Factorizing Polynomials.

[the dark knight]

its not even a "formula" cuz its not equal to 0 or anything...its a function

 

[the dark knight]

as far as i know polynomial functions are studied in calculus so you're either looking for the derivative (i already gave you all) or the integral (must know if its limited or not)

 

[John]

Maths is as boring as a Matt Morgan Interview! :shifty:

 

[the dark knight]

only if you're an idiot :shifty:

 

[Wrestling Station]

Mr Nerdy:

...lol, you guys still study this shit?

 

[C4]

Actually this is a part of my polynomials topic. And I figured it out the other day.

We have to substitute numbers until f(x) turns out to be zero. And with that value of x, the polynomial is factorized completely.

We could also divide f(x) by one of its factors (eg. (x+2)) and then list the quotient as the completely factorized polynomial.

 

[the dark knight]

what?

 

[C4]

Example:

f(x) = 3x^3 + 7x^2 + 2x + 9
We're now going substitute different values for x until f(x) becomes zero.
Then we're going to state that when x=whatever then the polynomial is factorized completely.

 

[the dark knight]

for how long are you gonna do it exactly? and whats the value of X, in that function

 

[Dylanfsd]

rofl.. I do this shit in my spare time, easyness.

 

[C4]

Ok, found the question, I'll partially solve it for you now.

A function f(x) is denoted by 3x^3 + ax^2 - 2x - 8

(i) Given that (x+2) is a factor of f(x), find the value of a.
(ii) Factorize the polynomial completely when a has this value.

SOLUTION:

As (x+2) is a factor of f(x) then it should be equal to zero when x= -2 (x+2=0)

Therefore, f(-2)= 3(-2)^3 + a(-2)^2 - 2(-2) - 8
And so, 0 = ^this
And then, a=7.

Now we're going to put this into our equation and get this:

f(x) = 3x^3 + 7x^2 - 2x - 8

Now we have to factorize this completely. Lol. I was wrong. If we substitute x=-2 then f(x) becomes zero but then everything becomes zero. Uh. How to factorize this?