Question

1. use substitution to expand $(-4x + 3y)^5$

$1a^5b^0 + 5a^4b + 10a^3b^2+ 10a^2b^3 + 5ab^4 +1a^0b^5$
$1(-4x)^5(3y)^0 + 5(-4x)^4(3y)^1 + 10(-4x)^3(3y)^2 + 10(-4x)^2(3y)^3 + 5(-4x)^1(3y)^4 + 1(-4x)^0(3y)^5$

Explanation:

Step1: Recall binomial expansion formula

The binomial expansion for $(a+b)^n$ is:
$$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^{k} = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \binom{n}{3}a^{n-3}b^3 + \binom{n}{4}a^{n-4}b^4 + \binom{n}{5}a^{0}b^5$$
For $n=5$, this is:
$$(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$$

Step2: Substitute $a=-4x$, $b=3y$

Replace $a$ and $b$ in the formula:
$$(-4x+3y)^5 = (-4x)^5 + 5(-4x)^4(3y) + 10(-4x)^3(3y)^2 + 10(-4x)^2(3y)^3 + 5(-4x)(3y)^4 + (3y)^5$$

Step3: Calculate each term

Term1: Compute $(-4x)^5$

$$(-4x)^5 = (-4)^5x^5 = -1024x^5$$

Term2: Compute $5(-4x)^4(3y)$

$$5(-4)^4x^4 \cdot 3y = 5 \cdot 256x^4 \cdot 3y = 3840x^4y$$

Term3: Compute $10(-4x)^3(3y)^2$

$$10(-4)^3x^3 \cdot 9y^2 = 10 \cdot (-64)x^3 \cdot 9y^2 = -5760x^3y^2$$

Term4: Compute $10(-4x)^2(3y)^3$

$$10(-4)^2x^2 \cdot 27y^3 = 10 \cdot 16x^2 \cdot 27y^3 = 4320x^2y^3$$

Term5: Compute $5(-4x)(3y)^4$

$$5(-4x) \cdot 81y^4 = 5 \cdot (-4x) \cdot 81y^4 = -1620xy^4$$

Term6: Compute $(3y)^5$

$$(3y)^5 = 243y^5$$

Step4: Combine all terms

Add the calculated terms together.

Answer:

$$-1024x^5 + 3840x^4y - 5760x^3y^2 + 4320x^2y^3 - 1620xy^4 + 243y^5$$