Perform the indicated operations and simplify completely?
[(x^2 - 9) / (x^2 - 5x)][(5x - x^2) / (x^2 - x - 12)][(x^2 - 8x + 16) / (x - 4)] [(x + 3)(x - 3) / (x^2 - 5x)][-(x^2 - 5x) / (x + 3)(x - 4)][(x - 4)^2 / (x - 4)] (x + 3)(x - 3)[-1 / (x + 3)][(x - 4) / (x - 4)] -(x - 3) 3 - x
Perform the indicated operations and simplify completely:?
Assume that the whole expression is... (x³ - 3x² + 6x - 18)/(x² - 4x) * (2x - 4)/(x^4 - 3x³) * (16 - x²)/(x² + 2x - 8) Simplify the whole expression via factoring out each expression and dividing the common factors in the numerator and denominator. You should get... (x²(x - 3) + 6(x - 3))/(x(x - 4)) * 2(x - 2)/(x³(x - 3)) * -(x - 4)(x + 4)/((x - 2)(x + 4)) = (x² + 6)(x - 3)/(x(x - 4)) * 2(x - 2)/(x³(x - 3)) * -(x - 4)(x + 4)/((x - 2)(x + 4)) = -2(x² + 6)/x^4 Good luck!
Please Perform the indicated operations and simplify completely?
FSD-- This will take a little time but try to stay with me. First, we should multiply by the inverse of the last term so we have all multiplication. It will look something like this: [x^2-9] * [5x-x^2] * [x^2-8x+16] --------------------------------------... [x^2-5x] * [x^2-x-12] * [x-4] Now we are ready to begin factoring each term in brackets. In the top, the first can be factored by difference of squares to The second term has a common factor of "x", x(5 - x). The third (a trinomial) is factored by trial and error to (x - 4)(x - 4). So now on the top we have, (x + 3)(x - 3) x(5 - x) (x - 4)(x - 4) --------------------------------------... [x^2-5x] * [x^2-x-12] * [x-4] Now let's work on factoring the bottom. The first term has a common factor of "x", x(x - 5). The second is a trinomial which factors to (x - 4)(x + 3). The third term is in simplest form. So now our problem looks like this: (x + 3)(x - 3) x(5 - x) (x - 4)(x - 4) --------------------------------------... x(x - 5) (x - 4)(x + 3) (x - 4) Now we are almost ready to do the fun stuff -- crossing out all the like factors! But wait!!!! In the top second factor we could have taken out a factor of "-x" and that term would have turned out to be -x(x - 5). This is a trick to remember when the difference is not quite what we wanted!!! So now we have (x + 3)(x - 3) -x(x - 5) (x - 4)(x - 4) --------------------------------------... x(x - 5) (x - 4)(x + 3) (x - 4) Now have at it and divide out the like factors. Remember, the "x's" will divide out and leave a factor of -1 in the numerator. We get: -1(x - 3) or -x - -3 or 3 - x. I hope you were able to stay with me through all of this!!! Good luck!
Algabar...Perform the indicated operation and simplify the result. Leave your answer in factored form.?
5x^2 - 23x - 10 = 0 = (5x + 2)(x - 5) = 0 Two answers 5x + 2 = 0 5x = -2 x = -2/5 And x - 5 = 0 x = 0 3x^2 - 7x - 20 = 0 (3x - 5)(x + 4) = 0 Two answers 3x - 5 = 0 3x = 5 x = 5/3 And x + 4 = 0 x = -4 25x^2 + 40x + 12 = 0 (5x + 6)(5x + 2) = 0 Two answers 5x + 6 = 0 5x = -6 x = -6/5 And 5x + 2 = 0 5x = -2 x = -2/5 3x^2 + 17x + 20 = 0 (3x + 5)(x + 4) = 0 Two answers 3x + 5 = 0 3x = -5 x = -5/3 And x + 4 = 0 x = -4 I hope this helped Kia
Perform the indicated operation and simplify the result. Leave in factored form.(8-(x+5/x))/(7+(x-5/x+5)?
Well, your parentheses are not quite right for what I think you mean. For example, 8-(x+5/x) means take five divided by x and add that value to x, then take that value and subtract it from 8. Here is the problem I think you meant: (8 - (x + 5)/x)/(7 + (x - 5)/(x + 5)) Now, to simplify this, first multiply the top and bottom by x(x + 5) to get rid of the complex fractions. (8x(x + 5) - x(x + 5)(x + 5)/x) / (7x(x + 5) + x(x + 5)(x - 5)/(x + 5)) (8x(x + 5) - (x + 5)(x + 5)) / (7x(x + 5) + x(x - 5)) (8x^2 + 40x - (x^2 + 10x + 25)) / (7x^2 + 35x + x^2 - 5x) (8x^2 + 40x - x^2 - 10x - 25) / (8x^2 + 30x) (7x^2 + 30x - 25) / (2x(4x + 15)) (7x - 5)(x + 5) / (2x(4x + 15)) If your complex fractions were supposed to only be 5/x then you ned to multiply the top and bottom by x to get rid of the complex fractions, but I will distribute the negative through the parentheses on top first and eliminate the extra set of perentheses on bottom.. (8 - (x + 5/x)) / (7 + (x - 5/x + 5)) (8 - x - 5/x) / (7 + x - 5/x + 5) (8x - x^2 - 5x/x) / (7x + x^2 - 5x/x + 5x) -(x^2 - 8x + 5) / (x^2 + 12x - 5) Neither of these factor. I recently created a new blog. The first posts are all on factoring and solving equations. I plan on posting to this blog regularly so it is likely that there will soon be several posts that could help you with your math. http://freemathhelponline.blogspot.com/ Please visit and let me know what you think. If you like it, do me a favor and visit the sponsoring adds because that is how I get paid. Also, tell your friends.
Perform the indicated operation & simplify completely. Please show work.?
(see work below)
Perform the indicated operations ((2x-10)/(8x^2-10x-3)) + ((1)/(3-2x))?
((2x-10)/(8x²-10x-3)) + ((1)/(3-2x)) Start by factoring out a Greatest Common Factor as shown: 2(x - 5) Now the problem can be written: ((2(x - 5))/(8x² - 10x - 3)) + ((1)/(3 - 2x)) Next, factor the trinomial 8x²-10x-3. Try various factors, and check the accuracy with the FOIL (First Outer Inner Last) method: 8x² - 10x - 3 = (4x + 1)(2x - 3) So now the problem is written as: ((2(x - 5))/((4x + 1)(2x - 3))) + ((1)/(3 - 2x)) = (2(x - 5))/((4x + 1)(2x - 3)) + (1)/(-2x + 3) Now you can multiply each term by a factor of 1 that will make all denominators equal. In this problem, what will make the denominators equal is (4x+1)(2x-3): = (2(x-5))/((4x+1)(2x-3)) + (1)/(-2x+3) * ((4x+1)(-1))/((4x+1)(-1)) = (2(x-5))/((4x+1)(2x-3)) + (1(4x+1)(-1))/((4x+1)(2x-3)) There are now numerators which have equal denominators that can be combined: = (2(x - 5) + (1(4x + 1)(-1))) / ((4x+1)(2x-3)) Once all the similar expressions are combined: = (-2x - 11)/((4x + 1)(2x - 3))