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February 21st, 2016, 12:26 PM   #1
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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?
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February 21st, 2016, 12:36 PM   #2
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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.
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February 21st, 2016, 12:41 PM   #3
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Perhaps he wanted to know whether you'd promptly offer a good reason why what he'd said was wrong.
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February 21st, 2016, 12:55 PM   #4
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Multiplying a polynomial by -1 will not change the location of function zeros, but it does reflect the graph over the x-axis
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February 21st, 2016, 01:05 PM   #5
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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
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February 21st, 2016, 06:03 PM   #6
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Quote:
Originally Posted by studiot View Post
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!
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