Answer :

Given, p(x) = x^{2} + x – 1

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ The given equation can be written as = x^{2} + x + (– 1) which is of the form x^{2} + (a + b)x + ab

∴ a + b = 1 and ab = – 1

⇒ we know that (a + b)^{2} –(a – b)^{2} = 4ab

∴ (a + b)^{2} – 4ab = (a – b)^{2}

⇒ we know a + b = 1 and ab = – 1

⇒ 1 – 4(– 1) = (a – b)^{2}

⇒ (a – b)^{2} = 5

⇒ a – b =

⇒ Solving the equation both equation a + b and a – b we get as follows

⇒ a + b + a – b = 1 + √5

⇒ a = (1 + √5)

⇒ (a + b) – (a – b) = 1 + √5

⇒ b = (1 – √5)

∴ x^{2} + x + (– 1) has factors (x + (1 + √5))(x + (1 – √5))

⇒ p(x) = 0 if (x + (1 + √5))is 0 and (x + (1 – √5))is 0

⇒ (x + (1 + √5)) = 0

⇒ x = – (1 + √5)

⇒ (– (1 + √5))^{2} + (– (1 + √5)) + (– 1) = 0

And

⇒ (x + (1 – √5)) = 0

⇒ x = – (1 – √5)

⇒ (– (1 – √5))^{2} + (– (1 – √5)) + (– 1) = 0

Hence, (x + (1 + √5))(x + (1 – √5)) are the first degree factors of the polynomial and – (1 + √5) and – (1 – √5) are the solution of the given polynomial x^{2} + x – 1

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