Question

Find all the zeroes of 2x^4-3x^3-3x^2+6x-2if two of it's zeroes are √2and -√2

Answer 1

Answer:

Step-by-step explanation:

p(x) = 2x^4-3x^3-3x^2+6x-2

Here, two zeroes are _/2 and -_/2

Let, g(x) be a polynomial whose zeroes are- alpha=_/2 and beeta= -_/2

g(x) = (x-alpha) (x-beeta)

= (x- _/2) [x-(-_/2)]

= (x-_/2) (x+_/2)

= (x)^2-(_/2)^2

= x^2-2

On dividing p (x) by g (x)

Here, q (x)= 2x^2-3x+1

= 2x^2-1x-2x+1

= x (2x-1)-1 (2x-1)

= (2x-1) (x-1)

q (x) will give other two zeroes when we put q (x)=0

==> (2x-1) (x-1)=0

==> 2x-1=0 and x-1=0

==> x=1/2 and x=1

Hence, other two zeroes of p (x) are 1/2 and 1