To factor the trinomial \(3x^2 + 22x + 7\), we need to find two numbers that multiply to the product of the coefficient of \(x^2\) (which is 3) and the constant term (which is 7). This product is \(3 \cdot 7 = 21\). We also need these two numbers to add up to the coefficient of \(x\), which is 22.
The two numbers that satisfy these conditions are 21 and 1, since \(21 \times 1 = 21\) and \(21 + 1 = 22\).
Now, we can rewrite the middle term (22x) using these two numbers:
\[ 3x^2 + 21x + 1x + 7 \]
Next, we group the terms:
\[ (3x^2 + 21x) + (1x + 7) \]
Now, we can factor by grouping:
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From the first group \(3x^2 + 21x\), we can factor out \(3x\): \[ 3x(x + 7) \]
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From the second group \(1x + 7\), we can factor out 1: \[ 1(x + 7) \]
Now we have: \[ 3x(x + 7) + 1(x + 7) \]
We can now factor out \((x + 7)\): \[ (3x + 1)(x + 7) \]
Thus, the expression equivalent to \(3x^2 + 22x + 7\) is:
\[ (3x + 1)(x + 7) \]
So the final answer is:
\[ 3x + 1 \quad \text{and} \quad x + 7 \]