Question: Factor the trinomial. \[ 21x^2y^2 + 2xy - 8 \] Select the correct choice below and, if…

Factor the trinomial.

\[ 21x^2y^2 + 2xy - 8 \]

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

A. \( 21x^2y^2 + 2xy - 8 = \boxed{\phantom{0}} \) (Factor completely.)

B. The polynomial is prime.

Solution

To factor the trinomial \(21x^2y + 2xy - 8\), we need to look for two numbers that multiply to \(21 \times -8 = -168\) and add to the middle coefficient, \(2\). The two numbers that satisfy these conditions are \(14\) and \(-12\), because \(14 + (-12) = 2\) and \(14 \times (-12) = -168\). Rewrite the middle term using these numbers: \[ 21x^2y + 14xy - 12xy - 8 \] Group the terms: \[ (21x^2y + 14xy) + (-12xy - 8) \] Factor out the greatest common factor from each group: \[ 7xy(3x + 2) - 4(3x + 2) \] Notice that \(3x + 2\) is a common factor: \[ (7xy - 4)(3x + 2) \] So, the trinomial \(21x^2y + 2xy - 8\) factors completely as: \[ (7xy - 4)(3x + 2) \]