Write each polynomial below as a product of first degree polynomials. Write also the solutions of the equation p(x) = 0 in each.

p(x) = 2x² – 5x + 2

Given, p(x) = 2x² – 5x + 2

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

⇒ 2x² – 5x + 2 can be written as 2(x² – x + 1)

⇒ Now, we will find out the factors for x² – x + 1

⇒ x² – x + 1 is in the form x² – (a + b)x + ab and (x – a)(x – b) are the factors

∴ coefficient on either side of the polynomial is same and a + b =

and ab = 1

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

∴ (a + b)² – 4ab = (a – b)²

⇒ – 4(1) = (a – b)²

⇒ – 4 = (a – b)²

⇒ = (a – b)²

⇒ = (a – b)²

⇒ a – b =

⇒ a – b =

⇒ We need to find out the values of a and b by solving (a + b) + (a – b) and (a + b) – (a – b)

⇒ we take a – b =

⇒ (a + b) + (a – b) = +

⇒ 2a =

⇒ a = 2

⇒ (a + b) – (a – b) = –

⇒ 2b =

⇒ b =

∴ we get a = 2 and b =

And if we take a – b = – we get as follows

⇒ (a + b) + (a – b) = + (– )

⇒ 2a =

⇒ a =

⇒ (a + b) – (a – b) = –

⇒ 2b =

⇒ b = 2

∴ a = and b = 2

⇒ So, we have 2(x – )(x – 2)

⇒ (2x – 1)(x – 2)

∴ (2x – 1)(x – 2) are the factors

⇒ p(x) = 0 if (2x – 1) is 0 and (x – 2) is 0

⇒ (2x – 1) = 0

⇒ 2x = 1

⇒ x =

⇒ Substitute, x = in 2x² – 5x + 2 = 2(² – 5( + 2 = – + 2 = 0

⇒ (x – 2) = 0

⇒ x = 2 substitute, 2x² – 5x + 2 = 2(2)² – 5(2) + 2 = 8 – 10 + 2 = 0

Hence, (2x – 1)(x – 2) are the first degree factors of the polynomial and and 2 are the solution of the given polynomial 2x² – 5x + 2