Question

2. Use the binomial theorem to expand the expression. (а — 2b)^5

Answer 1

Answer:

$(a-2b)^{5}=-32b^{5}+80ab^{4}-80a^{2}b^{3}+40a^{3}b^{2}-10a^{4}b+a^{5}$

Step-by-step explanation:

The binomial expansion is given by:

$(x+y)^{n}=_{0}^{n}\textrm{C}x^{^{0}}y^{n}+_{1}^{n-1}\textrm{C}x^{1}y^{n-1}+...+_{n}^{n}\textrm{C}x^{n}y^{0}$

In our case we have

$x=a\\y=-2b\\n=5$

Thus using the given terms in the binomial expansion we get

$(a-2b)^{5}=_{0}^{5}\textrm{C}a^{0}(-2b)^{5}+_{1}^{5}\textrm{C}a^{^{1}}(-2b)^{4}+{_{2}^{5}\textrm{C}}a^{2}(-2b)^{3}+_{3}^{5}\textrm{C}a^{3}(-2b)^{2}+_{4}^{5}\textrm{C}a^{4}(-2b)^{1}+_{5}^{5}\textrm{C}a^{5}(-2b)^{0}$

Upon solving we get

$(a-2b)^{5}=-32b^{5}+5\times a\times16b^{4}+10\times a^{2} \times (-8b^{3})+10\times a^{3}\times 4b^{2}+5\times a^{4}\times (-2b)+a^{5}\\\\(a-2b)^{5}=-32b^{5}+80ab^{4}-80a^{2}b^{3}+40a^{3}b^{2}-10a^{4}b+a^{5}$