Question

prove that ap+q=0if f(x)=x3-3px+2pis divisible by g(x), =x2+2ax+a2​

Answer 1

Answer:

Step-by-step explanation:

Prove that ap+p=0if f(x)=x3-3px+2pis divisible by g(x), =x2+2ax+a2​

f(x) = x³ - 3px + 2p

g(x) = x² + 2ax + a²

Let say f(x) = g(x)q(x)

q(x) = (bx + c)

g(x)q(x) =  (x² + 2ax + a²)(bx + c)

=> g(x)q(x) =  bx³ + cx² + 2abx² + 2acx + a²bx + a²c

=> g(x)q(x) =  bx³ + x²(c + 2ab) + x(2ac + a²b) + a²c

comparing with x³ - 3px + 2p

b = 1

c + 2ab = 0 => c + 2a = 0 => c = -2a

a²c = 2p  => -2a³ = 2p => p = -a³

2ac + a²b = -3p  => -4a² + a² = -3p  => p = a²

ap + p = a(a²) + (-a³) = a³ - a³ = 0

or

ap+q=0if f(x)=x³-3px+2q

f(x) = x³ - 3px + 2q

g(x) = x² + 2ax + a²

Let say f(x) = g(x)q(x)

q(x) = (bx + c)

g(x)q(x) =  (x² + 2ax + a²)(bx + c)

=> g(x)q(x) =  bx³ + cx² + 2abx² + 2acx + a²bx + a²c

=> g(x)q(x) =  bx³ + x²(c + 2ab) + x(2ac + a²b) + a²c

comparing with x³ - 3px + 2q

b = 1

c + 2ab = 0 => c + 2a = 0 => c = -2a

a²c = 2q  => -2a³ = 2q => q = -a³

2ac + a²b = -3p  => -4a² + a² = -3p  => p = a²

ap + q = a(a²) + (-a³) = a³ - a³ = 0

Answer 2

Answer:

Hence the Proof...

..........................

..............