Hi factorise x^2+2x+3 for me please (1 Viewer)

[Twickel]

Differentiate (x+3)(x^2+9)

I get 3x^2+6x+9
take out the three and I get x^2+2x+3 for some reason I CANT FACTORISE IT.


PLEASE HELP.

 

[L]

use the quadratic formular.

 

[Trebla]

(x + 1)² + 2

Next time, post maths questions in the maths forums.

 

[L]

treb dog

 

[Twickel]

all wrong.

 

[lyounamu]

all wrong.

Your question is confusing.

What are you asking?

Are you asking people to factorise or differentiate? There are two questions.

And you posted your question on a wrong forum.

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Differentiate (x+3)(x^2+9)


f(x) = x^3 + 9x + 3x^2 + 27 = x^3+3x^2 + 9x+27

f'(x) = 3x^2 + 6x+ 9

 

[Shoom]

Ok now factorise the fist derivative to find staionary points please.

 

[youngminii]

Differentiate (x+3)(x^2+9)

I get 3x^2+6x+9
take out the three and I get x^2+2x+3 for some reason I CANT FACTORISE IT.


PLEASE HELP.

What are you doing
You're not allowed cancel it down
If the answer is 3x^2 + 6x + 9, leave it like that

Oh wait, are you asked to find stat points?
In that case, complete the f*cking square
x^2 + 2x + 3 = 0
x^2 + 2x + 1 = -2
(x + 1)^2 = -2
x + 1 = root(-2)
Now obviously, you can't root a negative number (except in 4 unit) so there ARE NO SOLUTIONS, WHICH MEANS IT DOESN'T HAVE ANY STAT POINTS

OR you could quadratic formula it, let's find the discriminant
b^2 - 4ac
4 - 12 = -8
So obviously, no solutions for x

 

[lyounamu]

Ok now factorise the fist derivative to find staionary points please.

He didn't ask for that.

But ok, if he wanted to find the stationary point, we can just use quadratic formula as suggested earlier. But again, this question is highly ambiguous.

EDIT: Just found that there is no stationary point

 

[Twickel]

What the hell this from HSC 2005, 3 mark question, im sure their is a staionary point

question 4b
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_05.pdf

 

[lyounamu]

What the hell this from HSC 2005, 3 mark question, im sure their is a staionary point

question 4b
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_05.pdf

Yeah, that proves that you just made a typo on your question.

It should be f(x) = (x+3)(x^2-9) NOT f(x) = (x+3)(x^2+9)

f(x) = (x+3)(x^2-9) = (x+3)(x-3)(x+3)
so x=-3 or x=3 when the graph crosses the x=axis (i.e. when y=0)

 

[Twickel]

No wonder!!!!!


I kept reading it as (x^2+9)

thankyou.

 

[Twickel]

So what are the stat points?

( 1 , -32) and (-3 ,0 ) ?

and ( 1 , -32 ) is the min point and the ( -3 , 0 ) is max?

 

[youngminii]

So what are the stat points?

( 1 , -32) and (-3 ,0 ) ?

and ( 1 , -32 ) is the min point and the ( -3 , 0 ) is max?

(x + 3)(x^2 - 9)
x^3 + 3x^2 - 9x - 27
dy/dx = 3x^2 + 6x - 9
For stat pts, dy/dx = 0
0 = x^2 + 2x - 3
(x + 3)(x - 1) = 0
ie. x = -3, 1
To find nature, test x for d2y/dx2
d2y/dx2 = 6x + 6
Sub x = -3
-18 + 6 < 0 Therefore maximum
Sub x = 1
6 + 6 > 0 Therefore minimum
ie. (-3, 0) is a max
(1, -32) is a min
ie. You're right

Edit: Oops, don't forget
They're both relative turning points
ie. (-3, 0) is a relative maxima
and (1, -32) is a relative minima
Although I highly doubt they'd mark you down for not having that, just to make the answer a bit more elegant(lol)
I avoid this issue by using Concave up and Concave down =D