User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 February 21st, 2016, 12:26 PM #1 Newbie   Joined: Feb 2015 From: Brazil Posts: 13 Thanks: 0 Math Focus: Still figuring it out. Polynomials A professor of mine said that we cannot multiply a polynomial with a negative leading coefficient by minus one to yield a positive leading coefficient. Ex: -x³+5x²-9x+5, then I do (-1)(-x³+5x²-9x+5) to have x³-5x²+9x-5. He said that it produces a new graph and therefore is not like the original one. But in a textbook I have been using they seem to do it all the time. Can someone explain it to me? February 21st, 2016, 12:36 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra Your professor is right that you get a different graph. Multiplying by -1 reflects the graph in the x-axis. Perhaps if you give an example of when the textbook does it, we'll be able to explain why it doesn't matter in that context. February 21st, 2016, 12:41 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2273 Perhaps he wanted to know whether you'd promptly offer a good reason why what he'd said was wrong. February 21st, 2016, 12:55 PM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 3,044 Thanks: 1627 Multiplying a polynomial by -1 will not change the location of function zeros, but it does reflect the graph over the x-axis February 21st, 2016, 01:05 PM #5 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 A good tip to find out what is happening is to try some simple known values for x to see if multiplying the polynomial through by -1 makes any difference. So here goes put x = 0 Then P = +5 and (-1) P = -5, where P is the polynomial expression To show that x = 0 is not a special case try x = 2 Then P = -1 and (-1) P = 1 But what happens if x = 1? Well P = 0 so (-1)P = 0 = P and they are the same. So in the case where we convert our polynomial expression into an equation equal to zero we are justified and can multiply through by any convenient number That is solving the equation -x³+5x²-9x+5 = 0 is the same as solving the equation x³-5x²+9x-5 = 0 February 21st, 2016, 06:03 PM   #6
Newbie

Joined: Feb 2015
From: Brazil

Posts: 13
Thanks: 0

Math Focus: Still figuring it out.
Quote:
 Originally Posted by studiot A good tip to find out what is happening is to try some simple known values for x to see if multiplying the polynomial through by -1 makes any difference. So here goes put x = 0 Then P = +5 and (-1) P = -5, where P is the polynomial expression To show that x = 0 is not a special case try x = 2 Then P = -1 and (-1) P = 1 But what happens if x = 1? Well P = 0 so (-1)P = 0 = P and they are the same. So in the case where we convert our polynomial expression into an equation equal to zero we are justified and can multiply through by any convenient number That is solving the equation -x³+5x²-9x+5 = 0 is the same as solving the equation x³-5x²+9x-5 = 0
Thank you, that was just what I was looking for! Tags polynomials Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post omare702 Algebra 2 July 2nd, 2012 06:21 PM kunju Algebra 5 February 1st, 2012 01:38 PM ElMarsh Linear Algebra 3 October 15th, 2009 04:14 PM instict911 Algebra 2 November 9th, 2008 08:46 PM n777l Complex Analysis 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      