1. **Problem a:** Simplify the expression $ \frac{x^2 - 49}{x^2 + 2x - 35} \times \frac{2x^2 - 18x^2 + 40x}{3x^2 - 21x} $. 2. **Step 1:** Factor all polynomials. - $x^2 - 49 = (x - 7)(x + 7)$ (difference of squares). - $x^2 + 2x - 35 = (x + 7)(x - 5)$. - $2x^2 - 18x^2 + 40x = -16x^2 + 40x = -4x(4x - 10) = -4x(2x - 5)$. - $3x^2 - 21x = 3x(x - 7)$. 3. **Step 2:** Rewrite the expression with factors: $$\frac{(x - 7)(x + 7)}{(x + 7)(x - 5)} \times \frac{-4x(2x - 5)}{3x(x - 7)}$$ 4. **Step 3:** Cancel common factors: - Cancel $x + 7$ in numerator and denominator. - Cancel $x - 7$ in numerator and denominator. - Cancel $x$ in numerator and denominator. Intermediate step showing cancellation: $$\frac{\cancel{(x - 7)}\cancel{(x + 7)}}{\cancel{(x + 7)}(x - 5)} \times \frac{-4\cancel{x}(2x - 5)}{3\cancel{x}\cancel{(x - 7)}} = \frac{1}{x - 5} \times \frac{-4(2x - 5)}{3}$$ 5. **Step 4:** Multiply remaining factors: $$\frac{-4(2x - 5)}{3(x - 5)} = \frac{-8x + 20}{3(x - 5)}$$ 6. **Step 5:** Factor numerator if possible: $$-8x + 20 = -4(2x - 5)$$ but no further simplification with denominator. **Final simplified expression for a:** $$\boxed{\frac{-8x + 20}{3(x - 5)}}$$ --- 7. **Problem b:** Simplify the expression $$\frac{10x^2 - 9x}{2x^2 - 13x - 7} - \frac{x^2 - 36}{x^2 - 5x - 14} + \frac{3x^2 - 19x + 6}{3x^2 + 5x - 2}$$ 8. **Step 1:** Factor all denominators and numerators where possible. - Denominator 1: $2x^2 - 13x - 7$ - Factors of $2 \times -7 = -14$ that sum to $-13$ are $-14$ and $1$. - Rewrite: $2x^2 - 14x + x - 7 = 2x(x - 7) + 1(x - 7) = (2x + 1)(x - 7)$. - Denominator 2: $x^2 - 5x - 14 = (x - 7)(x + 2)$. - Denominator 3: $3x^2 + 5x - 2$ - Factors of $3 \times -2 = -6$ that sum to 5 are 6 and -1. - Rewrite: $3x^2 + 6x - x - 2 = 3x(x + 2) -1(x + 2) = (3x - 1)(x + 2)$. - Numerator 2: $x^2 - 36 = (x - 6)(x + 6)$. 9. **Step 2:** Rewrite expression with factors: $$\frac{10x^2 - 9x}{(2x + 1)(x - 7)} - \frac{(x - 6)(x + 6)}{(x - 7)(x + 2)} + \frac{3x^2 - 19x + 6}{(3x - 1)(x + 2)}$$ 10. **Step 3:** Factor numerator 1 and numerator 3 if possible. - Numerator 1: $10x^2 - 9x = x(10x - 9)$. - Numerator 3: $3x^2 - 19x + 6$ - Factors of $3 \times 6 = 18$ that sum to -19 are -18 and -1. - Rewrite: $3x^2 - 18x - x + 6 = 3x(x - 6) -1(x - 6) = (3x - 1)(x - 6)$. 11. **Step 4:** Rewrite expression: $$\frac{x(10x - 9)}{(2x + 1)(x - 7)} - \frac{(x - 6)(x + 6)}{(x - 7)(x + 2)} + \frac{(3x - 1)(x - 6)}{(3x - 1)(x + 2)}$$ 12. **Step 5:** Cancel common factors in the third fraction numerator and denominator: $$\frac{\cancel{(3x - 1)}(x - 6)}{\cancel{(3x - 1)}(x + 2)} = \frac{x - 6}{x + 2}$$ 13. **Step 6:** Find common denominator for all three fractions: $(2x + 1)(x - 7)(x + 2)$. 14. **Step 7:** Rewrite each fraction with common denominator: - First fraction multiply numerator and denominator by $x + 2$: $$\frac{x(10x - 9)(x + 2)}{(2x + 1)(x - 7)(x + 2)}$$ - Second fraction multiply numerator and denominator by $2x + 1$: $$\frac{(x - 6)(x + 6)(2x + 1)}{(2x + 1)(x - 7)(x + 2)}$$ - Third fraction multiply numerator and denominator by $2x + 1)(x - 7)$: $$\frac{(x - 6)(2x + 1)(x - 7)}{(2x + 1)(x - 7)(x + 2)}$$ 15. **Step 8:** Combine numerators over common denominator: $$\frac{x(10x - 9)(x + 2) - (x - 6)(x + 6)(2x + 1) + (x - 6)(2x + 1)(x - 7)}{(2x + 1)(x - 7)(x + 2)}$$ 16. **Step 9:** Expand each numerator term: - $x(10x - 9)(x + 2) = x(10x^2 + 20x - 9x - 18) = x(10x^2 + 11x - 18) = 10x^3 + 11x^2 - 18x$ - $(x - 6)(x + 6) = x^2 - 36$, so $(x - 6)(x + 6)(2x + 1) = (x^2 - 36)(2x + 1) = 2x^3 + x^2 - 72x - 36$ - $(x - 6)(2x + 1)(x - 7)$ expand stepwise: - First $(x - 6)(2x + 1) = 2x^2 + x - 12x - 6 = 2x^2 - 11x - 6$ - Then multiply by $x - 7$: $$ (2x^2 - 11x - 6)(x - 7) = 2x^3 - 14x^2 - 11x^2 + 77x - 6x + 42 = 2x^3 - 25x^2 + 71x + 42 $$ 17. **Step 10:** Substitute expansions back: $$\frac{10x^3 + 11x^2 - 18x - (2x^3 + x^2 - 72x - 36) + (2x^3 - 25x^2 + 71x + 42)}{(2x + 1)(x - 7)(x + 2)}$$ 18. **Step 11:** Remove parentheses and combine like terms: $$10x^3 + 11x^2 - 18x - 2x^3 - x^2 + 72x + 36 + 2x^3 - 25x^2 + 71x + 42$$ Combine: - Cubic terms: $10x^3 - 2x^3 + 2x^3 = 10x^3$ - Quadratic terms: $11x^2 - x^2 - 25x^2 = -15x^2$ - Linear terms: $-18x + 72x + 71x = 125x$ - Constants: $36 + 42 = 78$ 19. **Step 12:** Final numerator: $$10x^3 - 15x^2 + 125x + 78$$ 20. **Step 13:** Final simplified expression for b: $$\boxed{\frac{10x^3 - 15x^2 + 125x + 78}{(2x + 1)(x - 7)(x + 2)}}$$ --- 21. **Problem c:** Given $f(x) = \frac{1}{2}x^2 + 4x + 8$ and $g(x) = x - 2$, find: 22. **Part a:** Find $f(-4a)$. 23. **Step 1:** Substitute $-4a$ into $f(x)$: $$f(-4a) = \frac{1}{2}(-4a)^2 + 4(-4a) + 8$$ 24. **Step 2:** Calculate each term: - $(-4a)^2 = 16a^2$ - So, $\frac{1}{2} \times 16a^2 = 8a^2$ - $4 \times (-4a) = -16a$ 25. **Step 3:** Combine terms: $$f(-4a) = 8a^2 - 16a + 8$$ 26. **Part b:** Find $-2g[f(x) + 3] - 8$. 27. **Step 1:** Calculate $f(x) + 3$: $$f(x) + 3 = \frac{1}{2}x^2 + 4x + 8 + 3 = \frac{1}{2}x^2 + 4x + 11$$ 28. **Step 2:** Calculate $g[f(x) + 3]$: $$g\left(\frac{1}{2}x^2 + 4x + 11\right) = \left(\frac{1}{2}x^2 + 4x + 11\right) - 2 = \frac{1}{2}x^2 + 4x + 9$$ 29. **Step 3:** Multiply by $-2$ and subtract 8: $$-2 \times \left(\frac{1}{2}x^2 + 4x + 9\right) - 8 = -x^2 - 8x - 18 - 8 = -x^2 - 8x - 26$$ **Final answers:** - a) $f(-4a) = 8a^2 - 16a + 8$ - b) $-2g[f(x) + 3] - 8 = -x^2 - 8x - 26$