You could solve all the equations to see if a letter matches the question's request if you're into trail and error. Simply multiply each integer in one pair of parentheses by each individual in another, a box could help you with this (like a box in third grade showing what numbers shown on one side multiplied by numbers on the intersecting side are)

----------------------------------------------------------------------
The answer: If you're looking for the answer, continue below

The answer would be "B";
(x-2)(x^2+2x+4) could be rearranged into..
-2(x^2+2x+4) + x(x^2+2x+4) when, when distributed is..
(-2x^2-4x-8) + (x^3+2x^2+4x) which is then added..
x^3+8

There was some additional simplifying and adding, but you should understand it.

8 0

3 years ago

Read 2 more answers

From a deck of five cards numbered 2, 4, 6, 8, and 10, respectively, a card is drawn at random and replaced. this is done three

Serga [27]

Since the sum of the numbers on the three draws is 12, if we want the card numbered 2 to be drawn exactly two times, the third card can only be numbered 8. In fact, $2+2+8 = 12$ , and there are no other possibilities, unless you consider the various permutations of the terms.

So, we have three favourable cases: we can draw 2,2,8, or 2,8,2, or 8,2,2. This are the only three cases where the card numbered 2 is drawn exactly two times, and the sum of the number on the three draws is 12.

Now, the question is: we have three favourable cases over how many? Well, we have 5 possible outcomes with each draws, and the three draws are identical, because we replace the card we draw every time.

So, we have 5 possible outcomes for the first draw, 5 for the second and 5 for the third. This leads to a total of $5 \times 5 \times 5 = 5^3 = 125$ possible triplets.