Obtain all the zeroes of x4 -7x3 + 17x2 - 17x + 6, if two of its zeroes are 3 and 1 ....??
Let f(x) = x4 -7x3 + 17x2 - 17x + 6
Given: (x-3)(x-1) are the zeros of f(x).
product of zeros = (x-3)(x-1)=x2 -x-3x+3=x2 - 4x+3
therefore, on dividing the f(x) with x 2 -4x+3 , we get, x 2 -3x+2
now,
x 2 -3x+2=0
x 2 -2x-x+2=0
x(x-2)-1(x-2)=0
(x-1)(x-2)=0
either x-1= 0 or x-2=0
x=1 or x= 2
hence, the other zeros are 1 and 2.