To factor the quadratic expression \(2x^2 + 7x + 6\) completely, we can follow these steps:
-
Multiply the coefficient of \(x^2\) (which is 2) by the constant term (which is 6):
\(2 \times 6 = 12\). -
Find two numbers that multiply to 12 and add to 7:
The numbers 3 and 4 work because \(3 \times 4 = 12\) and \(3 + 4 = 7\). -
Rewrite the middle term using these numbers:
Rewrite \(7x\) as \(3x + 4x\): \[ 2x^2 + 3x + 4x + 6 \] -
Group the terms:
\[ (2x^2 + 3x) + (4x + 6) \] -
Factor out the common factors from each group:
- From the first group \(2x^2 + 3x\), factor out \(x\):
\(x(2x + 3)\) - From the second group \(4x + 6\), factor out 2:
\(2(2x + 3)\)
Now, we have: \[ x(2x + 3) + 2(2x + 3) \]
- From the first group \(2x^2 + 3x\), factor out \(x\):
-
Factor out the common binomial factor \((2x + 3)\):
\[ (2x + 3)(x + 2) \]
Thus, the completely factored form of \(2x^2 + 7x + 6\) is: \[ \boxed{(2x + 3)(x + 2)} \]