Answer:
[tex](x-2)^5 = x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32[/tex]
Step-by-step explanation:
Binomial Theorem
[tex](a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n[/tex]
Given expression:
[tex](x-2)^5[/tex]
Compare with the Binomial Theorem:
[tex]a=x, \quad b=-2, \quad n=5[/tex]
Substitute the values into the formula:
[tex]\begin{aligned}(x-2)^5 & = x^5+\dfrac{5!}{1!4!}x^{4}(-2)+\dfrac{5!}{2!3!}x^{3}(-2)^2+\dfrac{5!}{3!2!}x^{2}(-2)^3+\dfrac{5!}{4!1!}x^{1}(-2)^4+(-2)^5\\\\& =x^5+\dfrac{120}{24}x^{4}(-2)+\dfrac{120}{12}x^{3}(4)+\dfrac{120}{12}x^{2}(-8)+\dfrac{120}{24}x(16)-32\\\\& =x^5-\dfrac{240}{24}x^{4}+\dfrac{480}{12}x^{3}-\dfrac{960}{12}x^{2}+\dfrac{1920}{24}x-32\\\\& =x^5-10x^{4}+40x^{3}-80x^{2}+80x-32\end{aligned}[/tex]
Pascal's Triangle
Pascal's triangle is an arrangement of binomial coefficients in triangular form. Each row has 1 at both ends, and each remaining number in the row is found by adding together the two numbers directly above it.
[tex]\phantom{-b)))))))}1\\\phantom{-b)))))}1 \:\quad 1\\\phantom{-b)))}1 \:\quad 2\: \quad 1\\\phantom{-b)}1 \:\quad 3\: \quad 3\: \quad 1\\\phantom{))}1 \quad 4 \:\: \quad 6 \quad \;4 \: \quad 1\\1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1[/tex]
(Note: The initial row with the single number 1 is row zero)
Each term in the binomial expansion using Pascal's triangle is the product of a coefficient, a with an exponent, and b with an exponent.
As we move through each term from left to right, the exponent of a decreases from the exponent of the expression down to zero, and the exponent of b increases from zero up to the exponent of the expression.
The coefficients of each term are the numbers which appear in the number of the row that corresponds with the exponent of the expression.
From the given expression, the power that we are expanding the bracket to is 5, so we need to look at line 5 of Pascal's triangle, which is:
1 5 10 10 5 1
Therefore, the expansion using Pascal's triangle to the power 5 of is:
[tex](a+b)^5=1a^5b^0+5a^4b^1+10a^3b^2+10a^2b^3+5a^1b^4+1a^0b^5[/tex]
Substituting the values of a and b:
[tex]\begin{aligned}(x-2)^5 & = 1x^5(-2)^0 + 5x^4(-2)^1 + 10x^3(-2)^2 + 10x^2(-2)^3 + 5x^1(-2)^4 + 1x^0(-2)^5\\\\ & = x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32\end{aligned}[/tex]
