Question

use pascals triangle to expand (2y² + 2z)⁵. express your answer in simplest form.

Explanation:

Step1: Recall Pascal's Triangle for power 5

The coefficients for $(a + b)^n$ from Pascal's Triangle for $n = 5$ are 1, 5, 10, 10, 5, 1. Here $a=2y^{2}$ and $b = 2z$.

Step2: Apply the binomial - expansion formula

$(a + b)^5=a^{5}+5a^{4}b + 10a^{3}b^{2}+10a^{2}b^{3}+5ab^{4}+b^{5}$.
Substitute $a = 2y^{2}$ and $b = 2z$:

• For the first term: $(2y^{2})^{5}=2^{5}(y^{2})^{5}=32y^{10}$.
• For the second term: $5\times(2y^{2})^{4}\times(2z)=5\times16y^{8}\times2z = 160y^{8}z$.
• For the third term: $10\times(2y^{2})^{3}\times(2z)^{2}=10\times8y^{6}\times4z^{2}=320y^{6}z^{2}$.
• For the fourth term: $10\times(2y^{2})^{2}\times(2z)^{3}=10\times4y^{4}\times8z^{3}=320y^{4}z^{3}$.
• For the fifth term: $5\times(2y^{2})\times(2z)^{4}=5\times2y^{2}\times16z^{4}=160y^{2}z^{4}$.
• For the sixth term: $(2z)^{5}=32z^{5}$.

Answer:

$32y^{10}+160y^{8}z + 320y^{6}z^{2}+320y^{4}z^{3}+160y^{2}z^{4}+32z^{5}$