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 + 13x + 12

Given, p(X) = x² + 13x + 12

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

⇒ x² + 13x + 12 = 0

⇒ the given equation can be written as follows

⇒ (x + 12)(x + 1) = 0

⇒ Since, we know that x² + 12x + 13 can be written as (x + a)(x + b)

⇒ x² + 13x + 12 = x² + (a + b)x + ab

∴ coefficient on either side of the polynomial is same and a + b = 7 and ab = 12

⇒ we must find two numbers that satisfy a + b and ab we get, 12 and 1 as the numbers

∴ (x + 12)(x + 1) are the product of first degree polynomial

⇒ P(x) = 0 if (x + 12) is 0 and (x + 1) is 0

∴ x + 1 = 0

⇒ x = – 1

⇒ x² + 13x + 12 = (– 1)² + 13(– 1) + 12 = 1 – 13 + 12 = 0

And x + 12 = 0

⇒ x = – 12

⇒ x² + 13x + 12 = (– 12)² + 13(– 12) + 12 = 144 – 156 + 12 = 0

Hence, (x + 12)(x + 1) are the first degree factors of the polynomial and – 12, – 1 are the solutions of the given polynomial x² + 13x + 12