Chapter 12: Q 19. (page 836)

Expand ${(x - 2)}^{6}$ using the Binomial Theorem

Short Answer

Expert verified

The value of ${(x - 2)}^{6} = x^{6} - 12 x^{5} + 60 x^{4} - 160 x^{3} + 240 x^{2} - 192 x + 64$

Step by step solution

Step 1. Defination

Let $x$ and $a$ be real numbers for any positiveinteger $n$ , we have ${(x + a)}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) x^{n} + (\begin{matrix} n \\ 1 \end{matrix}) a x^{n - 1} + \dots \dots + (\begin{matrix} n \\ j \end{matrix}) a^{j} x^{n - j} + \dots \dots + (\begin{matrix} n \\ n \end{matrix}) a^{n} = \sum_{j = 0}^{n} (\begin{matrix} n \\ j \end{matrix}) x^{n - j} a^{j}$

Step 2. Explanation

$\Rightarrow {(x - 2)}^{6} = (\begin{matrix} 6 \\ 0 \end{matrix}) {(- 2)}^{0} {(x)}^{6} + (\begin{matrix} 6 \\ 1 \end{matrix}) {(- 2)}^{1} {(x)}^{5} + (\begin{matrix} 6 \\ 2 \end{matrix}) {(- 2)}^{2} {(x)}^{4} + (\begin{matrix} 6 \\ 3 \end{matrix}) {(- 2)}^{3} {(x)}^{3} + (\begin{matrix} 6 \\ 4 \end{matrix}) {(- 2)}^{4} {(x)}^{2} + (\begin{matrix} 6 \\ 5 \end{matrix}) {(- 2)}^{5} {(x)}^{1} + (\begin{matrix} 6 \\ 6 \end{matrix}) {(- 2)}^{6} {(x)}^{0}$

\Rightarrow {(x - 2)}^{6} = \frac{6!}{0! (6 - 0)!} x^{6} + \frac{6!}{1! (6 - 1)!} (- 2) x^{5} + \frac{6!}{2! (6 - 2)!} (4) x^{4} + \frac{6!}{3! (6 - 3)!} (- 8) x^{3} + \frac{6!}{4! (6 - 4)!} (16) x^{2} + \frac{6!}{5! (6 - 5)!} (- 32) x^{1} + \frac{6!}{6! (6 - 6)!} (64)

\Rightarrow {(x - 2)}^{6} = x^{6} - 12 x^{5} + 60 x^{4} - 160 x^{3} + 240 x^{2} - 192 x + 64

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Expand ${(x - 2)}^{6}$ using the Binomial Theorem

Short Answer

Step by step solution

Step 1. Defination

Step 2. Explanation

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Expand ${(x - 2)}^{6}$ using the Binomial Theorem