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3x^2 + 22x +7

1 answer

To factor the quadratic expression \(3x^2 + 22x + 7\), we will look for two binomials in the form \((ax + b)(cx + d)\).

  1. First, we check the product \(a \cdot c\) which is the coefficient of \(x^2\) (which is 3).
  2. We also need the product \(b \cdot d\) which is the constant term (which is 7).
  3. The middle term (22x) should be the result of adding the cross products of the binomials.

We will look for pairs of factors of \(3 \cdot 7 = 21\) that add up to 22.

The pairs of factors of 21 are:

  • (1, 21)
  • (3, 7)

Now, we can see that \(1\) and \(21\) add up to \(22\). Therefore, we can express 22x as \(1x + 21x\).

Now, we can rewrite the quadratic as follows:

\[ 3x^2 + 1x + 21x + 7 \]

Next, we group the terms:

\[ (3x^2 + 1x) + (21x + 7) \]

Now, we can factor each group:

\[ x(3x + 1) + 7(3x + 1) \]

We can see that \(3x + 1\) is a common factor:

\[ (3x + 1)(x + 7) \]

Thus, the factored form of \(3x^2 + 22x + 7\) is:

\[ (3x + 1)(x + 7) \]