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12 (a) Write \( \frac{4x^3 - 9}{6x + 9} \times \frac{-2x}{x^{-3} - 3x} \) in the form \( \frac{ax + b}{cx + d} \) where a, b, c, and d are integers - Edexcel - GCSE Maths - Question 13 - 2018 - Paper 2
Question 13
12 (a) Write \( \frac{4x^3 - 9}{6x + 9} \times \frac{-2x}{x^{-3} - 3x} \) in the form \( \frac{ax + b}{cx + d} \) where a, b, c, and d are integers. (b) Express \( ... show full transcript
Worked Solution & Example Answer:12 (a) Write \( \frac{4x^3 - 9}{6x + 9} \times \frac{-2x}{x^{-3} - 3x} \) in the form \( \frac{ax + b}{cx + d} \) where a, b, c, and d are integers - Edexcel - GCSE Maths - Question 13 - 2018 - Paper 2
Step 1
Write \( \frac{4x^3 - 9}{6x + 9} \times \frac{-2x}{x^{-3} - 3x} \) in the form \( \frac{ax + b}{cx + d} \)
Answer
To solve for the expression, first factor the numerator and the denominator where possible:
- Factor ( 4x^3 - 9 ) as ( (2x - 3)(2x^2 + 3x + 3) ).
- Factor ( 6x + 9 ) as ( 3(2x + 3) ).
- For the term ( x^{-3} - 3x ), rewrite it as ( \frac{1}{x^3} - 3x = \frac{1 - 3x^4}{x^3} ).
Thus, we can rewrite the entire expression:
[ \frac{(2x - 3)(2x^2 + 3x + 3)}{3(2x + 3)} \times \frac{-2x}{\frac{1 - 3x^4}{x^3}} ]
After simplifying, we can arrive at the required form ( \frac{ax + b}{cx + d} ) where specific integers can be deduced based on the final simplifications.
Step 2
Express \( \frac{3}{x + 1} + \frac{1}{x - 2} - \frac{4}{x} \) as a single fraction in its simplest form
Answer
To combine the fractions, we need to find a common denominator:
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The least common multiple of the denominators ( (x + 1)(x - 2)x ) will be used.
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Rewrite each term with the common denominator:
- ( \frac{3}{x + 1} = \frac{3x(x - 2)}{(x + 1)(x - 2)x} )
- ( \frac{1}{x - 2} = \frac{1 \cdot x(x + 1)}{(x + 1)(x - 2)x} )
- ( \frac{4}{x} = \frac{4(x + 1)(x - 2)}{(x + 1)(x - 2)x} )
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Combine the numerators: [ 3x(x - 2) + x(x + 1) - 4(x + 1)(x - 2) ]
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Simplify the entire expression to arrive at the single fraction in its simplest form.