1. **State the problem:** Simplify the product of two rational expressions: $$\frac{k^{12} + 19k^{6} + 88}{5k^{8} + 40k^{2}} \times \frac{k^{4} - 6k^{2}}{2k^{12} + 19k^{6} - 33}$$ 2. **Factor each polynomial where possible:** - Numerator of first fraction: $k^{12} + 19k^{6} + 88$ Let $x = k^{6}$, then it becomes $x^{2} + 19x + 88$. Factor $x^{2} + 19x + 88$: Factors of 88 that sum to 19 are 11 and 8. So, $x^{2} + 19x + 88 = (x + 11)(x + 8)$. Substitute back: $(k^{6} + 11)(k^{6} + 8)$. - Denominator of first fraction: $5k^{8} + 40k^{2}$ Factor out common factor $5k^{2}$: $$5k^{2}(k^{6} + 8)$$ - Numerator of second fraction: $k^{4} - 6k^{2}$ Factor out $k^{2}$: $$k^{2}(k^{2} - 6)$$ - Denominator of second fraction: $2k^{12} + 19k^{6} - 33$ Let $x = k^{6}$, then it becomes $2x^{2} + 19x - 33$. Factor $2x^{2} + 19x - 33$: Find two numbers that multiply to $2 \times (-33) = -66$ and add to 19: 22 and -3. Rewrite: $$2x^{2} + 22x - 3x - 33 = 2x(x + 11) - 3(x + 11) = (2x - 3)(x + 11)$$ Substitute back: $(2k^{6} - 3)(k^{6} + 11)$. 3. **Rewrite the entire expression with factored forms:** $$\frac{(k^{6} + 11)(k^{6} + 8)}{5k^{2}(k^{6} + 8)} \times \frac{k^{2}(k^{2} - 6)}{(2k^{6} - 3)(k^{6} + 11)}$$ 4. **Multiply the numerators and denominators:** Numerator: $$(k^{6} + 11)(k^{6} + 8) \times k^{2}(k^{2} - 6)$$ Denominator: $$5k^{2}(k^{6} + 8) \times (2k^{6} - 3)(k^{6} + 11)$$ 5. **Cancel common factors:** - Cancel $(k^{6} + 11)$ from numerator and denominator. - Cancel $(k^{6} + 8)$ from numerator and denominator. - Cancel $k^{2}$ from numerator and denominator. Intermediate step showing cancellation: $$\frac{\cancel{(k^{6} + 11)}\cancel{(k^{6} + 8)} \times \cancel{k^{2}}(k^{2} - 6)}{5\cancel{k^{2}}\cancel{(k^{6} + 8)} \times (2k^{6} - 3)\cancel{(k^{6} + 11)}} = \frac{k^{2} - 6}{5(2k^{6} - 3)}$$ 6. **Final simplified expression:** $$\boxed{\frac{k^{2} - 6}{5(2k^{6} - 3)}}$$ This is the simplified form of the product of the two rational expressions.