Question

13 multiple choice 1 point simplify: $-2k^{3}(-4k^{2}+5k + 2)$ $-6k^{6}+3k^{3}$ $8k^{5}-10k^{4}-4k^{3}$ $-6k^{5}+3k^{4}$ $8k^{6}-10k^{3}-4k^{3}$ 14 multiple choice 1 point factor: $16z^{2}-20z + 25$ $(4z + 5)(4z + 5)$ $(4z + 5)(4z - 5)$ cannot be factored $(8z - 25)(z + 1)$ 15 multiple choice 1 point simplify: $(a - b)-(b - a)$ 0 $2a-2b$ $2b$ $2a$

Explanation:

Step1: Distribute -2k³

\[

$$\begin{align*} -2k^{3}(-4k^{2}+5k + 2)&=(-2k^{3})\times(-4k^{2})+(-2k^{3})\times(5k)+(-2k^{3})\times2\\ \end{align*}$$

\]

Step2: Use exponent - rule \(a^{m}\times a^{n}=a^{m + n}\)

\[

$$\begin{align*} (-2k^{3})\times(-4k^{2})&=(-2)\times(-4)k^{3 + 2}=8k^{5}\\ (-2k^{3})\times(5k)&=(-2)\times5k^{3+1}=-10k^{4}\\ (-2k^{3})\times2&=-4k^{3} \end{align*}$$

\]
So, \(-2k^{3}(-4k^{2}+5k + 2)=8k^{5}-10k^{4}-4k^{3}\)

Step3: For factoring \(16z^{2}-20z + 25\)

The discriminant of a quadratic form \(az^{2}+bz + c\) is \(\Delta=b^{2}-4ac\). Here \(a = 16\), \(b=-20\), \(c = 25\). Then \(\Delta=(-20)^{2}-4\times16\times25=400 - 1600=- 1200<0\). So it cannot be factored.

Step4: Simplify \((a - b)-(b - a)\)

\[

$$\begin{align*} (a - b)-(b - a)&=a - b - b+a\\ &=(a + a)+(-b - b)\\ &=2a-2b \end{align*}$$

\]

Answer:

1. B. \(8k^{5}-10k^{4}-4k^{3}\)
2. C. cannot be factored
3. B. \(2a - 2b\)