Q.The polynomial $x^3+2x+3$ has

Step 1. Test $x=-1$. $P(-1)=(-1)^3+2(-1)+3=-1-2+3=0$ ✓ — a negative real root.

Step 2. Confirm via Descartes too. $P(x)=x^3+2x+3$: signs $+,+,+$, $0$ sign changes $\Rightarrow$ $0$ positive roots. $P(-x)=-x^3-2x+3$: signs $-,-,+$, $1$ sign change $\Rightarrow$ exactly $1$ negative root (parity forces it, since the difference $1-(\text{negative count})$ must be even, ruling out $0$).

Step 3. Divide by $(x+1)$. Quotient: $x^2-x+3$. …

Descartes' Rule of Signs pins down the real-root count exactly ($0$ positive, $1$ negative); dividing out t …

• Testing only Descartes' bounds without actually finding the concrete root $x=-1$ to confirm the discriminant of t …