scottchapman wrote:If #p# = ap³ + bp - 1, where a and b are constants, and #-5# = 3, what is the value of #5#?

5
0
-2
-3
-5

FOLLOW THE DIRECTIONS.

#p# = ap³ + bp - 1

implies

#-5# = a(-5)³ + b(-5) - 1 = -125a - 5b - 1.

Since #-5# = 3, we get:
-125a - 5b - 1 = 3.
-125a - 5b = 4
125a + 5b = -4.

#p# = ap³ + bp - 1

implies

#5# = a(5)³ + b(5) - 1 = 125a + 5b - 1.

Since 125a + 5b = -4, 125a + 5b - 1 = -4-1 = -5.

The correct answer is E.

In general, these symbol problems give you a definition first, then specific numbers.

For instance, suppose I said that

x$y = 3x + 2y

That means that for ANY values of x and y, the $ sign tells us to triple x, double y, then add these together. For instance

10$9 = 3*10 + 2*9

since in this case x = 10 (it's the number on the left) and y = 9 (it's the number on the right).

Your problem works the same way.

#p# = ap³ + bp - 1 says take p, then plug it into the equation on the right. We have p = -5, so

#-5# = a*(-5)³ + b*(-5) - 1

We're ALSO told that #-5# = 3. Since #-5# = 3 and = a*(-5)³ + b*(-5) - 1, we can set these equal, giving

3 = a*(-5)³ + b*(-5) - 1

or

4 = -125a -5b

(The color-coding will make sense in a sec.)

We're asked for the value of #5#, or 125a + 5b - 1.

We know that 125a + 5b - 1 = -(-125a - 5b) - 1 = -4 -1 = -5, so we're set!