Question

Write the binomial expansion of (x-3y)^5

Answer 1

(x−3y)5(x-3y)5

Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=n∑k=0nCk⋅(an−kbk)(a+b)n=∑k=0n⁡nCk⋅(an-kbk).

5∑k=05!(5−k)!k!⋅(x)5−k⋅(−3y)k∑k=05⁡5!(5-k)!k!⋅(x)5-k⋅(-3y)k

Expand the summation.

5!(5−0)!0!(x)5−0⋅(−3y)0+5!(5−1)!1!(x)5−1⋅(−3y)+…+5!(5−5)!5!(x)5−5⋅(−3y)55!(5-0)!0!(x)5-0⋅(-3y)0+5!(5-1)!1!(x)5-1⋅(-3y)+…+5!(5-5)!5!(x)5-5⋅(-3y)5

Simplify the exponents for each term of the expansion.

1⋅(x)5⋅(−3y)0+5⋅(x)4⋅(−3y)+…+1⋅(x)0⋅(−3y)51⋅(x)5⋅(-3y)0+5⋅(x)4⋅(-3y)+…+1⋅(x)0⋅(-3y)5

x5−15x4y+90x3y2−270x2y3+405xy4−243y5