Question

Find all the zeros of the polynomial (2x⁴ − 11x³ + 7x² + 13x), it being given that two if its zeros are $3+\sqrt{2}$ and $3-\sqrt{2}$.

Answer 1

$$\begin{gathered} \begin{array}{l}\text{The given polynomial is}f\left(x\right)=2x^4-11x^3+7x^2+13x-7.\\ \text{Since}(3+\sqrt{2})\text{and}(3-\sqrt{2})\text{are the zeroes of}f\left(x\right),\text{it follows that each one of}(x+3+\sqrt{2})\text{and}(x+3-\sqrt{2})\text{is a factor of}f\left(x\right).\\ \\ \text{Consequently,}[x-(3+\sqrt{2}\left)\right][x-(3-\sqrt{2}\left)\right]=\left[\right(x-3)-\sqrt{2}]\left[\right(x-3)+\sqrt{2}]\\ =\left[\right(x-3{)}^2-2]=x^2-6x+7,\mathrm{which}\text{is a factor of}f(x)\end{array} \\ \text{On dividing}f\left(x\right)\text{by}(x^2-6x+7)\text{, we get:} \end{gathered}$$


$\begin{array}{l}\therefore f\left(x\right)=0\\ \Rightarrow 2x^4-11x^3+7x^2+13x-7=0\\ \begin{array}{l}=>(x^2-6x+7)(2x^2+x-1)=0\\ =>(x+3+\sqrt{2})(x+3-\sqrt{2})(2x-1)(x+1)=0\\ =>x=-3-\sqrt{2}\text{or}x=-3+\sqrt{2}\mathrm{or}x=\frac{1}{2}\text{or}x=-1\\ \text{Hence, all the zeros are}(-3-\sqrt{2}),(-3+\sqrt{2}),\frac{1}{2}\mathrm{and}-1.\end{array}\\ \\ \end{array}$