Find the value of $(1+\alpha)(1+\beta)(1+r)(1+s)$

Question

If $\alpha,\beta,r,s$ are the roots of $x^{4}-x^{3}+x^{2}+x+3=0$, find the value of $(1+\alpha)(1+\beta)(1+r)(1+s)$.

This question appeared on one of the sample papers which I came across.

My first instinct was to graph the function and find its root using it. However, the function showed no real roots.

Please help me with this question. Thank you so much.

Answer 1

We have that $$(1 + \alpha)(1+\beta)(1 + r)(1 + s) = 1 + s + r + \alpha + \beta + \alpha\beta + \alpha r + \alpha s + \beta r + \beta s + rs + \alpha\beta r + \alpha\beta s+ \alpha rs + \beta r s + \alpha\beta r s$$ By Vieta's formulas, we know that $$\alpha + \beta + r + s = 1$$ $$\alpha\beta + \alpha r + \alpha s + \beta r + \beta s + rs = 1$$ $$\alpha\beta r + \alpha rs + \alpha \beta s + \beta r s = -1$$ $$\alpha\beta r s = 3$$ Hence, $(1 + \alpha)(1+\beta)(1 + r)(1 + s) = 1 + 1 + 1 -1 + 3 = 5$

Answer 2

If $\alpha, \beta, r, s$ are roots of $x^{4}-x^{3}+x^{2}+x+3=0$ then $1+\alpha,1+\beta,1+r,1+s$ are roots of $(x-1)^{4}-(x-1)^{3}+(x-1)^{2}+(x-1)+3=0$ and $(1+\alpha)(1+\beta)(1+r)(1+s)$ is the constant term in the polynomial. Now to expand the whole thing will be a headache but fortunately we don't need to do that. The constant term in $(x - 1)^n$ is always $(-1)^{n}$ for any $n$ so in our case the constant term will be $ 1 -(-1) + 1 + (-1) + 3 = 5$.