Complete the factorization shown.
$2 x^2-x-6=(2 x+\square)(x-\square)$
To factor the quadratic expression \(2x^2 - x - 6\), we want to write it in the form \((2x + A)(x - B)\), where \(A\) and \(B\) are integers. We start by expanding the expression \((2x + A)(x - B)\):
\[(2x + A)(x - B) = 2x^2 - 2Bx + Ax - AB = 2x^2 + (A - 2B)x - AB.\]
Comparing this result with the original quadratic \(2x^2 - x - 6\), we can set up a system of equations by matching the coefficients:
We have the system:
From the first equation, we can express \(A\) in terms of \(B\):
\[A = 2B - 1\]
Substituting this expression for \(A\) into the second equation:
\[(2B - 1)B = 6\]
\[2B^2 - B = 6\]
\[2B^2 - B - 6 = 0\]
We can solve this quadratic equation for \(B\) using the quadratic formula \(B = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[B = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-6)}}{2(2)}\]
\[B = \frac{1 \pm \sqrt{1 + 48}}{4}\]
\[B = \frac{1 \pm \sqrt{49}}{4} = \frac{1 \pm 7}{4}\]
The two potential values for \(B\) are \(B = \frac{8}{4} = 2\) and \(B = \frac{-6}{4} = -1.5\). Since \(B\) must be an integer, we choose \(B = 2\). Using \(A = 2B - 1\), we find \(A = 2(2) - 1 = 3\).
With \(A = 3\) and \(B = 2\), the factorization is:
\[2x^2 - x - 6 = (2x + 3)(x - 2)\]
We can verify this by expanding the factors:
\[(2x + 3)(x - 2) = 2x^2 - 4x + 3x - 6 = 2x^2 - x - 6\]
The factorization is correct. Therefore, the empty boxes are filled with 3 and 2. The expression is \((2x + 3)(x - 2)\), so the solution is \(\boxed{2x^2 - x - 6 = (2x + 3)(x - 2)}\).
Factoring quadratic expressions is a fundamental skill in algebra with applications in fields like physics and engineering. For example, if you are modeling the trajectory of a projectile where the height \(h\) at time \(t\) is given by a quadratic expression like \(h(t) = -5t^2 + 10t + 15\), you can factor the expression to find when the object hits the ground (i.e., when \(h(t) = 0\)). Factoring allows you to easily identify the roots of the equation and interpret the physical constraints of your model.