Question

If the zeroes of the polynomial x³ − 3x² + x + 1 are (a − b), a and (a + b), find the values of a and b.

Answer 1

The given polynomial $=x^3-3x^2+x+1$and its roots are $(a-b),a\mathrm{and}(a+b)$.

$$\begin{gathered} \mathrm{C}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{y}\mathrm{n}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{}A{x}^3+B{x}^2+Cx+D,\mathrm{we}\mathrm{have}: \\ A=1,B=-3,C=1\mathrm{and}D=1 \\ \mathrm{Now},(a-b)+a+(a+b)=\frac{-B}{A} \\ =>3a=-\frac{-3}{1} \\ =>a=1 \\ \mathrm{Also},(a-b)\times a\times (a+b)=\frac{-D}{A} \\ =>a\left(a^2-b^2\right)=\frac{-1}{1} \\ =>1(1^2-b^2)=-1 \\ =>1-b^2=-1 \\ =>b^2=2 \\ =>b=\pm \surd 2 \\ \therefore a=1\mathrm{and}b=\pm \surd 2 \end{gathered}$$