Thirty cards are numbered from 1 to 30, inclusive. if one card is drawn at random, what is the probability that the number on the card will be divisible by either 3 or 5?

Find the numbers between 1 and 30 that are divisible by 3 or 5:

Divisible by 3:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Divisible by 5:

5, 10, 15, 20, 25, 30

There are 10 numbers divisible by 3 and  6 numbers divisible by 5 for a total of 16 numbers.

The probability is 16/30, which reduces to 8/15 are divisible by 3 or 5.

Answers:

A = 3

B = -7

C = 3

D = -7

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Explanation:

You could use the AC method to factor, but I think the method described below is a more efficient way of factoring.

Set the given expression equal to zero and solve for x using the quadratic formula. So the goal is to find the roots for

We'll plug in a = 9, b = -42, and c = 49.

Because the root is x = 7/3, it leads to the factor 3x-7 = 0

Note that we go from x = 7/3 to 3x = 7 after multiplying both sides by 3. Then we subtract 7 from both sides.

It turns out that we have two copies of the identical factor (3x-7) because x = 7/3 is a double root.

So in reality we have the factorization (3x-7)(3x-7) which we can shorten into (3x-7)^2 if we wanted.

Use the box method or FOIL method to expand (3x-7)(3x-7) and you should get 9x^2-21x-21x+49 which simplifies to 9x^2-42x+49. This confirms we have the correct factorization.

Therefore, A and C are both 3, while B and D are both -7.