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

B. The polynomial is prime.

Solution

First, let’s identify the trinomial to factor: \[ 21x^2y^2 + 2xy - 8 \] Let’s treat \( xy \) as a single variable by setting \( z = xy \). The trinomial becomes: \[ 21z^2 + 2z - 8 \] Next, we need to factor the quadratic equation \( 21z^2 + 2z - 8 \). We look for two numbers that multiply to \( 21 \times (-8) = -168 \) and add up to \( 2 \). The numbers \( 14 \) and \( -12 \) satisfy these conditions because: \[ 14 \times (-12) = -168 \quad \text{and} \quad 14 + (-12) = 2 \] Now, rewrite the middle term using these two numbers: \[ 21z^2 + 14z - 12z - 8 \] Factor by grouping: \[ 7z(3z + 2) - 4(3z + 2) \] Factor out the common term \( (3z + 2) \): \[ (7z - 4)(3z + 2) \] Finally, substitute back \( z = xy \): \[ (7xy - 4)(3xy + 2) \] So, the factored form of the trinomial is: \[ 21x^2y^2 + 2xy - 8 = (7xy - 4)(3xy + 2) \]