Question

if two polynomials are x2+px+r and 3x2+p and both are divisible by x-a ,then prove that p3/27 +r2/4 = 0​

Answer 1

CORRECT QUESTION:

If two polynomials are x³ + px + r and 3x² + p and both are divisible by x - a ,then prove that p³/27 +r²/4 = 0

GIVEN:

x - a is the perfect divisor hence x = a

TO PROVE:

p³/27 + r²/4 = 0

PROOF:

Substitue x = a in 1st polynomial, we get

a³ + pa + r = 0

multipy by 3 on both sides

3a³ + 3pa + 3r = 0

Substitue x = a in 2nd polynomial, we get

3a² + p = 0

multiply by a on both sides

3a³ + ap = 0

3a³ = -ap

Substitute 3a³ = -ap in first polynomial.

-ap + 2ap + 3r = 0

ap + 3r = 0

a = -3r/2p

Substitue a = -3r/2p in 2nd polynomial

3( -3r/2p)² + p = 0

3 (9r²/ 4p²) = -p

27r² = -4p³

r²/4 = - p³/27

p³/27 + r²/4 = 0

HENCE PROVED.