Question

expression answer a. \\((32t^4 + 16t^2 - 8) \div (2t)\\) where \\(t \
eq 0\\) b. \\(\frac{18x^5y^4 + 8x^4y}{4x^4y^3}\\) where \\(x \
eq 0\\) and \\(y \
eq 0\\)

Explanation:

Part a

Step1: Divide each term by \(2t\)

We have \(\frac{32t^4}{2t} + \frac{16t^2}{2t} - \frac{8}{2t}\)

Step2: Simplify each term

For the first term: \(\frac{32t^4}{2t} = 16t^{4 - 1}=16t^3\)
For the second term: \(\frac{16t^2}{2t}=8t^{2 - 1} = 8t\)
For the third term: \(\frac{8}{2t}=\frac{4}{t}\)
So the result is \(16t^3 + 8t-\frac{4}{t}\)

Part b

Step1: Split the fraction

We get \(\frac{18x^5y^4}{4x^4y^3}+\frac{8x^4y}{4x^4y^3}\)

Step2: Simplify each term

For the first term: \(\frac{18x^5y^4}{4x^4y^3}=\frac{9}{2}x^{5 - 4}y^{4 - 3}=\frac{9}{2}xy\)
For the second term: \(\frac{8x^4y}{4x^4y^3}=\frac{2}{y^{3 - 1}}=\frac{2}{y^2}\)
So the result is \(\frac{9}{2}xy+\frac{2}{y^2}\)

Answer:

a. \(16t^3 + 8t-\frac{4}{t}\)

b. \(\frac{9}{2}xy+\frac{2}{y^2}\)