Question

Find the value of p and q at the polynomial x(power4) + p x(x power 3) + 2x(power 2)- 3x + q is divisible by (x-1) and (x+1).

Answer 1

p(x) = $x^{4} + px^{3} + 2 x^{2} - 3x + q$
Here, x-1 and x + 1 are its factors.
so, x - 1 = 0, x + 1 = 0
x = 1, x= -1
p(1) = 0
p(-1) = 0
p(1) = $1^{4} + p 1^{3} + 2 *1^{2} - 3*1 +q$
0 = 1 + p + 2 - 3 + q
0 = p +q
-p = q eqn 1
p(-1) = $(-1)^{4} + p (-1)^{3} + 2 *(-1)^{2} - 3*(-1) +q$
0 = 1 - p +2 +3 +q
0 = 6 - p + q
from eqn 1 -p = q
0 = 6 + q + q
6 = 2q
3 = q
-p = q
p = -q
p = -3

Answer 2

p(x) = x⁴ + px³+ 2x² -3x +q -----------------(1)
and (x-1) & (x+1)
x-1 = 0
therfore x = 1
X+1 = 0
x = -1
therfore, x = (1,-1)
Substituting x= 1 in (1)
therfore
1⁴+px³+2x²-3x+q = 0
1 + p + 2 - 3 +q =0
p+q = 0 -----------------(2)
Substituting x=(-1)
(-1)⁴+p(-1)³+2(-1)²-3(-1)+q = 0
1 -p+ 2 + 3 + q =0
-p+q =  --6---------------------(3)
(2) + (3) =
2q= --6
q = -3
substituting q = -3 in (2)
p+q=0
p+(-3)=0
p=3
q=-3