question help - thanks (1 Viewer)

[muttiah]

if f(x) = 3x^4 - 16x^3 - 18x^2 + 216x + 40, find f'(x) in factored form.

what does factored form mean?

i got it up to here

f'(x) = 12x^3 - 48x^2 - 36x + 216

= 12(x^3 - 4x^2 - 3x + 18)

how can i simplify this further without know polynomials.. is there any other way?

(b)Find all stationery pts

so x^3 -4x^2 -3x +18 = 0

and how can i solve this?

thanks for ur help!!!

 

[Affinity]

12(x^3 - 4x^2 - 3x + 18)
= 12(x+2)(x-3)^2

think that solves the question (both a and b)

 

[muttiah]

syr but could u plz show how u done this one?

thanks bye

 

[SoulSearcher]

With this question, you would have to use the factor theorem, and then use long division to get the equation fully factorised.

 

[ronaldinho]

do u have to know polynomials for this question?

 

[SoulSearcher]

Well, that's the only way I can see you solving this equation, for now.

 

[muttiah]

there MUST be another way.. thanks neways
goodbye

 

[airie]

Unless you wanna do it the "longer" way

Since the coefficient of x^3 is 1, you know that x^3 - 4x^2 - 3x + 18, it will be factorised into the form (x+a)(x+b)(x+c) where a, b, c are some numbers. Then you can either expand this out and equate coefficients, or just be using Vieta's Theorem (that's the one on relationships between roots and coefficients of a polynomial - hmm, that's part of the polynomial topic then ), you get:
a+b+c=-4
ab+bc+ca=-3
abc=18
Then solve for a, b, and c Oh and, since abc=18 which is a relatively small number here, you can just try all its factors that fit into the two other equations

 

[SoulSearcher]

Both equally good From that, you can see that the roots have to be multiples of 18, which are +/- 1, +/- 2, +/- 3, +/- 6, +/- 9 or +/- 18. Well integer roots anyway.

 

[muttiah]

k

 

[Affinity]

or you can use the quartic formula

 

[airie]

^Cubic, rather And imo that should be the last resort