Question

Can somebody please explain the question and the answer I’m so, so lost in this unit.

Answer 1

Answer:

Undefined

Step-by-step explanation:

1. In order to even begin to answer this question, we'll need to find the definition of a few terms:

Coefficient: A constant quantity placed before a variable in an equation.

• ex: the 3 in the expression 3x^3 is the coefficient.

The expansion of an equation is the unfactored version of an equation.

• ex: x^2 +2x + 1 is the expansion of (x + 1)^2

2. Based on these definitions, the question is asking:

"What number is in front of of x^3 in the expanded version of (6x + 5y)^4?"

It still sounds super confusing, but at least its comprehensible.

3. Now that we know what we're being asked for, we can begin to solve... but how? Expanding (6x + 5y)^4 seems like it would take forever right?

This is when it gets complicated. You'll to have a couple concepts in your mind:

• The equation $nC_r=n!/r!*(n!-r!)$. This is called the Combination Formula
• Pascal's Triangle. I recommend finding an image of it.

If what I'm doing right now seems confusing, look up how to, "Find the Coefficient of an Expansion." I checked and there are videos on this.

4. Here is the order of how we solve this problem from this point forward:

1. Define our variables
2. Translate our variables into the format that our question is asking for.
3. Plug the translated variable into the Combination Formula.
4. Solve

5. We are given two different variables to expand. (6x) and (5y). The question is asking for the coefficient in front of x^3, so our format is (6x)^3(5y)^0. The 5y should still be shown although its exponent is 0.

I'm going to have to again stress looking for videos on what exactly I'm doing. I'm sorry that I can't properly teach Combinations. From this point on you may have trouble understanding what's going on.

Math:

• (6x)^3(5y)^0
• $4C_0$(6x)^3(5y)^0

Change $4C_0$ into the format of the Combination Formula

• $4C_0$ = 4!/ 4! - (4! - 0!)

Now plug this back into our equation

• = (4!/(4! - 4!)) (6x)^3(5y)^0
• = (4!/0) (6x)^3
• undefined

There is no term containing x^3 based on the fact that $4C_0$ is undefined.

I hope that you can figure out Combinations. I remember learning about them, but it is a very conceptual concept. Once you do figure out how they work though, they become very easy to use. Good luck on your math.