Q.Construct a cubic equation with roots

Step 1. Recall the formula. A cubic with roots $\alpha,\beta,\gamma$ is $x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma=0$.

Step 2. Part (i), roots $1,2,3$. Sum $=6$; pairwise sum $=1(2)+2(3)+3(1)=2+6+3=11$; product $=6$. Equation: $x^3-6x^2+11x-6=0$.

Step 3. Part (ii), roots $1,1,-2$. Sum $=0$; pairwise sum $=1(1)+1(-2)+(-2)(1)=1-2-2=-3$; product $=1(1)(-2)=-2$. Equation: $x^3-0x^2-3x-(-2)=0$, i.e. $x^3-3x+2=0$.

Step 4. Part (iii), roots $2,\tfrac12,1$. Sum $=2+\tfrac12+1=\tfrac72$; pairwise sum $=2(\tfrac12)+\tfrac12(1)+1(2)=1+\tfrac12+2=\tfrac72$; product $=2\cdot\tfrac12\cdot1=1$. Equation: $x^3-\tfrac72x^2+\tfrac72x-1=0$; multiplying by $2$: $2x^3-7x^2+7x-2=0$.

Direct construction from given roots using the sum / pairwise-sum / product formula.

• Forgetting a repeated root still contributes to every symmetric sum separately (part (ii))
• Leaving a fractional-coefficient answer instead of clearing denominators (part (iii))