Question

obtain all other zeroes of the polynomial 2x4+4x3_22x2_32x+48 if two of its zeroes are +2root 2 and -2root2

Answer 1

Answer: the other two zeroes are -3 &1

Step-by-step explanation: when the zeroes are  2√2 &-2√2

we take the zeroes as (x-2√2)    (x+2√2) {expand using (a+b) (a-b)} identity

then we get the answer as x²-8 where we have an an (x) term missing so therefore we add 0 to it and we get the answer as x²+0x-8 which is a complete equation

and now for obtaining all the zeroes we have to divide the polynomial 2x4+4x³-22x²-32x+48 by the divisor x²+0x-8

after dividing we get the quotient as 2x²+4x-6

and the reminder as 0

so therefore we have to factorize the quotient 2x²+4x-6

2x²+4x-6      (sum=4 produt=12)

2x²-2x +6x-6

2x(x-1) 6(x-1)

(2x+6) (x-1)      

2x+6=0    or    x-1=0

2x=-6          or   x=1

x=-6/2

x=-3

so the zeroes r -3,1,2√2,-2√2