Question

please answer questions no 18

Answer 1

Let p (x) = x^4+px^3 +2x^2 -3x +q


(x-1) -----factor 1
(x+1) ----- factor 2

By substituting the value of x (from factor 1) in polynomialp (x)

p (1) = (1)^4 +p (1)^3 +2 (1)^2 -3 (1) +q =0

[ this polynomial must be writen equal to 0 as (x-1) is a factor of p (x) ]
= 1 +p+2-3+q=0
= p+q= 0 ------- equation 1.

Again by substituting the value of x (from factor 2 ) in polynomial p (x)

p (-1) = (-1)^4+p (-1)^3+2 (-1)^2-3 (-1)+q=0
L = 1 -p +2 +3 +q=0
= 6-p+q =0
= -p+q = -6
= p- q =6 ------ equation 2.


By adding equation 1 and 2
p+q + p-q= 0 +6
2p = 6
p = 3

Again by substituting the value of p in any of the two equations
p+q =0
3+q=0
q =-3

So, the value of p and q are respectively 3 and -3.