To factor the expression \(2x^2 + 21x + 49\), we will look for two binomials of the form \((ax + b)(cx + d)\).
We start by finding two numbers that multiply to \(2 \times 49 = 98\) (the coefficient of \(x^2\) multiplied by the constant term) and add up to \(21\) (the coefficient of \(x\)).
The numbers \(14\) and \(7\) meet these criteria because:
- \(14 \times 7 = 98\)
- \(14 + 7 = 21\)
Next, we can rewrite the middle term \(21x\) by breaking it into \(14x + 7x\):
\[ 2x^2 + 14x + 7x + 49 \]
Now we can group the terms:
\[ (2x^2 + 14x) + (7x + 49) \]
Factor out the common factors in each group:
\[ 2x(x + 7) + 7(x + 7) \]
Now we can factor out the common binomial \((x + 7)\):
\[ (2x + 7)(x + 7) \]
Thus, the completely factored form of \(2x^2 + 21x + 49\) is:
\[ (2x + 7)(x + 7) \]
1 answer