Answer:
See Explanation
Step-by-step explanation:
The question is incomplete, as Hernando and Rachel's solution are not provided. So, I will just solve the question directly.
Given
Required
Factor
Group into 2
![2mp-6p+27-9m = [2mp-6p]+[27-9m]](https://tex.z-dn.net/?f=2mp-6p%2B27-9m%20%3D%20%5B2mp-6p%5D%2B%5B27-9m%5D)
Factor each group
![2mp-6p+27-9m = 2p[m-3]+9[3-m]](https://tex.z-dn.net/?f=2mp-6p%2B27-9m%20%3D%202p%5Bm-3%5D%2B9%5B3-m%5D)
Rewrite 3 - m as -(m-3)
So, we have:
![2mp-6p+27-9m = 2p[m-3]-9[m-3]](https://tex.z-dn.net/?f=2mp-6p%2B27-9m%20%3D%202p%5Bm-3%5D-9%5Bm-3%5D)
Factor out m - 3
![2mp-6p+27-9m = [2p-9][m-3]](https://tex.z-dn.net/?f=2mp-6p%2B27-9m%20%3D%20%5B2p-9%5D%5Bm-3%5D)
Answer:
There could be many answers, please be more clearer.
The answer is 70 miles.
If we express distances as following:
T - total distance
a - distance from <span>San Antonio to Austin
b - </span>distance from <span>Austin to Waco
c - </span>distance from <span>Waco to Dallas
Then:
T = 280 mi
b = a - 30 mi </span>⇒ a = b + 30<span>
b = c + 20 mi </span>⇒ c = b - 20<span>
T = a + b + c
</span>⇒ a + b + c = 280
⇒ b + 30 + b + b - 20 = 280
⇒ 3b +10 = 280
⇒ 3b = 280 - 10
⇒ 3b = 270
⇒ b = 90 mi.
Distance from <span>Waco to Dallas is c.
</span><span>c = b - 20
</span>⇒ c = 90 - 20
⇒ c = 70 mi
Therefore, distance from <span>Waco to Dallas is 70 miles.</span>
In this equation, p=27 because it is the last term and q=3 because it is the coefficient of the first term. First find all divisors of these numbers:
27: 1, 3, 9, 27
3:1, 3
Then use +/- p/q to find all possible combinations
+/-1, +/-3, +/-9, +/-27, +/- 1/3
Or your answer can be written as: 1, -1, 3, -3, 9, -9, 27,-27, 1/3, -1/3
Hope this helps!
