Question

use pascals triangle to expand $(4 - z)^{5}$. express your answer in simplest form.

Explanation:

Step1: Find the coefficients from Pascal's Triangle

For the power of \(5\), the coefficients from Pascal's Triangle are \(1,5,10,10,5,1\).

Step2: Apply the binomial expansion formula \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\)

Here \(a = 4\), \(b=-z\), \(n = 5\)

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Step3: Calculate each term

• For \(k = 0\): \(\binom{5}{0}(4)^{5}(-z)^{0}=1\times1024\times1 = 1024\)
• For \(k = 1\): \(\binom{5}{1}(4)^{4}(-z)^{1}=5\times256\times(-z)=- 1280z\)
• For \(k = 2\): \(\binom{5}{2}(4)^{3}(-z)^{2}=10\times64\times z^{2}=640z^{2}\)
• For \(k = 3\): \(\binom{5}{3}(4)^{2}(-z)^{3}=10\times16\times(-z^{3})=-160z^{3}\)
• For \(k = 4\): \(\binom{5}{4}(4)^{1}(-z)^{4}=5\times4\times z^{4}=20z^{4}\)
• For \(k = 5\): \(\binom{5}{5}(4)^{0}(-z)^{5}=1\times1\times(-z^{5})=-z^{5}\)

Step4: Combine all the terms

\((4 - z)^{5}=1024-1280z + 640z^{2}-160z^{3}+20z^{4}-z^{5}\)

Answer:

$1024 - 1280z + 640z^{2}-160z^{3}+20z^{4}-z^{5}$