Q.Solve $18x^3+81x^2+121x+60=0$ given that one root is equal to half the sum of the remaining roots.

Use the given relation among the roots together with the sum of roots to identify one root directly, then divide it out to get a quadratic.

Let the roots of $18x^3+81x^2+121x+60=0$ be $a,b,c$, with $a$ equal to half the sum of the other two: $a=\dfrac{b+c}{2}$, i.e. $b+c=2a$.

Step 1 — find $a$ using the sum of roots. For $18x^3+81x^2+121x+60=0$:

$a+b+c = -\frac{81}{18} = -\frac{9}{2}$

Substituting $b+c=2a$:

$a+2a = -\frac92 \Rightarrow 3a=-\frac92 \Rightarrow a=-\frac32$

Verify $x=-\dfrac32$ is a root: $18\left(-\frac32\right)^3+81\left(-\frac32\right)^2+121\left(-\frac32\right)+60 = -60.75+182.25-181.5+60=0$ ✓.

Step 2 — divide out $(x+\frac32)$, equivalently $(2x+3)$. Using synthetic division of $18,81,121,60$ by root $-\frac32$:

$18 \;\to\; 81+18(-\tfrac32)=54 \;\to\; 121+54(-\tfrac32)=40 \;\to\; 60+40(-\tfrac32)=0$

Quotient: $18x^2+54x+40=0$, i.e. (dividing by $2$) $9x^2+27x+20=0$.

Step 3 — solve the quadratic: …