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 – 7x + 12

Given, p(X) = x² – 7x + 12

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

⇒ x² – 7x + 12 = 0

⇒ The given equation can be written as follows

⇒ (x – 4)(x – 3) = 0

⇒ Since, we know that x² – 7x + 12 can be written as (x – a)(x – b)

⇒ x² – 7x + 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, 4 and 3 as the numbers

∴ (x – 4)(x – 3) are the product of first degree polynomial

⇒ Substitute x = 4 and x = 3, We get p(x) as 0

⇒ x² – 7x + 12 = 4² – 7(4) + 1 2 = 16 – 28 + 12 = 0

⇒ x² – 7x + 12 = 3² – 7(3) + 12 = 9 – 21 + 12 = 0

Hence, (x – 4)(x – 3) are the first degree factors of the polynomial and 4, 3 are the solutions of the given polynomial x² – 7x + 12