4.1.5

Binomial Expansion & Rational Powers 2 (A2 only)

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Binomial Expansion for Rational Powers

We can use the binomial expansion of (1+x)n(1+x)^n to expand expressions of the form (a+bx)n(a+bx)^n.

Illustrative background for Taking out factorsIllustrative background for Taking out factors ?? "content

Taking out factors

  • In order to expand (a+bx)n(a+bx)^n using the binomial expansion, we need to take out a factor of ana^n and write it in front of the brackets:
    • (a+bx)n=((a×1)+(a×ba)x)n=an(1+bax)n(a+bx)^n = ((a\times 1) + (a\times\frac{b}{a})x)^n = a^n\left(1+\frac{b}{a}x\right)^n
  • We can then expand (1+bax)n(1+\frac{b}{a}x)^n using the binomial expansion equation in the formula book, remembering to replace any xx terms with bax\frac{b}{a}x.
  • Don't forget to multiply every term in the expansion by ana^n at the end to give the final answer.
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Example

  • What are the first four terms of the binomial expansion of (3+2x)2(3+2x)^{-2}?
    • What are the range of values of xx for which the expansion is valid?
Illustrative background for Take out factorsIllustrative background for Take out factors ?? "content

Take out factors

  • In order to use the expansion of (1+x)n(1+x)^n, we need to take out a factor of 323^{-2} from (3+2x)2(3+2x)^{-2}"
    • (3+2x)2=((3×1)+(3×23)x)2=32(1+23x)2(3+2x)^{-2}=((3\times 1)+(3\times \frac{2}{3})x)^{-2} = 3^{-2}\left(1+\frac{2}{3}x\right)^{-2}
Illustrative background for Expand $$(1+\frac{2}{3}x)^{-2}$$Illustrative background for Expand $$(1+\frac{2}{3}x)^{-2}$$ ?? "content

Expand (1+23x)2(1+\frac{2}{3}x)^{-2}

  • The first term of the expansion of (1+23x)2\left(1+\frac{2}{3}x\right)^{-2} is equal to 1.
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Second term

  • The second term of the binomial expansion of is nxnx.
  • Substituting in n=2n = -2 and noticing that we must change x23xx \rightarrow \frac{2}{3}x, we get
    • nx43xnx \rightarrow -\frac{4}{3}x
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Third term

  • The third term of the binomial expansion is n(n1)2!x2\frac{n(n-1)}{2!}x^2.
  • Substituting in n=2n = -2 and changing x23xx \rightarrow \frac{2}{3}x, we get:
    • n(n1)2!x2(2)×(3)2×1(23x)2=43x2\frac{n(n-1)}{2!}x^2 \rightarrow \frac{(-2)\times(-3)}{2\times1}\left(\frac{2}{3}x\right)^2 = \frac{4}{3}x^2
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Fourth term

  • The third term of the binomial expansion is n(n1)(n2)3!x3\frac{n(n-1)(n-2)}{3!}x^3.
  • Substituting in n=2n = -2 and changing x23xx \rightarrow \frac{2}{3}x, we get:
    • n(n1)(n2)3!x3(2)×(3)×(4)3×2×1(23x)3=3227x3\frac{n(n-1)(n-2)}{3!}x^3 \rightarrow \frac{(-2)\times(-3)\times(-4)}{3\times2\times1}\left(\frac{2}{3}x\right)^3 = -\frac{32}{27}x^3
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Add terms

  • The expansion of (1+23x)2(1+\frac{2}{3}x)^{-2} is equal to:
    • (1+23x)2=143x+43x23227x3+...\left(1+\frac{2}{3}x\right)^{-2}=1-\frac{4}{3}x+\frac{4}{3}x^2 -\frac{32}{27}x^3+...
Illustrative background for Multiply each term by $$3^{-2}$$Illustrative background for Multiply each term by $$3^{-2}$$ ?? "content

Multiply each term by 323^{-2}

  • To find the final answer, we need to multiply the whole expansion by 32=193^{-2}=\frac{1}{9}:
    • (3+2x)2=19=427x+427x232243x3+...(3+2x)^{-2} = \frac{1}{9}=\frac{4}{27}x+\frac{4}{27}x^2-\frac{32}{243}x^3+...
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Validity

  • The binomial expansion is valid when 23x<1\left|\frac{2}{3}x\right|<1, which means:
    • x<1.5\left|x\right|<1.5

Jump to other topics

1Proof

2Algebra & Functions

2.1Powers & Roots

2.2Quadratic Equations

2.3Inequalities

2.4Polynomials

2.5Graphs

2.6Functions

2.7Transformation of Graphs

2.8Partial Fractions (A2 Only)

3Coordinate Geometry

4Sequences & Series

5Trigonometry

6Exponentials & Logarithms

7Differentiation

8Integration

9Numerical Methods

10Vectors

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