Question

if 5x⁶-9x²+1 (px⁴+qx³+rx²+3x+1)(ax²+bx+c) , then the value of a p is​

Answer 1

Answer:

Step-by-step explanation:

x

2

−1=0

⟹=(x+1)(x−1)=0

⟹x=0,1

Now x

2

−1 is a factor of px

4

+qx

3

+rx

2

+sx+u

Putting x=1

we get,

p(1)

4

+q(1)

3

+r(1)

2

+s(1)+u=0

⟹p+q+r+s+u=0

Putting x=−1

we get,

p(−1)

4

+q(−1)

3

+r(−1)

2

+s(−1)+u=0

⟹p−q+r−s+u=0

Hence p+r+u=q+s=0