Question

1. (5 - 6q - 8q²) - (14q² + 2)
2. (d - 4d² + 8d³) + 3/4(-16d + 32d²)
3. (2/3h - 3h³ + 12) + (2/5h³ - 6)
4. (6 - 3b²) - (5b⁵ - 9b + 2b²) - (2b - b² + 7)
5. write a polynomial expression in standard form for the perimeter of the triangle, where x > 2/3.

Explanation:

11.

Step1: Remove parentheses

\[

$$\begin{align*} (5 - 6q-8q^{2})-(14q^{2}+2)&=5 - 6q-8q^{2}-14q^{2}-2 \end{align*}$$

\]

Step2: Combine like - terms

\[

$$\begin{align*} &5 - 2-6q+(-8q^{2}-14q^{2})\\ &=3-6q - 22q^{2} \end{align*}$$

\]

Step1: Distribute the fraction

\[

$$\begin{align*} (d-4d^{2}+8d^{3})+\frac{3}{4}(-16d + 32d^{2})&=d-4d^{2}+8d^{3}+\frac{3}{4}\times(-16d)+\frac{3}{4}\times32d^{2}\\ &=d-4d^{2}+8d^{3}-12d + 24d^{2} \end{align*}$$

\]

Step2: Combine like - terms

\[

$$\begin{align*} &8d^{3}+(-4d^{2}+24d^{2})+(d - 12d)\\ &=8d^{3}+20d^{2}-11d \end{align*}$$

\]

Step1: Remove parentheses

\[

$$\begin{align*} (\frac{2}{3}h-3h^{3}+12)+(\frac{2}{5}h^{3}-6)&=\frac{2}{3}h-3h^{3}+12+\frac{2}{5}h^{3}-6 \end{align*}$$

\]

Step2: Combine like - terms

\[

$$\begin{align*} &-3h^{3}+\frac{2}{5}h^{3}+\frac{2}{3}h+(12 - 6)\\ &=(-3+\frac{2}{5})h^{3}+\frac{2}{3}h + 6\\ &=(-\frac{15}{5}+\frac{2}{5})h^{3}+\frac{2}{3}h+6\\ &=-\frac{13}{5}h^{3}+\frac{2}{3}h + 6 \end{align*}$$

\]

Answer:

\(-22q^{2}-6q + 3\)

12.