Factor x^3+(2a+1)x^2+(a^2+2a-1)x+(a^2-1)
\[x^3+(2a+1)x^2+(a^2+2a-1)x+(a^2-1)\]
Here's a somewhat roundabout hint: when you have a polynomial with roots \(a\pm bi\) and need to construct the polynomial, you multiply \[(x-a-bi)(x-a+bi) = x^2-2ax +a^2+b^2\]the sign of the \(b^2\) term changes because of the \(i^2=-1\) so let's consider the related multiplication \[(x+a-b)(x+a+b) = x^2 + ax + bx + ax + a^2 + a b - bx - ab - b^2\]\[\qquad=x^2+2ax + a^2 -b ^2\] Notice that the constant term here is \(a^2 -b^2\), and the constant term in our polynomial to be factored could be written the same way, if \(b =1\) This suggests that two of our factors might be \((x+a-1)(x+a+1)\) Now multiplying those two terms together only gives you a polynomial with an \(x^2\), not an \(x^3\), but hopefully it is clear upon consideration that if we multiply by another binomial, we'll get that \(x^3\) term we need, and we could find that binomial by polynomial long division, if necessary.
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