polynomial question (1 Viewer)

[Jaydels]

I have this question which I really can't work out. Can someone please help?

If P(x)=x^3 -3x^2 -9x +c has a double zero, find c and factorise P(x) over the real numbers.

Thanks

 

[jumb]

I have this question which I really can't work out. Can someone please help?

If P(x)=x^3 -3x^2 -9x +c has a double zero, find c and factorise P(x) over the real numbers.

Thanks

A double 0 means you can differentiate it and still get a factor.

This will remove the c.

 

[Jaydels]

oh thanks i'll try that

 

[jumb]

So if i'm right (probably not) then c = -5 or 27

edit: oops

 

[Jaydels]

the answer is c=-5 or 27

 

[Jaydels]

thanks i've got it now

 

[jumb]

Shit I'm wrong! A double zero means it has 2 roots the same. I'm thinking of things having a multiplicty of 2.

Edit: Oh wait, They're the same thing.

 

[Jaydels]

lol all these different names for things gets confusing

 

[jumb]

God actually theyre not the same thing.

Just ignore me. Seriously.

 

[Jaydels]

they are the same thing, my text book says " a zero of multiplicity 1 called a single zero"

 

[jumb]

Okay, i remember maths now.

Doubles: indicate a where the graph touches the x axis, include a (x + a)^2 in the polynomial.

Multiplicity, means you can diff. and the same root will be there.

Therefore, the answer is 27. This will result in the full factorisation being (x+3)(x - 3)^2

 

[dawso]

oh no.....duffman is out of control....

 

[Jaydels]

Okay, i remember maths now.

Doubles: indicate a where the graph touches the x axis, include a (x + a)^2 in the polynomial.

Multiplicity, means you can diff. and the same root will be there.

Therefore, the answer is 27. This will result in the full factorisation being (x+3)^2(x - 3)

and (x+1)^2(x-5)

 

[Slidey]

Jumb: Or c=-5, P(x)=(x+1)^2(x-5)

We don't know which since we don't know the value of c.

 

[jumb]

Oh yeah!

It's only been 3 months and i've already forgotten all of my maths :\

 

[Slidey]

Haha. It's OK, man. You don't have to know it for a second HSC or anything.

 

[jumb]

Nah but I like maths. I also like helping people. It fuels my selfish desire for being selfless.

 

[Slidey]

Sounds like me.

 

[TimeAndTide]

Blah! This isnt an EXT2 question!

 

[Slidey]

It is, actually. Roots of muliplicity.

P(x)=x^3 -3x^2 -9x +c
P'(x)=3x^2-6x-9=3(x-3)(x+1)
So either P(-1)=0 or P(3)=0
i.e. c=-5 or 27,
P(x)=x^3 - 3x^2 - 9x - 5=(x+1)^2(x-5), or
P(x)=x^3 - 3x^2 - 9x +27=(x-3)^2(x-3)

.'. P(x)=(x-3)^2(x-3) or (x+1)^2(x-5)