Can I use a factoring company to buy a profitable business ?
Q. I am interested in purchasing a profitable online business but I do not have the available funds to do it at the moment. So i'd like to know if a factoring company would be able to help me out.
Asked by avee - Wed Feb 21 23:07:57 2007 - - 2 Answers - 0 Comments
A. You are talking about buying a business based upon its receiveables. Borrowing against the receivables to finance the business. I highly doubt it. Keep in mind factoring is going to be like getting 70% of the value of the invoice.It might be possible, call a factoring company, be prepared to have a resume that shows why you can run this company. Good Luck
Answered by batwanda - Wed Feb 21 23:44:53 2007
Q. I am interested in purchasing a profitable online business but I do not have the available funds to do it at the moment. So i'd like to know if a factoring company would be able to help me out.
Asked by avee - Wed Feb 21 23:07:57 2007 - - 2 Answers - 0 Comments
A. You are talking about buying a business based upon its receiveables. Borrowing against the receivables to finance the business. I highly doubt it. Keep in mind factoring is going to be like getting 70% of the value of the invoice.It might be possible, call a factoring company, be prepared to have a resume that shows why you can run this company. Good Luck
Answered by batwanda - Wed Feb 21 23:44:53 2007
Why is it a good idea to factor out the GCF first and then use other methods of factoring?
Q. Use 3x^2-18x+15 as an example. What happens if one first uses trial and error to factor as two binomials rather than first factoring out the GCF? Thank you for your help!
Asked by Photogurl - Fri Oct 2 09:49:07 2009 - - 1 Answers - 0 Comments
A. Factoring out the 3 and giving you 3(x^2-6x+5) decreases the amount of time and effort you have to use in trial and error to find the factors which can easily be seen now to be (x+6)(x-1)
Answered by JLP - Fri Oct 2 09:55:39 2009
Q. Use 3x^2-18x+15 as an example. What happens if one first uses trial and error to factor as two binomials rather than first factoring out the GCF? Thank you for your help!
Asked by Photogurl - Fri Oct 2 09:49:07 2009 - - 1 Answers - 0 Comments
A. Factoring out the 3 and giving you 3(x^2-6x+5) decreases the amount of time and effort you have to use in trial and error to find the factors which can easily be seen now to be (x+6)(x-1)
Answered by JLP - Fri Oct 2 09:55:39 2009
In order to be able factoring polynomial equations. What would you recommend to bone up more?
Q. What's the easiest way to learn factoring polynomials?
Asked by ranmat_88 - Sun Apr 2 10:57:44 2006 - - 2 Answers - 0 Comments
A. Be familiar with the divisors of small numbers. Learn to quickly look at two numbers and know if one is the sum or difference of two numbers that multiply to the other. Be able to do arithmetic in your head. Learn how to do synthetic division. Practice on many examples.
Answered by ymail493 - Sun Apr 2 11:01:38 2006
Q. What's the easiest way to learn factoring polynomials?
Asked by ranmat_88 - Sun Apr 2 10:57:44 2006 - - 2 Answers - 0 Comments
A. Be familiar with the divisors of small numbers. Learn to quickly look at two numbers and know if one is the sum or difference of two numbers that multiply to the other. Be able to do arithmetic in your head. Learn how to do synthetic division. Practice on many examples.
Answered by ymail493 - Sun Apr 2 11:01:38 2006
How would I go about factoring this expression?
Q. Hi, I need help on factoring this expression. I tried on my own, but I'm pretty bad at factoring, so I need help on how to factor this type of expression. 6x^2 + 3xy + 2xz + yz thanks.
Asked by mschuhlein - Sun Aug 17 17:26:10 2008 - - 2 Answers - 0 Comments
A. Group. (6x^2 + 2xz) + (3xy+ yz) Factor. 2x(3x + z) + y(3x+ z) Due to distributive properties, you can rewrite as: (2x + y)(3x + z)
Answered by Lucy - Sun Aug 17 17:34:58 2008
Q. Hi, I need help on factoring this expression. I tried on my own, but I'm pretty bad at factoring, so I need help on how to factor this type of expression. 6x^2 + 3xy + 2xz + yz thanks.
Asked by mschuhlein - Sun Aug 17 17:26:10 2008 - - 2 Answers - 0 Comments
A. Group. (6x^2 + 2xz) + (3xy+ yz) Factor. 2x(3x + z) + y(3x+ z) Due to distributive properties, you can rewrite as: (2x + y)(3x + z)
Answered by Lucy - Sun Aug 17 17:34:58 2008
When factoring how do you know which order to place the numbers in the parenthesis?
Q. For example, if you are factoring: 2x^2 - x - 6 how do you know if it should be (2x+3)(x-2) or if it the last two numbers should be reversed like (2x-2)(x+3)? Also, how do you know which numbers are positive and which negative, what are the rules of the signs?
Asked by b_crazy_101 - Fri Jan 2 16:05:46 2009 - - 2 Answers - 0 Comments
A. There's a very quick way, let's try an expanded binomial square: x^2 - 5x + 6 First make the two binomials --> x^2- 5x + 6 = (x ) (x ) with the srqt of first term The signs are like this always: -) In the first binomial goes the same sign that the second term so (x - ) (x ) -) In the second binomial goes the multiplication of second and third term signs, that means - * + = - Finally you get --> (x - ) (x - ) Remember to put first the bigger number in absolute value, of course. Good Luck!!! P.D: you'll get (x - 3)(x - 2) at the end
Answered by pelirrojo2007 - Fri Jan 2 16:16:51 2009
Q. For example, if you are factoring: 2x^2 - x - 6 how do you know if it should be (2x+3)(x-2) or if it the last two numbers should be reversed like (2x-2)(x+3)? Also, how do you know which numbers are positive and which negative, what are the rules of the signs?
Asked by b_crazy_101 - Fri Jan 2 16:05:46 2009 - - 2 Answers - 0 Comments
A. There's a very quick way, let's try an expanded binomial square: x^2 - 5x + 6 First make the two binomials --> x^2- 5x + 6 = (x ) (x ) with the srqt of first term The signs are like this always: -) In the first binomial goes the same sign that the second term so (x - ) (x ) -) In the second binomial goes the multiplication of second and third term signs, that means - * + = - Finally you get --> (x - ) (x - ) Remember to put first the bigger number in absolute value, of course. Good Luck!!! P.D: you'll get (x - 3)(x - 2) at the end
Answered by pelirrojo2007 - Fri Jan 2 16:16:51 2009
How do you use factoring to solve this?
Q. Originally a rectangle was twice was long as it was wide. When 4 m were added to its length and 3 m subtracted from its width, the resulting rectangle had an area of 600 m^2. Find the dimensions of the new rectangle. The directions were to use factoring to solve, as mentioned before. Please explain each step. Thanks in advance!
Asked by Brown eyed girl. - Sun Jan 3 19:56:43 2010 - - 2 Answers - 0 Comments
A. (x - 4)(x - 3) = 600 after solving, 2 solutions come up: x = -21, or x = 28 the solution can't be negative, so x = 28 Plug this value in and the dimensions = 24 by 25
Answered by i^i iZ ReaL - Sun Jan 3 20:05:31 2010
Q. Originally a rectangle was twice was long as it was wide. When 4 m were added to its length and 3 m subtracted from its width, the resulting rectangle had an area of 600 m^2. Find the dimensions of the new rectangle. The directions were to use factoring to solve, as mentioned before. Please explain each step. Thanks in advance!
Asked by Brown eyed girl. - Sun Jan 3 19:56:43 2010 - - 2 Answers - 0 Comments
A. (x - 4)(x - 3) = 600 after solving, 2 solutions come up: x = -21, or x = 28 the solution can't be negative, so x = 28 Plug this value in and the dimensions = 24 by 25
Answered by i^i iZ ReaL - Sun Jan 3 20:05:31 2010
How do you find the X-intercept of a qaudratic function while factoring?
Q. My question wants me to find the X-intercept of the qaudratic eqaution y=x^2+4x+4 by factoring. How do i do this?
Asked by Macamoo - Fri Apr 3 00:40:18 2009 - - 2 Answers - 0 Comments
A. Ok, so you are going to have to complete the square. First: y-4=x^2+4x Then, you need to add the square of half of the non squared x term. So (4/2)^2 to both sides y-4+4=x^2+4x+4 Third, you need to factor the right side. So: y=(x+2)^2 Because the standard form of a parabola is y-k=a(x-h)^2, this shows you that the h term is -2 which means the x intercept is (-2,0)
Answered by Alex S - Fri Apr 3 00:47:33 2009
Q. My question wants me to find the X-intercept of the qaudratic eqaution y=x^2+4x+4 by factoring. How do i do this?
Asked by Macamoo - Fri Apr 3 00:40:18 2009 - - 2 Answers - 0 Comments
A. Ok, so you are going to have to complete the square. First: y-4=x^2+4x Then, you need to add the square of half of the non squared x term. So (4/2)^2 to both sides y-4+4=x^2+4x+4 Third, you need to factor the right side. So: y=(x+2)^2 Because the standard form of a parabola is y-k=a(x-h)^2, this shows you that the h term is -2 which means the x intercept is (-2,0)
Answered by Alex S - Fri Apr 3 00:47:33 2009
Good websites for explaining rights re factoring charges in Scotland?
Q. I live in Scotland, am an owner occupier in an ex-council house and have to pay factoring charges every month which I am really annoyed about. Are there any websites which explain the rights of owner occupiers regarding this? Thanks.
Asked by Gianna - Mon Jun 2 14:35:02 2008 - - 1 Answers - 0 Comments
A. I would check it out with Citizens advice but I assume that this was explained before u bought the house.
Answered by badcat - Tue Jun 3 17:17:43 2008
Q. I live in Scotland, am an owner occupier in an ex-council house and have to pay factoring charges every month which I am really annoyed about. Are there any websites which explain the rights of owner occupiers regarding this? Thanks.
Asked by Gianna - Mon Jun 2 14:35:02 2008 - - 1 Answers - 0 Comments
A. I would check it out with Citizens advice but I assume that this was explained before u bought the house.
Answered by badcat - Tue Jun 3 17:17:43 2008
when would factoring not work to solve a quadratic equation?
Q. I an find the zeros, roots, or solutions to a quadratic by taking the square root of both sides of the equation (when there is not a bx term), factoring, and completing the square.
Asked by frannieogreen - Wed Nov 19 21:41:55 2008 - - 3 Answers - 0 Comments
A. When the roots are irrational. This happens whenever the sqrt of the discriminant is irrational. In other words whenever b^2 - 4ac is not the square of a rational number.
Answered by xtempore - Wed Nov 19 21:45:27 2008
Q. I an find the zeros, roots, or solutions to a quadratic by taking the square root of both sides of the equation (when there is not a bx term), factoring, and completing the square.
Asked by frannieogreen - Wed Nov 19 21:41:55 2008 - - 3 Answers - 0 Comments
A. When the roots are irrational. This happens whenever the sqrt of the discriminant is irrational. In other words whenever b^2 - 4ac is not the square of a rational number.
Answered by xtempore - Wed Nov 19 21:45:27 2008
NEED TO KNOW STAT! When factoring a trinomial by grouping it is necessary to write it in 4 terms because why?
Q. When factoring a trinomial by grouping it is necessary to write it in 4 terms because why?? Thank you! Need to know today! Sorry! thank you sooo much!
Asked by Amy - Wed Feb 24 15:56:40 2010 - - 1 Answers - 0 Comments
Q. When factoring a trinomial by grouping it is necessary to write it in 4 terms because why?? Thank you! Need to know today! Sorry! thank you sooo much!
Asked by Amy - Wed Feb 24 15:56:40 2010 - - 1 Answers - 0 Comments
Is there a quicker way to factoring trinomials without a leading coefficient of 1 rather than grouping?
Q. For example show your method by factoring this: 2x^2-5x+3
Asked by yugioh - Wed Sep 30 20:23:32 2009 - - 4 Answers - 0 Comments
A. In some cases (like perhaps this one) trial and error may be faster. Here's my mental process (roughly) as I try to factor this trinomial. We know that 2x^2 - 5x + 3 has to factor as something like (ax + b)(cx + d). Since 2 is prime, then the "2x^2" term can only break apart into x and 2x, which means we have to have (2x + ?)(x + ?) The number 3 is prime, so it can only break into 1 and 3 or -1 and -3. Since the coefficient of x is negative, it looks like we have to break the 3 into -1 and -3. Let's try (2x - 1)(x - 3) But this gives -7 as the coefficient of x, so let's try the other way: (2x - 3)(x - 1) This works. --- I find that trial and error tends to be faster when the coefficient of x^2 and the constant term are prime… [cont.]
Answered by TheMathemagician - Wed Sep 30 20:34:39 2009
Q. For example show your method by factoring this: 2x^2-5x+3
Asked by yugioh - Wed Sep 30 20:23:32 2009 - - 4 Answers - 0 Comments
A. In some cases (like perhaps this one) trial and error may be faster. Here's my mental process (roughly) as I try to factor this trinomial. We know that 2x^2 - 5x + 3 has to factor as something like (ax + b)(cx + d). Since 2 is prime, then the "2x^2" term can only break apart into x and 2x, which means we have to have (2x + ?)(x + ?) The number 3 is prime, so it can only break into 1 and 3 or -1 and -3. Since the coefficient of x is negative, it looks like we have to break the 3 into -1 and -3. Let's try (2x - 1)(x - 3) But this gives -7 as the coefficient of x, so let's try the other way: (2x - 3)(x - 1) This works. --- I find that trial and error tends to be faster when the coefficient of x^2 and the constant term are prime… [cont.]
Answered by TheMathemagician - Wed Sep 30 20:34:39 2009
Why is factoring important in finding a least common denominator?
Q. 1) why is factoring important in finding a least common denominator? 2) A common student error is to confuse an equation with an expresstion. How do you distinguish between them? 3) Whether you have an expression containing fractions or a fractional equation, the first step is to find the LCD. Contrast what you do with the LCD in each situation. 4) Compare and contrast the two methods of simplifying complex fractions.
Asked by sw_joey - Thu Mar 22 20:42:50 2007 - - 3 Answers - 0 Comments
A. 1) It's part of a surefire technique for getting to the right answer. 2) An equation sets two expressions equal to each other. If there's no = sign, there's no equation. 3) and 4) require actually understanding the material you're going to have to teach. We can't give you quick answers here. And frankly, I don't WANT to help somebody get the job of teaching math if they don't respect the material enough to try to understand it themselves.
Answered by Curt Monash - Thu Mar 22 23:43:04 2007
Q. 1) why is factoring important in finding a least common denominator? 2) A common student error is to confuse an equation with an expresstion. How do you distinguish between them? 3) Whether you have an expression containing fractions or a fractional equation, the first step is to find the LCD. Contrast what you do with the LCD in each situation. 4) Compare and contrast the two methods of simplifying complex fractions.
Asked by sw_joey - Thu Mar 22 20:42:50 2007 - - 3 Answers - 0 Comments
A. 1) It's part of a surefire technique for getting to the right answer. 2) An equation sets two expressions equal to each other. If there's no = sign, there's no equation. 3) and 4) require actually understanding the material you're going to have to teach. We can't give you quick answers here. And frankly, I don't WANT to help somebody get the job of teaching math if they don't respect the material enough to try to understand it themselves.
Answered by Curt Monash - Thu Mar 22 23:43:04 2007
How do you use a graph, synthetic division, and factoring to find all the roots of an equation?
Q. Heres my problem I am given: x^3 + 5x^2 + 3x -9 = 0 It says: Use a graph, synthetic division, and factoring to find all of the roots of each equation So if you could show me how to do this using a calculator by explaining all the steps in order to get the answer I would very much appreciate it and you will receive 10 points from me quaranteed. I have an exam tomorrow and I dont get this at all
Asked by Brandon - Wed Feb 25 22:53:18 2009 - - 1 Answers - 0 Comments
A. use the synthetic division it! test +1 see work by use synthetic division it (x-1)(x^2+6x+9) (x-1)(x+3)(x+3) power highest tell you how many answer already it will be 3 answer it. you tell which graphing calculator i can help you it contact by email or something help put in the graphing caluculator.
Answered by Windy N - Wed Feb 25 22:59:43 2009
Q. Heres my problem I am given: x^3 + 5x^2 + 3x -9 = 0 It says: Use a graph, synthetic division, and factoring to find all of the roots of each equation So if you could show me how to do this using a calculator by explaining all the steps in order to get the answer I would very much appreciate it and you will receive 10 points from me quaranteed. I have an exam tomorrow and I dont get this at all
Asked by Brandon - Wed Feb 25 22:53:18 2009 - - 1 Answers - 0 Comments
A. use the synthetic division it! test +1 see work by use synthetic division it (x-1)(x^2+6x+9) (x-1)(x+3)(x+3) power highest tell you how many answer already it will be 3 answer it. you tell which graphing calculator i can help you it contact by email or something help put in the graphing caluculator.
Answered by Windy N - Wed Feb 25 22:59:43 2009
Does anybody know how to complete factoring puzzle?
Q. I am in algebra 2 advanced and my teacher gave us factoring puzzle for extra credit. I factored all the polynomials but i cant fix the puzzle! What should i do?
Asked by cutie pie - Sat Nov 22 16:06:22 2008 - - 1 Answers - 0 Comments
A. You need to give more detail.
Answered by Curt Monash - Sat Nov 22 19:21:11 2008
Q. I am in algebra 2 advanced and my teacher gave us factoring puzzle for extra credit. I factored all the polynomials but i cant fix the puzzle! What should i do?
Asked by cutie pie - Sat Nov 22 16:06:22 2008 - - 1 Answers - 0 Comments
A. You need to give more detail.
Answered by Curt Monash - Sat Nov 22 19:21:11 2008
When or in what situations do we use factoring, taking the square root, graphing, and completing the square?
Q. Factoring, Taking the square root, graphing, and completing the square are four methods to solve quadratic equations. When or in what situations do we use them?
Asked by Tay - Mon Apr 27 16:17:52 2009 - - 1 Answers - 0 Comments
Q. Factoring, Taking the square root, graphing, and completing the square are four methods to solve quadratic equations. When or in what situations do we use them?
Asked by Tay - Mon Apr 27 16:17:52 2009 - - 1 Answers - 0 Comments
How would I show the order to group terms when factoring a four term polynomial does not make a difference?
Q. Show that the order you group terms when factoring a four-term polynomial does not make any difference. Give one example to illustrate this, showing the steps of both situations
Asked by J - Mon Aug 11 22:01:44 2008 - - 2 Answers - 0 Comments
A. give an example and factor by grouping: x^3 + 4x^2 - 2x - 8 x^2(x + 4) - 2(x + 4) (x^2 - 2)(x + 4) or reorder: x^3 - 2x + 4x^2 - 8 x(x^2 - 2) + 4(x^2 - 2) (x + 4)(x^2 - 2)
Answered by notthejake - Mon Aug 11 22:10:37 2008
Q. Show that the order you group terms when factoring a four-term polynomial does not make any difference. Give one example to illustrate this, showing the steps of both situations
Asked by J - Mon Aug 11 22:01:44 2008 - - 2 Answers - 0 Comments
A. give an example and factor by grouping: x^3 + 4x^2 - 2x - 8 x^2(x + 4) - 2(x + 4) (x^2 - 2)(x + 4) or reorder: x^3 - 2x + 4x^2 - 8 x(x^2 - 2) + 4(x^2 - 2) (x + 4)(x^2 - 2)
Answered by notthejake - Mon Aug 11 22:10:37 2008
Need some help with the steps for factoring a polynomial?
Q. Have a algebra class and the question I am asked it What would you recommend should be the steps for factoring a polynomial? Anyone have any suggestions??
Asked by Mick - Fri Jul 31 14:43:10 2009 - - 1 Answers - 0 Comments
A. Steps for factoring... 1. Always factor out a GCF first 2. Look for special forms, like the difference of perfect squares and perfect square trinomials 3. Factor when the leading coefficient is 1 4. Factor when the leading coefficient is not 1 To factor ax^2 + bx + c, find two numbers such that they multiply to ac and add to b at the same time. When the leading coefficient is 1, you can go right to the factored form. Example: x^2 + 10x + 24 ac = 24 b = 10 The two numbers are 6,4. Since a = 1, go right to factored form (x +6)(x + 4) When a>1 find two numbers such that they multiply to ac and add to b. Then replace the middle term with those two numbers and factor by grouping. Example: 2x^2 + 9x - 5 ac = -10 b = -9 … [cont.]
Answered by unknown - Fri Jul 31 14:53:46 2009
Q. Have a algebra class and the question I am asked it What would you recommend should be the steps for factoring a polynomial? Anyone have any suggestions??
Asked by Mick - Fri Jul 31 14:43:10 2009 - - 1 Answers - 0 Comments
A. Steps for factoring... 1. Always factor out a GCF first 2. Look for special forms, like the difference of perfect squares and perfect square trinomials 3. Factor when the leading coefficient is 1 4. Factor when the leading coefficient is not 1 To factor ax^2 + bx + c, find two numbers such that they multiply to ac and add to b at the same time. When the leading coefficient is 1, you can go right to the factored form. Example: x^2 + 10x + 24 ac = 24 b = 10 The two numbers are 6,4. Since a = 1, go right to factored form (x +6)(x + 4) When a>1 find two numbers such that they multiply to ac and add to b. Then replace the middle term with those two numbers and factor by grouping. Example: 2x^2 + 9x - 5 ac = -10 b = -9 … [cont.]
Answered by unknown - Fri Jul 31 14:53:46 2009
why is factoring important in finding a least common denominator?
Q. 1) why is factoring important in finding a least common denominator? 2)when will you need to apply the squaring property of equality twice in the process of solving an equation containing radicals? 3) what kind of number will always result when we multiply complex conjugates?
Asked by sw_joey - Fri Apr 6 13:12:49 2007 - - 1 Answers - 0 Comments
A. 1) Not clear what you want. Do you really mean LCM or Least Common Mutiple? Factors help you find this but the process is rather long to explain. 2) Take the example sqrt(a) + sqrt(b) = c If you square the whole line you get a + 2*sqrt(a)*sqrt(b) + b = c^2 This needs to be rearranged to 2*sqrt(a)*sqrt(b) = c^2 - a - b Only when you square the whole line again will the radicals disappear. 3) Complex conjugates always multiply to a real number.
Answered by mathsmanretired - Fri Apr 6 13:34:53 2007
Q. 1) why is factoring important in finding a least common denominator? 2)when will you need to apply the squaring property of equality twice in the process of solving an equation containing radicals? 3) what kind of number will always result when we multiply complex conjugates?
Asked by sw_joey - Fri Apr 6 13:12:49 2007 - - 1 Answers - 0 Comments
A. 1) Not clear what you want. Do you really mean LCM or Least Common Mutiple? Factors help you find this but the process is rather long to explain. 2) Take the example sqrt(a) + sqrt(b) = c If you square the whole line you get a + 2*sqrt(a)*sqrt(b) + b = c^2 This needs to be rearranged to 2*sqrt(a)*sqrt(b) = c^2 - a - b Only when you square the whole line again will the radicals disappear. 3) Complex conjugates always multiply to a real number.
Answered by mathsmanretired - Fri Apr 6 13:34:53 2007
When do yo use partial factoring and completing the square in a quadratic equation?
Q. Im not sure when to use partial factoring with quadratic equations...is it when ax^2 + bx are the only ones that can be factored? Also for completing the square do you use that if you want to find the vertex of the quadratic equation right away ?? im confused on how to use them since our teacher has been away for 4 days and she's giving us a test on this tomorrow which im still confused about..HELPP
Asked by xoxo_peridot05_xoxo - Wed Nov 21 21:52:33 2007 - - 1 Answers - 0 Comments
A. "Partial factoring" or setting up the equation in vertex form is for the most part used when you want to graph the equation's vertex. To write the equation in vertex form it will end up as y = a(x-h)^2 + k, where (h, k) = -b/(2a) , (4ac - b^2)/(4a) = f(-b/(2a)) You use the quadratic equation when you want to find the zeros or x-intercepts of the equation (which also helps with the graphing). I hope that helps sort out things a little. ;)
Answered by AAAAF - Wed Nov 21 22:27:10 2007
Q. Im not sure when to use partial factoring with quadratic equations...is it when ax^2 + bx are the only ones that can be factored? Also for completing the square do you use that if you want to find the vertex of the quadratic equation right away ?? im confused on how to use them since our teacher has been away for 4 days and she's giving us a test on this tomorrow which im still confused about..HELPP
Asked by xoxo_peridot05_xoxo - Wed Nov 21 21:52:33 2007 - - 1 Answers - 0 Comments
A. "Partial factoring" or setting up the equation in vertex form is for the most part used when you want to graph the equation's vertex. To write the equation in vertex form it will end up as y = a(x-h)^2 + k, where (h, k) = -b/(2a) , (4ac - b^2)/(4a) = f(-b/(2a)) You use the quadratic equation when you want to find the zeros or x-intercepts of the equation (which also helps with the graphing). I hope that helps sort out things a little. ;)
Answered by AAAAF - Wed Nov 21 22:27:10 2007
Can you please help me with this math involving partial factoring?
Q. For this quadratic equation a) use partial factoring to find points that are equidistant from the axis of symmetry b)find the coordinates of the vertex c)express the relation in vertex form y=-2x^2 + 12x - 11 Thanks in advance.
Asked by GK27V6 - Thu Apr 26 17:54:54 2007 - - 1 Answers - 0 Comments
A. (a) to find the axis of symmetry:-12/-2(2)=3 y=3 (b)To find the co-ordinates of the vertex; -12/-2(2)=3 x=3, since you know x, you can just plug in the 3 and find y. -2(3)^2+12(3)-11=7 V=(3,7) (c) you now write the equation in vertex form... The formular is always: a(x-h)+k -2(x-3)^2+7...
Answered by petite fille - Thu Apr 26 19:56:41 2007
Q. For this quadratic equation a) use partial factoring to find points that are equidistant from the axis of symmetry b)find the coordinates of the vertex c)express the relation in vertex form y=-2x^2 + 12x - 11 Thanks in advance.
Asked by GK27V6 - Thu Apr 26 17:54:54 2007 - - 1 Answers - 0 Comments
A. (a) to find the axis of symmetry:-12/-2(2)=3 y=3 (b)To find the co-ordinates of the vertex; -12/-2(2)=3 x=3, since you know x, you can just plug in the 3 and find y. -2(3)^2+12(3)-11=7 V=(3,7) (c) you now write the equation in vertex form... The formular is always: a(x-h)+k -2(x-3)^2+7...
Answered by petite fille - Thu Apr 26 19:56:41 2007
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