Can someone explain this to me?

Question

I've had the flu for the past few days so I haven't been studying for my GED test like I should of been. I need somoene to in depth explain to me how to do this algebra problem which involves using distributive property and combining coefficients and like terms.

5(2+x)+3(5x+4)-(x^2)^2

please be as descriptive as you can. Even going over distributive property and exactly where to multiply what in what order would be a big help!

I need the problem simplified, not really solved I guess. This is the page I got it from

http://math.com/school/subject2/lessons/S2U2L5DP.html

Answer 1

It might help to remember that the full name of the property is the distributive property of multiplication over addition. You only use it when you are multiplying a sum (or difference). For instance:

3(x + y) = 3x + 3y

You distribute the 3 over both the x and the y.

Think of it like this: suppose there's a bag lunch containing a sandwich and an apple. Someone comes along and gives you three of them. How many sandwiches and apples do you have? Obviously, you have three of each. When you got three of the bags, you multiplied everything inside the bag, both the sandwiches and the apples. That's more or less why the distributive property works the way it does.

It gets a little trickier when you're multiplying a sum by a sum. For instance,

(a + b)(x + y)

There's a rule for this called the FOIL method. It stands for "First Outside Inside Last". What it means is to multiply the first two term (ax), the outside terms (ay), the inside terms (bx) and the last terms (by). So the result would be

(a + b)(x + y) = ax + ay + bx + by

It's a simplification, though. The actual rule is to multiply each term in the first sum by each term in the last one. FOIL only works if it's a two-term sum times a two-term sum, but you can multiply sums of any number of terms by remembering that: multiply everything in the first sum by everything in the last. So:

(a + b + c)(x + y + z) = ax + ay + az + bx + by + bz + cx + cy + cz

The order in which you multiply them is totally up to you. You just want to be sure you don't leave anything out.

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Like terms: terms are considered to be "like" if they have exactly the same variables with exactly the same powers. For instance,

3x, 12x and ax are all like terms. I could add them as

3x + 12x + ax = (3 + 12 + a)x = (15 + a)x

If you look at it, combining like terms is basically the reverse of the distributive property. It's easy to check to see if you did it right - just multiply the thing out again.

These terms would not be like terms:

3x, 5y, 2x^2, ax^3

"Combining coefficient", as I suspect you see, is simply combining like terms.

In the problem you have stated, you'd distribute first:

5(2 + x) + 3(5x + 4) - (x^2)^2

= 10 + 5x + 15x + 12 - (x^2)^2

Then combine like terms

= (10 + 12) + (5x + 15x) - (x^2)^2

= 22 + 20x - (x^2)^2

A quick note about that last term: there are rules for dealing with powers of powers, but they all go back to the original definitions. Consider:

(x^2)^2 = (x^2)(x^2) (because that's what "^2" means)

= (x * x)(x * x) (again, that's what "^2" means

= x * x * x * x = x^4

So your problem simplifies to

22 + 20x + x^4

Rules for combining exponents:

(x^a)^b = x^(ab)

So (x^3)^4 = x^(3 * 4) = x^12

If that doesn't make sense, write it out as I did the (x^2)^2. You'll see it pretty quickly.

Also,

(x^a)(x^b) = x^(a + b)

Consider: (x^2)(x^4)

= (x * x)(x * x * x * x)

= x * x * x * x * x * x = x^6

So (x^2)(x^4) = x^(2 + 4) = x^6

People - me included - constantly get those two rules mixed up. If you forget, you can derive them easily just by doing what I did above. Just remember what x^2 means in terms of how many x's and you'll be fine.

Long-winded, I know, but I hope it helps. Email me if there's anything else I can do, and good luck with the test. I'm sure you'll do just fine.

Answer 2

This eqn can be split into 3 parts: 5(2 + x), 3(5x + 4) and -(x^2)^2

Let's take a look at the 1st part, 5(2 + x). Since 5 is outside the bracket, you have to multiply everything inside the bracket by 5. Hence, you get : (10 + 5x).

Same thing for the 2nd part, you multiply everything inside the bracket by 3 to get: (15x + 12).

It's a little trickier for the last part. Take note that it's a NEGATIVE term since there is a negative sign outiside the bracket. This is important as there is a (major) difference between having the negative sign inside and outside the bracket. For this qns, it is outside, so u know that the 3rd part gives you a negative term.

To solve the 3rd part, you see that there's a power 2 outside the bracket. Hence, you have to mulyiply 2 to the power of the term inside the bracket. You will get : (- x^4).

Combining all 3 parts will give you:

5(2+x)+3(5x+4)-(x^2)^2
= (10 + 5x) + (15x + 12) + (- x^4)
= -x^4 + 20x + 22
[add all like terms together (i.e numbers, x terms) before tidying up your answer by ranking them according to their power.]

Answer 3

remember: MDAS (multiply, divide, add, subtract)

when distributing a number, first multiply all those you need to
like in your example:
5(2 + x) + 3 (5x + 4) - (x^2)^2
= (10 + 5x) + (15x + 12) - (x^2 * x^2)
= 10 + 5x + 15x + 12 - x^4
= 20x + 22 - x^4

Answer 4

First get everything out of parentheses:

10+5X+15X+12-x^4
20X + 22-x^4

After that point, I am confused. Since there is only one side of the equation, there is nothing to solve.

Answer 5

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