teach me how to solve part ii---B4
(i) Express each of the following in partial fractions.
(a) $\frac{2x^2 - 7x + 14}{(x + 2)(x - 4)}$
(b) $\frac{42}{(2x - 5)(x + 1)^2}$
(ii) Using the results in part (i)(a), or otherwise, express $\frac{2x^4 - 7x^2 + 14}{(x^2 + 2)(x^2 - 4)}$ in partial fractions.
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To solve part ii, we need to decompose the fraction with x to the fourth power. The key insight is recognizing that this expression has a special structure. If we substitute y equals x squared, our expression becomes exactly the same form as part i a. This substitution transforms our complex quartic expression into the simpler quadratic form we already solved.
Now we apply the substitution by replacing x with x squared in our result from part i a. This gives us our preliminary decomposition. We have the constant term 2, a fraction with x squared plus 2 in the denominator, and a fraction with x squared minus 4 in the denominator. However, we notice that x squared minus 4 can be factored as x minus 2 times x plus 2, so this term needs further decomposition.
Now we need to decompose the term with x squared minus 4 in the denominator. We factor this as x minus 2 times x plus 2, then set up partial fractions with unknown constants C and D. Multiplying through by the denominator gives us 3 equals C times x plus 2 plus D times x minus 2. Setting x equals 2 gives us C equals three fourths. Setting x equals negative 2 gives us D equals negative three fourths.
Now we combine all our results. Starting with our preliminary decomposition, we substitute the partial fraction form we found for the term with x squared minus 4. This gives us the constant term 2, minus 6 over x squared plus 2, plus 3 over 4 times x minus 2, minus 3 over 4 times x plus 2. This is our final answer and matches the given solution perfectly.
Let's summarize the key steps we used to solve this problem. First, we recognized the substitution pattern where x to the fourth becomes y squared. Second, we applied the substitution y equals x squared. Third, we used our result from part i a. Fourth, we decomposed the reducible quadratic terms. Finally, we combined all terms for our final answer. The key insight is that substitution can transform complex problems into simpler forms we've already solved.