Question

find the zeroes of the polynomial  2x⁴ - 3x³ -3x +6x-2 if two of its zeroes are √2 and -√2​

Answer 1

Answer:

The zeros of the polynomial $2x^{4}-3x^{3}-3x^{2} +6x-2$ are $\frac{1}{2},$ 1 , $\sqrt{2}$ and $-\sqrt{2}$.

Step-by-step explanation:

Since $\sqrt{2}$ and $-\sqrt{2}$ are zeros of the polynomial $2x^{4}-3x^{3}-3x^{2} +6x-2$,$\\ (x-\sqrt{2})$ and $(x+\sqrt{2})$ are factors of the given polynomial.

Thus the product $(x-\sqrt{2})(x+\sqrt{2})=(x^{2} -2)$ is a factor of the given polynomial.

On dividing $2x^{4}-3x^{3}-3x^{2} +6x-2$ by $(x^{2} -2)$. we obtain

$2x^{4}-3x^{3}-3x^{2} +6x-2$÷$(x^{2} -2)=2x^{2} -3x+1$

Factorizing the expression on the right hand side of the above equation, we get

$2x^{2} -3x+1$$=(2x-1)(x-1)$

Thus

$(2x-1)(x-1)=0$   ⇒ $2x-1=0$  or $x-1=0$

which implies

$x=\frac{1}{2}$  or x = 1.

So the zeros  of the polynomial are $\frac{1}{2},$ 1 , $\sqrt{2}$ and $-\sqrt{2}$.