To factor the quadratic expression \(-12x^2 + 16xy + 3y^2\), we can use the method of factoring by grouping or looking for two binomials that multiply to give the original expression.
First, we can rearrange the expression and consider it in standard form: \[ -12x^2 + 16xy + 3y^2 = 3y^2 + 16xy - 12x^2. \] Now, let's look for two numbers that multiply to \(a \cdot c\) (where \(a = -12\) and \(c = 3\)) and add up to \(b\) (where \(b = 16\)): \[ a \cdot c = -12 \cdot 3 = -36, \] and we need two numbers that multiply to \(-36\) and add to \(16\). The numbers \(18\) and \(-2\) fit this requirement because: \[ 18 \times (-2) = -36, \] \[ 18 + (-2) = 16. \]
Next, we rewrite the middle term \(16xy\) using \(18xy\) and \(-2xy\): \[ -12x^2 + 18xy - 2xy + 3y^2. \]
Now, we can factor by grouping: \[ (-12x^2 + 18xy) + (-2xy + 3y^2). \] Factoring out the common factors from each group: \[ 6x(-2x + 3y) - y(2x - 3y). \] Notice that we can rearrange the second group: \[ 6x(-2x + 3y) - y(-2x + 3y). \] Now factor out \((-2x + 3y)\): \[ (-2x + 3y)(6x - y). \]
Thus, the factorization of the expression \(-12x^2 + 16xy + 3y^2\) is: \[ (-2x + 3y)(6x - y). \]