find the coefficient of x^5 in the expansion (x+2)(x+1)^8
x* (x+1)^8
+
2(x+1)^8
so
find coef of x^4 in (x+1)^8
find 2*coef of x^5 in (x+1)^*
add them
from binomial coefficient table
(x+1)^8 = x^8 ..... 56x^5+70x^4....+1
so 70 + 2(56) = 182
182 is the final answer? Thank you for your help.
To find the coefficient of \(x^5\) in the expansion of \((x+2)(x+1)^8\), we need to expand the polynomial using the binomial theorem.
The binomial theorem states that for any real numbers \(a\) and \(b\) and any positive integer \(n\), the expansion of \((a+b)^n\) can be written as:
\((a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \ldots + \binom{n}{n-1}a^1 b^{n-1} + \binom{n}{n}a^0 b^n\)
where \(\binom{n}{k}\) is the binomial coefficient, which represents the number of ways to choose \(k\) elements from a set of \(n\) elements.
In our case, we have \((x+2)(x+1)^8\), which can be expanded as:
\((x+2)(x+1)^8 = \binom{8}{0}x^8 2^0 + \binom{8}{1}x^7 2^1 + \binom{8}{2}x^6 2^2 + \ldots + \binom{8}{7}x^1 2^7 + \binom{8}{8}x^0 2^8\)
To find the coefficient of \(x^5\), we need to identify the term in the expansion that contains \(x^5\):
\(\binom{8}{3}x^5 2^3\)
Therefore, the coefficient of \(x^5\) in the expansion is \(\binom{8}{3} \cdot 2^3\):
\(\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56\)
\(\binom{8}{3} \cdot 2^3 = 56 \cdot 2^3 = 56 \cdot 8 = 448\)
Therefore, the coefficient of \(x^5\) in the expansion \((x+2)(x+1)^8\) is 448.