Expansion of (2x-3) to the fifth

Statement

Expand $(2x-3)^5$ using Newton’s binomial.

Solution

We apply $(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$ with $a=2x$, $b=-3$, $n=5$; the coefficients $\binom{5}{k}$ are $1,5,10,10,5,1$ (row of Tartaglia’s triangle).

(2x-3)^5 &= (2x)^5 + 5(2x)^4(-3) + 10(2x)^3(-3)^2 \\ &\quad + 10(2x)^2(-3)^3 + 5(2x)(-3)^4 + (-3)^5 \\[4pt] &= \boxed{32x^5 - 240x^4 + 720x^3 - 1080x^2 + 810x - 243} \end{aligned}$$

Links

Topics: Combinatorics
Concepts: Newton’s binomial · Binomial coefficient
Methods: Newton binomial
Skills: Combinatorial calculus · Using formulae
People: Isaac Newton
Exercise type: Combinatorial problem