question_answer

Find all the zeroes of the polynomial \[8{{x}^{4}}+8{{x}^{3}}-18{{x}^{2}}-20x-5,\] if it is given that two of its zeroes are \[\sqrt{\frac{5}{2}}\] and \[-\sqrt{\frac{5}{2}}\].

Answer:

Given polynomial is\[8{{x}^{4}}+8{{x}^{3}}-18{{x}^{2}}-20x-5\].

Since two zeroes are \[\sqrt{\frac{5}{2}}\] and \[-\sqrt{\frac{5}{2}}\]

\[\therefore \left( x-\sqrt{\frac{5}{2}} \right)\left( x+\sqrt{\frac{5}{2}} \right)={{(x)}^{2}}-{{\left( \sqrt{\frac{5}{2}} \right)}^{2}}\]

\[={{x}^{2}}-\frac{5}{2}\]

Dividing the polynomial with \[={{x}^{2}}-\frac{5}{2}\]

\[\therefore \,\,8{{x}^{4}}+8{{x}^{3}}-18{{x}^{2}}-20x-5\]

\[=\left( {{x}^{2}}-\frac{5}{2} \right)(8{{x}^{2}}+8x+2)\]

\[=\left( {{x}^{2}}-\frac{5}{2} \right).2(4{{x}^{2}}+4x+1)\]

\[=2\left( {{x}^{2}}-\frac{5}{2} \right)(4{{x}^{2}}+2x+2x+1)\]

\[=2\left( {{x}^{2}}-\frac{5}{2} \right)[2x(2x+1)+1(2x+1)]\]

\[=2\left( {{x}^{2}}-\frac{5}{2} \right)(2x+1)(2x+1)\]

All the zeroes are \[\sqrt{\frac{5}{2}},-\sqrt{\frac{5}{2}},\frac{-1}{2}\] and \[\frac{-1}{2}\].