Question

if the polynomial x⁶+px⁵+qx⁴-x²-x-3 divisible by x⁴-1 then the value of p²+q² is (a)1 (b)5 (c)10 (d)13​

Answer 1

Answer:

==f(x) = x^6+P x^5+qx^4--x^2 -x-3.

x^4 --1 divides f(x)

(x^2 ---1) ( x^2 +1)

x^2--1 =( x+1)(x--1)

f(--1) = (--1)^6 +(-1)5 +q(-1)^4. --(-1)^2 --(-1) --3

=1 --p+q --1 +1 -3 =0

--p+q =2. (1)

f(1) = 1^6 + P 1^5+q1^4 --1^2 —1 --3 =0

=1 +p+q --1 --1 -3 =0

P+q= 4 (2)

Adding 1 and 2

2q=6

q=3

p=4-3=1

p^2+q^2= 1+3^2=10

The required answer is 10

Answer 2

Answer:

The divisor is x4−1=(x−1)(x+1)(x2+1)

By factor theorem, f(1)=f(−1)=0

Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1

i.e., p+q=4 and p−q=−2

Adding the two, 2p=2 i.e. p=1 and ∴ q=3.

∴ p2+q2=1+9=10