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Algebra
checking problem Use FOIL to multiply: (6x – 1)(3x – 4) 6x*3x + 6x*(4) + (1)*3x + (1)*(4) = 6x*3x = 18x^2 6x*4 = 24x 1*3x = 3x 1* 4 = 4 18x^2 – 24x 3x +4 = 18x^2 – 27x + 4 
Math
(2x1)(3x7)(4x+3) =6x^214x33x+7 =(6x^219x+7)(4x+3) =24x^3+18x^2126x+28x+21 =24x^3+18x^288x+21 
BobPursley..here it is:
Problem 1: (x2)/(3x+4)  6x/(x+1) What I did: Common denominator is: (3x+4)(x+1) (x2)(x+1)/(3x+4)(x+1)  6x(3x+4)/(x+1)(3x+4) [(x2)(x+1)  6x(3x+4)] / (x+1)(3x+4) (x^2  x  2  18x^2  24x) / (x+1)(3x+4) x^2  18x^2 = 17x^2 
Math
Find the Least Common multiple of the two expressions: 24x^296 & 18x^2+18x36 
Math
Find the HCF of f(x) = 6x^3 + 24x^2 + 18x, g(x) = 4x^4 + 4x^3  24x^2 
algebra
12X^6+24x^518x^2/6x^2=2x^4+24x^33x could some one check this for me please 
Algebra (Re.: Reiny)
The fact that there was that nice symmetry to the question, makes it easier than it first appears. I had it as: (a/b + b/a)÷(b/a  a/b) look at (a/b + b/a) the common denominator for this would be ab, so a/b + b/a = (a^2 + 
algebra
Did I do this problem the correct way if so is there a shorter way to explain this? x(x + 6)(x + 4) = 24x x(6 + x)(x + 4) = 24x x(6 + x)(4 + x) = 24x (6 + x) * (4 + x x(6(4 + x) + x(4 + x)) = 24x x((4 * 6 + x * 6) + x(4 + 
partial fraction
if f(x) ax^2+bx+c , g(x) :1/3x^2+2 and fg(x)= 18x^4+24x^2+11/(3x^2+2)^" 
math
24x10=18x4