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) = x² – 8x + 12

Given, p(X) = x² – 8x + 12

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

⇒ x² – 8x + 12 = 0

⇒ The given equation can be written as follows

⇒ (x – 6)(x – 2) = 0

⇒ Since, we know that x² – 8x + 12 can be written as (x – a)(x – b)

⇒ x² – 8x + 12 = x² – (a + b)x + ab

∴ coefficient on either side of the polynomial is same and a + b = 8 and ab = 12

⇒ we must find two numbers that satisfy a + b and ab we get, 6 and 2 as the numbers

∴ (x – 6)(x – 2) are the product of first degree polynomial

⇒ P(x) = 0 if (x – 6) is 0 and (x – 2) is 0

∴ x – 6 = 0

⇒ x = 6

⇒ x² – 8x + 12 = 6² – 8(6) + 12 = 36 – 48 + 12 = 0

And x – 2 = 0

⇒ x = 2

⇒ x² – 8x + 12 = 2² – 8(2) + 12 = 4 – 16 + 12 = 0

Hence, (x – 6)(x – 2) are the first degree factors of the polynomial and 6, 2 are the solutions of the given polynomial x² – 8x + 12