# How do you factor the trinomial x^2+8x+24=0?

${x}^{2} + 8 x + 24 = \left(x + 4 + i 2 \sqrt{2}\right) \left(x + 4 - i 2 \sqrt{2}\right)$
As the determinant for equation ${x}^{2} + 8 x + 24 = 0$ is ${8}^{2} - 4 \times 1 \times 24 = 64 - 96 = - 32$, the factors will be complex
Using quadratic formula, zeros of ${x}^{2} + 8 x + 24$ are $\frac{- 8 \pm \sqrt{- 32}}{2}$ or $- 4 \pm \frac{4}{2} \sqrt{- 2}$ or $- 4 \pm i 2 \sqrt{2}$
Hence factors of ${x}^{2} + 8 x + 24$ are $\left(x + 4 + i 2 \sqrt{2}\right) \left(x + 4 - i 2 \sqrt{2}\right)$