SOLUTION: Hello, sorry to be a bother!!! But I need help a usual!! Here are my problems!! 1. Simplify the rational expression: x^2-3x-10 ________

SOLUTION: Hello, sorry to be a bother!!! But I need help a usual!! Here are my problems!! 1. Simplify the rational expression: x^2-3x-10 _________ x^2-x-20 2. Divide: 2x^2+5x-12 / 2

Question 33091: Hello, sorry to be a bother!!! But I need help a usual!! Here are my problems!!
1. Simplify the rational expression:
x^2-3x-10
_________
x^2-x-20
2. Divide:
2x^2+5x-12 / 2x^2-7x+6
_______________________
9x^2-16 / 3x^2-x-4
3. Simplify:
square root[144x^10y^12z^18]
4. Perform the indicated operations:
square root[50]+2 square root[32]- 2 square root[8]
5. Multiply:
(5 square root[2]+3)(square root[2]-2 square root[3])
6. Rationalize the denominator:
2
______________
square root[6]- square root[5]
I spelled out the square root because there is nothing on key board for symbol of the square root check. I will appreciate any help that I can get. Thanks!!!
Answer by sarah_adam(201) (Show Source): You can put this solution on YOUR website!
a)x^2-3x-10
x^2-5x+2x-10
x(x-5)+2(x-5)
(x+2)(x-5)
b)x^2-x-20
X^2-5x+4x-20
x(x-5)+4(x-5)
(x+4)(x-5)
c)(2x^2+5x-12) /( 2x^2-7x+6)
(2x^2+8x-3x-12)/(2x^2-4x-3x+6)
{2x(x+4)-3(x+4)}/{2x(x-2)-3(x-2)}
{(2x-3)(x+4)}/{(2x-3)(x-2)}
cancelling (2x-3)on both numerator and denominator we get
(x+4)/(x-2)
d)9x^2-16 / 3x^2-x-4
{(3x)^2 - (4)^2}/(3x^2 - 4x-+3x -4)
{(3x-4)(3x+4)}/{x(3x-4)+(3x-4)} numerator: according to a^2 - b^2 = (a+b)*(a-b)
{(3x-4)(3x+4)}/{(x+1)(3x-4)}
cancelling (3x-4) on both numerator and denominator we get
(3x+4)/(x-1)
e)square root[144x^10y^12z^18]

f)square root[50]+2 square root[32]- 2 square root[8]
+2-2
+2-2
+-

g) Rationalize the denominator:

2/
rationalising the denominator by multipling with
2 * []/
2 * []

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