IF x3-3x +2 is factor of x4-px2 +q
then
the value of p and q are res.
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Step-by-step explanation:
SInce x2−3x+2 is factorizable as (x-2)(x-1).
So, two of the zeroes of the polynomial, x4−px2+q are 1 and 2.
Putting x = 1 and equating the polynomial to zero, we have ,
1 - p + q = 0 ——— (1)
Putting x = 2 and equating the polynomial to zero, we have ,
16 - 4p + q = 0 ———- (2)
Solving for p and q from (1) and (2) , we have,
1 - p + q = 16 - 4p + q
or, 3 p = 15 , or p = 5
Putting p = 5 in (1) gives , q = 4.
Thus (p,q)ϵ{(5,4)}
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