1.

Find all the zeroes of `2x^4-3x^3-3x^2+6x-2`, if you know that two of its zeroes are `sqrt(2)`and `-sqrt(2)`.

Answer» `q(x) = 2x^4 - 3x^3 - 3x^2 + 6x - 2`
`q(x) = p(x) *(x- sqrt2)(x + sqrt2)`
`q(x) = p(x)*(x^2-2)`
bu using identity `(x+a)(x-a) = x^2 - a^2`
`p(x) = (q(x))/(x^2-2)`
`= (2x^4 - 3x^3 - 3x^2 + 6x-2)/(x^2 - 2)`
by dividing , we get `p(x) = 2x^2 - x+1`
`= 2x^2 - 2x-x+1`
`= 2x^2 - 2x-x +1`
`= 2x[x-1]-1[x-1]`
`p(x)= (x-1)(2x-1)`
`x = 1,1/2`
answer


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