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

The given root-relation pins down one root via the sum-of-roots formula; dividing it out of the cubic leaves a quadratic that gives the remaining two roots.

Step 1 — Use the given relation.

Let the roots be $a,b,c$ with $a=\dfrac{b+c}{2}$, i.e. $b+c=2a$.

Step 2 — Sum of roots (from the cubic's coefficients).

For $18x^3+81x^2+121x+60=0$: sum of roots $=-\dfrac{81}{18}=-\dfrac92$.

$a+b+c=a+2a=3a=-\dfrac92 \implies a=-\dfrac32$.

Step 3 — Verify $x=-\dfrac32$ is a root, then divide it out.

Substituting $x=-\tfrac32$ into $18x^3+81x^2+121x+60$ gives $0$ (direct check), confirming it's a root, i.e. $(2x+3)$ is a factor.

Dividing: $18x^3+81x^2+121x+60 = (2x+3)(9x^2+27x+20)$.

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