I have many equations?

Question

I have taken this test and bombed it, one thing i do not understand is rational expressions, so i am hoping that if I put my math problems in here ppl can help me, explain it to me.


4c^2+14c-8 / c^2+6c+8


(a^2-3a-10 / a+5)(a^2+10a+25 / a^2-25)

help would be much appreciated

SIMPLIFY.........

Answer 1

Rational expressions can be simplified by factoring them and cancelling common factors or binomials.

For the first problem, on the top, you have:
4c² + 14c - 8 = 2(2c² + 7c - 4) = 2(2c - 1)(c + 4)
On the bottom, you have:
c² + 6c + 8 = (c + 4)(c + 2)

When simplifying, cancel the common factor of (c + 4)
2(2c - 1) / (c + 2), where c ≠ -2 or c ≠ -4.

For the second problem, you have on top:
(a² - 3a - 10)·(a² + 10a + 25) = (a - 5)(a + 2)·(a + 5)²
On bottom, you have:
(a + 5)·(a² - 25) = (a + 5)·(a + 5)(a - 5)

Cancel common factors of (a + 5)²·(a - 5)
(a + 2) / 1 = a + 2, where a ≠ ±5.

Answer 2

(4c^2 + 14c - 8) / (c^2 + 6c + 8)
2(2c^2 + 7c - 4) / (c + 2)(c + 4)
2(2c - 1)(c + 4) / (c + 2)(c + 4)
2(2c - 1) / (c + 2)

(a^2 - 3a - 10) / (a + 5) * (a^2 + 10a + 25) / (a^ 2 - 25)
(a - 5)(a + 2) / (a + 5) * (a + 5)(a + 5) / (a + 5)(a - 5)
(a - 5)(a + 2)(a + 5)^2 / [(a + 5)^2(a - 5)]
a + 2

There are some asymptotes to deal with in each expression, but that is how they each can be simplified.

Answer 3

I believe you mean (4c^2+14c-8)/(c^2+6c+8)

(4c^2+14c-8) becomes (2c-1)(2c+8) = 2*(2c-1)(c+4)

(c^2+6c+8) becomes (c+4)(c+2)

2*(2c-1)(c+4)/(c+4)(c+2) = 2(2c-1)/(c+4)

(a^2-3a-10) becomes (a-5)(a+2)
(a^2+10a+25) becomes (a+5)(a+5)
(a^2-25) becomes (a-5)(a+5)
If think your original problem is:
(a^2-3a-10)/(a+5) * (a^2+10a+25)/(a^2-25)
this becomes
(a-5)(a+2)/(a+5) * (a+5)(a+5)/(a-5)(a+5)
(a-5)(a+2)/(a+5) * (a+5)/(a-5)
(a-5)(a+2)/(a-5)
a+2

Answer 4

1. factors into 2(2c-1)(c+4)/(c+4)(c+2). (c+4) cancels out because its a common factor of the numerator and denominator. so the answer is: (4c-2 )/ (c+2).

2. factors into (a-5)(a+2) / (a+5) times (a+5)(a+5) / (a+5)(a-5).
When common factors (a-5) , (a+5), and (a+5) are cancelled out from the numerator and denominator, all that's left is: a+2

Good luck!

Answer 5

I haven't done these types of problems in a while so I went to this site and it helped me to remember how to do them. Check it out.

http://www.purplemath.com/modules/rtnldefs.htm

The answer to number 2 is (a+2). I did the math, but I don't have time to write it out or explain right now. Send me an e-mail if you'd like a detailed explanation.

Answer 6

Do you know the formula for solving 2nd degree equations?
x1= [- b + sqrt(b^2-4ac)]/(2a)
x2=[-b-sqrt(b^2-4ac)]/(2a)
Use this to factor all numerators and all denominators.
Then divide by the common factors, and it would be better to say for instance when dividing by a+5 that you can do this if a<> -5.

And I think you are missing some parentheses.

Answer 7

These are expressions, not equations. Equations must have an equals symbol,"=". What are you asked to *do* with these expressions?

Answer 8

1. The answer is 1883

2.The answer is -3570.

Hope that helps for ya.

Answer 9

http://www.khake.com/page47