To factorize the given trinomial [tex]\( x^2 + 14x + 49 \)[/tex], we need to find two binomials that, when multiplied together, produce this trinomial.
Let's denote the trinomial as [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 14 \)[/tex], and [tex]\( c = 49 \)[/tex].
### Step-by-Step Solution:
1. Identify the coefficients:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 14 \)[/tex]
- [tex]\( c = 49 \)[/tex]
2. Find the roots:
We need to find two numbers that multiply to [tex]\( a \cdot c = 1 \cdot 49 = 49 \)[/tex] and add up to [tex]\( b = 14 \)[/tex].
Those two numbers are [tex]\( 7 \)[/tex] and [tex]\( 7 \)[/tex], because:
- [tex]\( 7 \times 7 = 49 \)[/tex]
- [tex]\( 7 + 7 = 14 \)[/tex]
3. Express the middle term using these numbers:
Rewrite [tex]\( 14x \)[/tex] as [tex]\( 7x + 7x \)[/tex]:
[tex]\[
x^2 + 7x + 7x + 49
\][/tex]
4. Group the terms and factor by grouping:
[tex]\[
(x^2 + 7x) + (7x + 49)
\][/tex]
Factor out the greatest common factor (GCF) in each group:
[tex]\[
x(x + 7) + 7(x + 7)
\][/tex]
5. Factor out the common binomial:
Both terms have a common binomial factor [tex]\((x + 7)\)[/tex]:
[tex]\[
(x + 7)(x + 7)
\][/tex]
6. Simplify the expression:
Since it is the same binomial repeated, it can be written as:
[tex]\[
(x + 7)^2
\][/tex]
Therefore, the factorization of the trinomial [tex]\( x^2 + 14x + 49 \)[/tex] is:
[tex]\[
(x + 7)^2
\][/tex]
### Conclusion:
The correct answer is [tex]\( \boxed{(x + 7)^2} \)[/tex], which corresponds to option B.
