Finding $3r^2 + rs - 2s^2 + 2r - 3s$ given that $r$ and $s$ are the roots of $x^2 + x + 7$

Question

Suppose $r$ and $s$ are the roots of $x^2 + x + 7.$ Then, find the value of $3r^2 + rs - 2s^2 + 2r - 3s.$

---

I wanted to try and find a relationship between $3r^2 + rs - 2s^2 + 2r - 3s$ and perhaps $r + s$ and $rs,$ but I can't seem to find a factorization that works. Can someone give me a hint please?

Answer 1

Why not just substitute $r = a+ib, s = a-ib$ into your expression?

Answer 2

Answer is $1$ $$ (x^2+x+7)=0~~~.…...\tag1$$ and $r,s$ is roots so $rs=7$, $r+s=-1$ this implies $s=(-1-r)$ Now \begin{align} & 3r^2+rs-2s^2+2r-3s \\&=3r^2+7-2(-1-r)^2+2r-3(-1-r) \\&=3r^2+7-2(1+2r+r^2)+2r+3+3r \\&=3r^2+7-2-4r-2r^2+2r+3+3r \\&=r^2+r+8 \\&=(r^2+r+7)+1 \\&=0+1 \end{align} Since $r$ is root of the equation so putting $x=r$ in equation (1) we got $r^2+r+7=0$.