Question

Find all the zeros of the polynomial $(2x^4-11x^3+7x^2+13x-7)$ , it being given that two of its zeros are $(3+\sqrt{2})and(3-\sqrt{2})$ .

Answer 1

The given polynomial is
$f\left(x\right)=(2x^4-11x^3+7x^2+13x-7)$

Since $(3+\sqrt{2})and(3-\sqrt{2})$ are the zeros of f(x),
It follows that each one of $(x+3+\surd 2)and(x+3–\surd 2)$ is a factor of f(x).

Consequently,
$[x–(3+\sqrt{2}\left)\right][x–(3–\sqrt{2}\left)\right]=\left[\right(x-3)–\sqrt{2}]\left[\right(x-3)+\sqrt{2}]=\left[\right(x–3{)}^2–2\left)\right]=x^2–6x+7$, which is a factor of f(x)

On dividing f(x) by $(x^2–6x+7)$ we get:

Therefore, $f\left(x\right)=0$

⇒ $2x^4–11x^3+7x^2+13x–7=0$
⇒ $(x^2–6x+7)(2x^2+x–1)=0$
Hence, all the zeros are ⇒ $x=(-3–\sqrt{2}),(-3+\sqrt{2}),\frac{1}{2}and-1$