Simplified Binomial Expansion Using Partial Fractions
Learn how to simplify binomial expansions of complex expressions using partial fractions. Explore step-by-step examples and techniques to find the expansion terms efficiently up to x^3. Master the process of expressing the expansions as partial fractions for better understanding and problem-solving skills.
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Presentation Transcript
Teachings for section 4C
Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions 4 5? Find the expansion of: up to and including the term in x3 (1 + ?)(2 ?) Express as Partial Fractions ? ? 4 5? + = (2 ?) (1 + ?) (1 + ?)(2 ?) Cross-multiply and combine ? 2 ? + ?(1 + ?) (1 + ?)(2 ?) = The numerators must be equal 4 5? = ? 2 ? + ?(1 + ?) If x = 2 6 2 = 3? = ? If x = -1 9 3 = 3? = ? Express the original fraction as Partial Fractions, using A and B 3 2 4 5? = (1 + ?) (2 ?) (1 + ?)(2 ?) 4C
Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions 4 5? Find the expansion of: up to and including the term in x3 (1 + ?)(2 ?) 3 2 4 5? = Both fractions can be rewritten (1 + ?) (2 ?) (1 + ?)(2 ?) = 3(1 + ?) 1 2(2 ?) 1 Expand each term separately 3(1 + ?) 1 Write out the general form: + ?? +?(? 1) +?(? 1)(? 2) ?2 ?3 1 + ?? = 1 2! 3! Sub in: x = x n = -1 Work out each term carefully + ( 1)(?) +( 1)( 2) +( 1)( 2)( 3) (?)2 (?)3 1 + ? 1 = 1 2 6 + ?2 ?3 = 1 ? Remember that this expansion is to be multiplied by 3 3 1 + ? 1 3? + 3?2 3?3 = 3 4C
Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions 4 5? Find the expansion of: up to and including the term in x3 3 1 + ? 1 3? + 3?2 3?3 = 3 (1 + ?)(2 ?) 3 2 4 5? = Both fractions can be rewritten (1 + ?) (2 ?) (1 + ?)(2 ?) = 3(1 + ?) 1 2(2 ?) 1 Expand each term separately 2(2 ?) 1 Take a factor 2 out of the brackets (and keep the current 2 separate ) 1 2 2 1 1 2? Both parts in the square brackets are raised to -1 1 2 2 11 1 2? Work out 2-1 1 1 2 1 1 2 2? This is actually now cancelled by the 2 outside the square bracket! 1 1 1 2? 4C
Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions 4 5? Find the expansion of: up to and including the term in x3 3 1 + ? 1 3? + 3?2 3?3 = 3 (1 + ?)(2 ?) 3 2 4 5? = Both fractions can be rewritten (1 + ?) (2 ?) (1 + ?)(2 ?) = 3(1 + ?) 1 2(2 ?) 1 Expand each term separately 1 1 1 2(2 ?) 1 = 2? Write out the general form: + ?? +?(? 1) + ?(? 1)(? 2) 3! ?2 ?3 1 + ?? = 1 Sub in: x =-1/2 x n = -1 Work out each term carefully 2! 2 3 1 1 ? + ( 1) 1 2? +( 1)( 2) 1 +( 1)( 2)( 3) 1 = 1 2? 2? 2 2 6 1 1 ? + 1 2? +1 4?2+1 = 1 8?3 2 4C
Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions 4 5? Find the expansion of: up to and including the term in x3 3 1 + ? 1 3? + 3?2 3?3 = 3 (1 + ?)(2 ?) 3 2 4 5? +?3 +?2 1 = 1 ? + ? Both fractions can be rewritten (1 + ?) (2 ?) (1 + ?)(2 ?) = 1 2 2 8 4 = 3(1 + ?) 1 2(2 ?) 1 Replace each bracket with its expansion 1 + 1 2? + 1 4?2+ 1 8?3 = (3 3? + 3?2 3?3) Subtract the second from the first (be wary of double negatives in some questions) = 2 7 2? +11 4?2 25 8?3 4C
