To factor the trinomial \(3x^2 + 22x + 7\), we look for two numbers that multiply to \(3 \times 7 = 21\) (the product of the leading coefficient and the constant) and add up to \(22\) (the coefficient of the linear term).
The numbers \(21\) and \(1\) fit these conditions since:
- \(21 \cdot 1 = 21\)
- \(21 + 1 = 22\)
Now, we rewrite the trinomial using these numbers to break apart the middle term:
\[ 3x^2 + 21x + 1x + 7 \]
Next, we group the terms:
\[ (3x^2 + 21x) + (1x + 7) \]
Now, we factor out the common coefficients from each group:
\[ 3x(x + 7) + 1(x + 7) \]
Now, we can factor out the common factor \((x + 7)\):
\[ (3x + 1)(x + 7) \]
Thus, the expression \(3x^2 + 22x + 7\) can be written in factored form as:
\[ (3x + 1)(x + 7) \]
This is the equivalent expression to the original trinomial.
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