Question

find the zeros of 2x^4-3x^3-3x^2+6x-2 if you know that it's zeroes are √2 and -√2​

Answer 1

Solution: Perform division by x²-2.

Let $P(x)=2x^4-3x^3-3x^2+6x-2$ and let it be the dividend.

And, let $g(x)=x^2-2$ and let it be the divisor.

→ By the division algorithm

→ $P(x)=g(x)\times Q(x)$ (Since it has x²-2 as a factor, it is divisible.)

→ ∴$Q(x)=2x^2-3x+1$ and it is the quotient.

Therefore, two other roots of $Q(x)$ are x=1/2 or x=1.

Therefore, all the zeroes will be 1/2, 1, √2, and -√2.