Prove that if x<s

Given, ax³ + bx² + cx + d

Need to show a = – c and b = – d if x² – 1 is a factor

⇒ consider, x² – 1 is a factor of ax³ + bx² + cx + d

⇒ then x = + 1, x = – 1

⇒ substitute x value in the equation ax³ + bx² + cx + d

We get as follows

⇒ for x = 1 we get a(1)³ + b(1)² + c(1) + d

= a + b + c + d ……..eq(1)

⇒ for x = – 1 we get a(– 1)³ + b(– 1)² + c(– 1) + d

= – a + b – c + d ………eq(2)

⇒ Solving the two equations we get

⇒ (a + b + c + d) + (– a + b – c + d) = 0

⇒ b + d = 0

∴ b = – d

And if (a + b + c + d) – (– a + b – c + d) = 0

⇒ a + c = 0

∴ a = – c