Question

Factorize the following expressions
I. $ {p^2} + 6p + 8 $
II. $ {q^2} - 10q + 21 $
III. $ {p^2} + 6p - 16 $

Answer 1

Hint: Factorization is the process of writing the numbers of the function or any mathematical expression in the simplest form such that no further simplification can take place after the factorization of any polynomial.
Here, in the question we need to factorize all the given expressions. As we can see that all the given expressions have the degree two, which implies that the given expressions are quadratic. Now, to factorize the quadratic equation, splitting the middle term is the best method that we can opt for. In this method, the coefficient of $ x $ is splinted into two-parts such that the product of those coefficients results in the product of the coefficients of $ {x^2} $ and $ {x^0} $

Complete step-by-step answer:
I. $ {p^2} + 6p + 8 $
The given polynomial has the degree of 2 and so, we can say that the polynomial is a quadratic polynomial. Now, a quadratic polynomial can be factorized by splitting the middle term method.
So, the given quadratic equation can be factorized as:
 $
  {p^2} + 6p + 8 = {p^2} + 4p + 2p + 8 \\
   = p(p + 4) + 2(p + 4) \\
   = (p + 2)(p + 4) \\
  $
Hence, the polynomial $ {p^2} + 6p + 8 $ can be factorized as $ (p + 2)(p + 4) $ .

II. $ {q^2} - 10q + 21 $
The given polynomial has the degree of 2 and so, we can say that the polynomial is a quadratic polynomial. Now, a quadratic polynomial can be factorized by splitting the middle term method.
So, the given quadratic equation can be factorized as:
 $
  {q^2} - 10q + 21 = {q^2} - 7q - 3q + 21 \\
   = q(q - 7) - 3(q - 7) \\
   = (q - 3)(q - 7) \\
  $
Hence, the polynomial $ {q^2} - 10q + 21 $ can be factorized as $ (q - 3)(q - 7) $ .

III. $ {p^2} + 6p - 16 $
The given polynomial has the degree of 2 and so, we can say that the polynomial is a quadratic polynomial. Now, a quadratic polynomial can be factorized by splitting the middle term method.
So, the given quadratic equation can be factorized as:
 $
  {p^2} + 6p - 16 = {p^2} + 8p - 2p - 16 \\
   = p(p + 8) - 2(p + 8) \\
   = (p - 2)(p + 8) \\
  $
Hence, the polynomial $ {p^2} + 6p - 16 $ can be factorized as $ (p - 2)(p + 8) $ .

Note: Students should be aware of the concept of the splitting the middle term to factorize the quadratic equation. Moreover, candidates should be careful while splitting the middle term as many times the coefficient of $ {x^2} $ is not 1 as here in the question.