Binomial Expansion Calculator

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Expand $\left(2 x + 5\right)^{3}$.

Solution

The expansion is given by the following formula: $\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$, 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, $\left(2 x + 5\right)^{3} = \sum_{k=0}^{3} {\binom{3}{k}} \left(2 x\right)^{3 - k} 5^{k}$.

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

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

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

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

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

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

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