yes, I knew the answer before I tried, and no, I don't know how I can solve this in other ways. It would be great if someone can tell me if my reasoning can be made more clear.

• Well, rewriting the polynomial as $a^4-2a^2+1-a^2-2ab-b^2=(a^2-1)^2-(a+b)^2$ is simpler. – egreg Jan 21 '17 at 0:08

Hint: consider it as a quadratic in $b\,$:

$$-b^2-2ab + a^4-3a^2+1$$

Its discriminant is:

$$\frac{1}{4}\Delta = a^2 +(a^4-3a^2+1)=a^4-2a^2+1=(a^2-1)^2$$

Then calculate the roots $b_{1,2}$ as expressions in $a$, and the polynomial factors as $-(b-b_1)(b-b_2)$.

• I like this answer because of its purely mechanical nature. Things won't always work out this cleanly, of course, but this is an excellent tool for this problem. – Steven Stadnicki Jan 21 '17 at 0:12
• Thanks, everything's clear. – edmont123-carla Jan 21 '17 at 1:24

Here is probably the most simple method to factor it.

Note that $$a^2-b^2=(a-b)(a+b)$$So $$a^4-3a^2-2ab+1-b^2=a^4-2a^2+1-a^2-2ab-b^2=(a^2-1)^2-(a+b)^2$$ So we can factor it into $$(a^2+a+b-1)(a^2-a-b-1)$$