Question

Find the value of the expression $2x^3+2x^2-7x+72$ where $x=\frac{3+5\sqrt{1}}{2}$.

Answer 1

It is given that $x=\frac{3+5\sqrt{-1}}{2}$ which is same as $x=\frac{3+5i}{2}$ because $i^2=-1$.

Now, let us substitute $x=\frac{3+5i}{2}$ in the given polynomial $2x^3+2x^2-7x+72$ as follows:

$2x^3+2x^2-7x+72=2{\left(\frac{3+5i}{2}\right)}^3+2{\left(\frac{3+5i}{2}\right)}^2-7\left(\frac{3+5i}{2}\right)+72=2\left(\frac{3^3+(5i{)}^3+(3\times 3^2\times 5i)+(3\times 3\times \left(5i{)}^2\right)}{8}\right)+2\left(\frac{3^2+(5i{)}^2+(2\times 3\times 5i)}{2}\right)-\left(\frac{21+35i}{2}\right)+72(\because (a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2,(a+b{)}^2=a^2+b^2+2ab)=2\left(\frac{27-125i+135i-225}{8}\right)+2\left(\frac{9-25+30i}{4}\right)-\left(\frac{21+35i}{2}\right)+72(\because i^2=-1)$

$=\left(\frac{-198+10i}{4}\right)+\left(\frac{-16+30i}{2}\right)-\left(\frac{21+35i}{2}\right)+72=\left(\frac{2(-99+5i)}{4}\right)+\left(\frac{30i-16}{2}\right)-\left(\frac{35i+21}{2}\right)+72=\frac{5i-99}{2}+\frac{30i-16}{2}-\left(\frac{35i+21}{2}\right)+\left(\frac{72\times 2}{2}\right)=\frac{5i-99+30i-16-35i-21+144}{2}=\frac{35i-35i-136+144}{2}=\frac{8}{2}=4$

Hence, $2x^3+2x^2-7x+72=4$ when $x=\frac{3+5\sqrt{-1}}{2}$.