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Question
2. Perform the indicated operation and simplify completely
    5x+5/5x+9 + 5x+13/5x+9
                                                                 
                                                   
     3.   Perform the indicated operation and simplify completely
  (x+11) (x-11)    

   

5. Factor 4x-40 completely              

6.   Perform the indicated operation and simplify completely
(x-2)²          

7.    factor x²+10x+21 completely        

8.   factor x² - 8x +16 completely           


9.   factor 4x²-9y²



10.   perform the indicated operation completely
(x+10) (x²+9x-8)    
     
11.    perform the indicated operation completely  
(x-3)³

Answer
Hello Laurie

Problem #2
You see that the denominators are the same, so you just add the numerators and put with the same denominator as before.

   5x+5 + 5x+13 = 10X + 18 = 2(5X + 9)
Now divide it by the denominator

2(5X + 9)/ 5x+9 = the expression cancels out and 2 will be the most simplified answer.



Problem #3

(x+11) (X-11)    

Just multiply the two polynomials and it is
X2 – 121 (its just multiplication)


Problem #5

To factor the equation, look out for common numbers or algebraic expressions everywhere.
Look, 4 is common for 4 and 40. X can’t be because it is only found in the first term.

4(X – 10)

Problem #6

In order to carry out this operation, you have to split it into two equal equations and multiply them

(X – 2) * (X – 2) = X2 - 4X + 4 (multiply and collect like terms)

Problem #7

To tackle this problem you have too look for two numbers of which their product is 21 and their sum is 10.

They are 7 and 3 aren’t they? Laurie
So the factored equation will be
(X +7) * (X + 3)



Problem #8

The same principle applies here.

Look for two numbers of which their product is 16 and their sum is -8

They are both -4

Therefore, (X – 4)*(X – 4) is the factored equation.

Problem #9

Now this is a little bit baffling, but simple if you observe it carefully. It is a product of two functions. You know that (a + b)(a – b) = a2 – b2

Following the above formula

4x²-9y² is the product of 2x +3y and 2x – 3y

(2x + 3y)*(2x – 3y)

Problem #10

Laurie, this is a matter of multiplying.
Carefully multiply each term in the bracket by every member of the equation in the other bracket.

X*( x²+9x-8) = X3 + 9X2 – 8X
10*( x²+9x-8) = 10X2 + 90X -80

Now collect like terms and add them

Finally you’ll get

X3 + 19X2 + 82X – 80

Final Problem

First find (X – 3)2 and multiply it by (X – 3)

Try it yourself and let me give you the answer

X3 – 3X2 – 9X + 27  

Algebra

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Michael Tadesse

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I can answer questions regarding Relations, linear functions and equations, Squares and square roots, Translation and Similarity, Special Properties of Right Angled Triangles, Measurement of Area and Volume, algebra. I can also answer questions related to Pre calculus, Limits and Continuity, Derivatives Applications of Derivatives, High school math in general. I can’t answer questions of the following type. Power Series, Multiple Integrals, Fourier series, Vector valued Functions.

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I love mathematics as it is the mother of all sciences. I had an outstanding math results while I was in school. The tradition continues while I am in university. I have been helping elementary school Students for the past five years with mathematics problems. In fact, I have a brother who is four years younger than me I usually help him to understand what the crux of the question is before trying to solve it. He loves math very much, he is indeed one of the top scorers form his class in Mathematics. I didn’t stop helping my younger relatives after I joined the Engineering campus and I am more than ready to help anyone with my skill.

Education/Credentials
have a high school diploma with a 3.97 GPA; I also have many merit certificates granted for my outstanding results in high school math. I won a merit certificate for my participation in the Black Month Essay Contest. I took two different National Examinations in 2006 and 2007. In the former Exam I got 9As which includes Mathematics. I also have a score of 90 out of hundred in the later Examination which is the toughest known in Ethiopia. So you don’t have to worry about my ability to help my clients.

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I have never had an online client but I was helping students coming to me for assistance in Mathematics.

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