Question

binomial theorem (pascals triangle)
score: 4/5 penalty: none
question
use pascals triangle to expand $(y^{2}+2x)^{4}$. express your answer in simplest form.
answer attempt 1 out of 2
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Explanation:

Step1: Recall Pascal's Triangle for power 4

The 5th row (since we start counting rows from 0) of Pascal's Triangle gives the coefficients 1, 4, 6, 4, 1 for \((a + b)^4=a^{4}+4a^{3}b + 6a^{2}b^{2}+4ab^{3}+b^{4}\). Here \(a = y^{2}\) and \(b = 2x\).

Step2: Calculate each term

For the first - term: \(1\times(y^{2})^{4}=y^{8}\).
For the second - term: \(4\times(y^{2})^{3}\times(2x)=4\times y^{6}\times2x = 8xy^{6}\).
For the third - term: \(6\times(y^{2})^{2}\times(2x)^{2}=6\times y^{4}\times4x^{2}=24x^{2}y^{4}\).
For the fourth - term: \(4\times(y^{2})\times(2x)^{3}=4\times y^{2}\times8x^{3}=32x^{3}y^{2}\).
For the fifth - term: \(1\times(2x)^{4}=16x^{4}\).

Step3: Combine the terms

\((y^{2}+2x)^{4}=y^{8}+8xy^{6}+24x^{2}y^{4}+32x^{3}y^{2}+16x^{4}\).

Answer:

\(y^{8}+8xy^{6}+24x^{2}y^{4}+32x^{3}y^{2}+16x^{4}\)