Question

pascals triangle to expand $(2x^{2}-z^{2})^{4}$. express your answer simplest form.

Explanation:

Step1: Find the coefficients from Pascal's Triangle

For the power of \(4\), the coefficients from Pascal's Triangle are \(1,4,6,4,1\).

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

Here \(a = 2x^{2}\), \(b=-z^{2}\) and \(n = 4\).

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Step3: Calculate each term

• For \(k = 0\): \(\binom{4}{0}(2x^{2})^{4}(-z^{2})^{0}=1\times16x^{8}\times1 = 16x^{8}\)
• For \(k = 1\): \(\binom{4}{1}(2x^{2})^{3}(-z^{2})^{1}=4\times8x^{6}\times(-z^{2})=-32x^{6}z^{2}\)
• For \(k = 2\): \(\binom{4}{2}(2x^{2})^{2}(-z^{2})^{2}=6\times4x^{4}\times z^{4}=24x^{4}z^{4}\)
• For \(k = 3\): \(\binom{4}{3}(2x^{2})^{1}(-z^{2})^{3}=4\times2x^{2}\times(-z^{6})=-8x^{2}z^{6}\)
• For \(k = 4\): \(\binom{4}{4}(2x^{2})^{0}(-z^{2})^{4}=1\times1\times z^{8}=z^{8}\)

Step4: Combine all the terms

\((2x^{2}-z^{2})^{4}=16x^{8}-32x^{6}z^{2}+24x^{4}z^{4}-8x^{2}z^{6}+z^{8}\)

Answer:

$16x^{8}-32x^{6}z^{2}+24x^{4}z^{4}-8x^{2}z^{6}+z^{8}$