View Full Version : Multiplying and Factoring Algebraic Expressions

I am suppose to factor and simplify each expression. I need help with the following problem:

(x^3+1) Square root (x+1)- (x^3+1)^2 Square root (x+1)

What gives me trouble in this problem is the square root. I dont know how to start with this problem.

mmm4444bot

08-04-2011, 05:44 PM

The factor sqrt(x + 1) cannot be simplified, so treat it as a single number.

Instead, focus on the minus sign.

What factors appear on both sides of the minus sign? Factor those out, to start.

Then simplify.

^ I wanted to see what was the answer and it gave me this -x^3 (Square root of X+1) (x+1) (x^2-x+1). The only thing I wonder is why is -x^3 added?

I have another question regarding these problems. I am asked that once I factor them to set them to 0 and solve the equation in other words solve for the unknown.

I had this problem x^2+4xy+4y^2. Once you factor it, it comes out to (x+2y)^2. My question is if I set it to 0 would it be possible for me to get the unknown or do I just leave it like that.

mmm4444bot

08-04-2011, 09:01 PM

I wanted to see what was the answer and it gave me this -x^3 (Square root of X+1) (x+1) (x^2-x+1). The only thing I wonder is why is -x^3 added?

Work the exercise, and you'll find out why there is a factor of -x^3. If not, then please show your work.

I am asked that once I factor them to set them to 0 and solve the equation in other words solve for the unknown.

I had this problem x^2+4xy+4y^2. Once you factor it, it comes out to (x+2y)^2. My question is if I set it to 0 would it be possible for me to get the unknown or do I just leave it like that.

Huh? If you are asked to solve for x, then do it. If you are not asked to solve for x, then do not do it.

In other words, just follow the specific instructions that come with each exercise.

By the way, if you were to set (x+2y)^2 equal to zero, then you would be saying that x+2y=0.

There are an infinite number of xy-pairs that make x+2y equal to zero.

Here are three such pairs: (0, 0) (1, -1/2) (-4, 2)

.

Okay here is what I did so far:

(x^3+1) Square root (x+1)-(x^3+1)^2 Square root (x+1)- The original problem

Step 1: I gather the main factors as instructed in what the book tells me

(x^3+1) Square root (x+1)-[x^3+1+Square root (x+1)]

(x^3+1) Square root (x+1)-x^3-1-Squareroot (x+1) I don’t know if I did this step right and this is where I stayed stuck.

As far as the second part goes, I am thinking that the answer is infinite number of xy pairs. I got the second part pretty much thanks :).

mmm4444bot

08-05-2011, 05:36 PM

(x^3+1) Square root (x+1) - (x^3+1)^2 Square root (x+1)

(x^3+1) Square root (x+1) - [x^3+1 + Square root (x+1)] This line is wrong

If we factor out (x^3 + 1) sqrt(x + 1), then we get:

(x^3 + 1) sqrt(x + 1) [1 - (x^3 + 1)]

Simplify the expression inside square brackets, and you're left with -x^3.

x^3 + 1 factors as a sum of cubes.

.

I had to edit my post because I didnt see the one in the brackets, but my question is [1 - (x^3 + 1)] where does the 1 infront of x^3+1 come from?

mmm4444bot

08-06-2011, 02:12 AM

The 1 is needed to take the place of (x^3+1) sqrt(x+1) on the LEFT side of the subtraction sign, when we factor out (x^3+1) sqrt(x+1) from both sides of the subtraction sign.

Let's look at a simpler expression. The initial factorization in your exercise takes this form:

ab - a^2b = ab[1 - a]

In other words, the given expression in your exercise has the form ab - a^2b.

We see that the factors a and b appear on each side of the subtraction sign, so we factor them out: ab[1 - a]

The factorization ab[1 - a] must multiply out to give the expression ab-a^2b; otherwise, it's not a good factorization.

When we multiply out the factorization ab[1 - a], each of the two terms inside the square brackets get multiplied by ab, like this:

ab times 1 is ab

ab times -a is -a^2b

So, the multiplication of the factorization gives ab - a^2b, and we see that the factorization works.

If the 1 were not inside the square brackets, then the factorization would not multiply out to be ab-a^2b.

Again, your exercise is the same form. In your exercise, symbol a above represents the factor x^3+1, and symbol b above represents the factor sqrt(x+1).

(x^3+1) sqrt(x+1) [1 - (x^3+1)]

Multiply it out:

(x^3+1) sqrt(x+1) times 1 is (x^3+1) sqrt(x+1)

(x^3+1) sqrt(x+1) times -(x^3+1) is -(x^3+1)^2 sqrt(x+1)

So, the multiplication gives (x^3+1) sqrt(x+1) - (x^3+1)^2 sqrt(x+1)

And that's your original expression.

If the 1 were not inside the square brackets, then the factorization would not multiply out to give the original expression.

So I am assuming that in my next problem which is (x^2+1) square root (x+1) minus square root (x+1)^3 is just (x^2+1) square root (x+1) as my answer then. Am I correct?

mmm4444bot

08-10-2011, 07:10 PM

So I am assuming that in my next problem which is (x^2+1) square root (x+1) minus square root (x+1)^3 is just (x^2+1) square root (x+1) as my answer then. Am I correct?

Rather than assuming, it is better to simply work the exercise.

We type "minus square root (x+1)^3" like this:

- sqrt[(x + 3)^3]

No, your result is not correct. Try working the exercise, and show us what you tried, if you would like more help.

Note: We prefer that you start a new thread for each new exercise. (I seem to remember telling you this, already, but maybe not.)

Oh ok, I will start a new thread for each exercise. I figured I would post it on the same thread since it had to deal with the same theme. Well, the thing is I would work on the problem, but I am completely clueless. But I will try:

-sqrt[(x + 1)^3]

sqrt[(x + 1)] sqrt[(x + 1)] sqrt[(x + 1)]

If we multiply two of those then they would turn to (x+1)-sqrt[(x + 1)] so my answer is:

(x^2+1)sqrt[(x + 1)^3-1(x+1)-sqrt[(x + 1)]

(x^2+1)sqrt[(x + 1)^3-x-1+sqrt[(x + 1)] this would be my answer.

mmm4444bot

08-11-2011, 03:25 PM

I figured I would post it on the same thread since it had to deal with the same theme.

That's reasonable thinking, but that approach causes some threads to become quite long, containing many exercises over two or more screens. We've found that this confuses some people (like me).

I am completely clueless. But I will try

Good. Trying is how we all learn; understanding mathematics is a process of making mistakes, followed by finding them, understanding them, and fixing them. :D

(x^2 + 1) sqrt(x + 1)^3 - 1(x + 1) - sqrt(x + 1)

In the original expression, the cubed term appears only on the right side of the subtraction sign; therefore, the ^3 above should not be there.

Also, it is standard to not write the factor of 1 between the subtraction sign and the factor (x + 1).

(x^2 + 1) \sqrt{x + 1} - \sqrt{(x + 1)^3}

(x^2 + 1) \sqrt{x + 1} - \sqrt{x + 1} \sqrt{(x + 1)^2}

We see that the factor sqrt(x + 1) appears on both sides of the subtraction sign, so we factor it out.

\sqrt{x + 1} \left( (x^2 + 1) - \sqrt{(x + 1)^2} \right)

\sqrt{x + 1} \bigg ( (x^2 + 1) - (x + 1) \bigg )

Now simplify the expression inside the outer parentheses, and factor that result to finish.

^Actually, its square root (x+1)^3

My answer is square root (x+1) (x^2-x+2)

mmm4444bot

08-11-2011, 06:11 PM

My answer is square root (x+1) (x^2-x+2)

You sure like typing "square root (x + 1)" instead of sqrt(x + 1) ;)

You did not correctly simplify the expression inside the outer parentheses (that I posted).

I note that Jeff got a different result; I'm too rushed right now to check either his work or mine.

I'm leaving the check up to you. Remember, your factorization must multiply out to get back the original expression.

Sqrt(x+1) (x^2+1)- sqrt(x+1)^3 Original Problem

Okay so from what I remember sqrt(x+1) ((x^2+1) -(x+1))

sqrt(x+1) ((x^2+1-x-1))

sqrt(x+1) (x^2-x)

My answer is sqrt(x+1)*x(x-1)

Oh ok, thats good. I know where I made the mistake I wrote the problem wrong as well lol. Thanks :).

Powered by vBulletin® Version 4.2.2 Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.