Question

Factorize $ {p^2} + 6p + 8 $
A) $ (p + 2)(p - 2) $
B) $ (p + 4)(2p - 2) $
C) $ (2p + 4)(p + 2) $
D) $ (p + 4)(p + 2) $

Answer 1

Hint:
A quadratic equation in the variable p is an equation of the form of $ a{p^2} + bp + c = 0 $ . In fact, any equation of the form $ x(p) = 0 $ where $ x(p) $ is polynomial of degree 2 is a quadratic equation.
When we factorize an algebraic equation, we write it as a product of factors. These factors may be numbers, algebraic variables, or expressions. Consider expressions like $ {x^2} + 5x + 6 $ . It is not obvious what their factors are. If the form is not a perfect square. So, factorize this expression $ (x + a)(x + b) = {x^2} + (a + b)x + ab $

Complete step by step solution:
Step 1: The given equation is $ {p^2} + 6p + 8 $ . This is a quadratic equation. Here, the coefficient of $ {p^2} $ is 1, and the coefficient of p is 6, 8 is a constant term. In this expression, the term 8 is not a perfect square.
Step 2. They however seem to be the type of $ (p + a)(p + b) = {p^2} + (a + b)p + ab $ . We have to look at the coefficient of p and a constant term. If we compare the right-hand side of identity $ {p^2} + (a + b)p + ab $ with $ {p^2} + 6p + 8 $ . We find ab = 8 and a+b = 6.
From this, we must obtain a and b . the factors will be $ (p + a)(p + b) $ . If ab = 8 it means that a and b are factors of 8. Let us try a = 8 and b = 1
For these values a+b = 9 not 6. This is incorrect.
Let us try a = 4 and b = 2 for these values a+b = 6 this the correct .
So, the given expression can be written as $ {p^2} + (4 + 2)p + 4 \times 2 $ . So, we get the factorization form of the given expression then $ (p + 4)(p + 2) $ .

So, option D is correct.

Note:
In general, for factoring an algebraic expression of the type $ a{p^2} + bp + c = 0 $ , we find two factors l and m (i.e. constant term) such that lm = c and l+m =b. we can use this type of factorization in only one type of variable. If we have the constant term perfect square, we can use some other suitable identities.