Question

BINOMIAL EXPANSION

Expand (2x+5)³​​

Answer 1

The expansion is given by the following formula:

${\bold \red{\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}, where {\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!} \: and \: n! = 1 \cdot 2 \cdot \ldots \cdot n}}$

We have that a = 2 x, b = 5, and n = 3.

Therefore,

${\bold \red{\left(2 x + 5\right)^{3} = \sum_{k=0}^{3} {\binom{3}{k}} \left(2 x\right)^{3 - k} 5^{k}}}$

Now, calculate the product for every value of k from 0 to 3.

${ \bold \red{k = 0: {\binom{3}{0}} \left(2 x\right)^{3 - 0} \cdot 5^{0} = \frac{3!}{\left(3 - 0\right)! 0!} \left(2 x\right)^{3 - 0} \cdot 5^{0} = 8 x^{3}}}$

${\bold \red{k = 1: {\binom{3}{1}} \left(2 x\right)^{3 - 1} \cdot 5^{1} = \frac{3!}{\left(3 - 1\right)! 1!} \left(2 x\right)^{3 - 1} \cdot 5^{1} = 60 x^{2}}}$

${ \bold \red{k = 2: {\binom{3}{2}} \left(2 x\right)^{3 - 2} \cdot 5^{2} = \frac{3!}{\left(3 - 2\right)! 2!} \left(2 x\right)^{3 - 2} \cdot 5^{2} = 150 x}}$

${ \bold \red{k = 3: {\binom{3}{3}} \left(2 x\right)^{3 - 3} \cdot 5^{3} = \frac{3!}{\left(3 - 3\right)! 3!} \left(2 x\right)^{3 - 3} \cdot5^{3} = 125}}$

Thus,

$\bold \red{\left(2 x + 5\right)^{3} = 8 x^{3} + 60 x^{2} + 150 x + 125.}$