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Expand the following using either the Binomial Theorem or Pascal’s Triangle. You must show your work for credit.


(x - 5)⁵

Verified Answer :

I have my notes here that might help you solve the problem on your own:

We sometimes need to expand binomials as follows:

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

Clearly, doing this by direct multiplication gets quite tedious and can be 

rather difficult for larger powers or more complicated expressions.

Answer:

\(x^5 -25x^4+250x^3 - 1250x^2 + 3125x-3125\)

Step-by-step explanation:

Given expression,

\((x-5)^5\)

\(=(x+(-5))^5\)

∵ By the binomial expansion,

\((a+b)^n=\sum_{r=0}^{n}^nC_r a^{n-r} b^{r}\)

Where,

\(^nC_r=\frac{n!}{r!(n-r)!}\)

Thus,

\((x+(-5))^5=^5C_0 (x)^5 (-5)^{0}+^5C_1 (x)^4 (-5)^{1}+^5C_2 (x)^3 (-5)^{2}+^5C_3 (x)^2 (-5)^{3}+^5C_4 (x)^1 (-5)^{4}+^5C_0 (x)^0 (-5)^{5}\)

\(=x^5 + 5(x)^4(-5)+10(x)^3(25)+10x^2(-125)+(5)(x)(625)+(-3125)\)

\(=x^5 -25x^4+250x^3 - 1250x^2 + 3125 x-3125\)

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