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Vesna [10]
3 years ago
13

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

Mathematics
1 answer:
spayn [35]3 years ago
5 0

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}

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