Configuring a Cubic

Algebra Level 3

Let $f(x)=x^{3}+ax^{2}+bx+c$ be a cubic which vanishes at $1$ and $2$ . If it is given that $f(0)=4$ , find the value of $a+b+c.$

3 solutions

B.S.Bharath Sai Guhan
Aug 4, 2014

Actually, there is a lot of extra information given (probably to throw you off. Nice one, Krishna! :) )

We are given that $f(x) = x^{3} + ax^{2} + bx + c = 0$ for $x=1$

Hence, $f(1) = 1 + a + b + c = 0$

Therefore, $a + b + c = \boxed{-1}$

Excellent....Bharat...really good :)....The original problem actually asked for the values of a, b,c separately. I made it too easy :(

Krishna Ar - 6 years, 10 months ago

Amples of Information!!

Anik Mandal - 6 years, 10 months ago

did the same way

Aman Gautam - 6 years, 10 months ago

Excelent solution. I've found the values of a,b,c. Great problem!

Carlos David Nexans - 6 years, 10 months ago

Good....actual problem had that...I messed up and made this too easy (sob..)

Krishna Ar - 6 years, 10 months ago

@B.S.Bharath Sai Guhan -Which grade are ya in?

Krishna Ar - 6 years, 10 months ago

The 11th..Why d'ya ask?

B.S.Bharath Sai Guhan - 6 years, 10 months ago

Just wanted to know ....co-Tamilian :P

Krishna Ar - 6 years, 10 months ago

Nivedit Jain
Dec 14, 2016

Put x=1 as it vanishes there being a polynomial it has domain and range all defined everywhere . So we get 0=1+a+b+c a+b+c=-1

Vishal S
Jan 13, 2015

Given that

$f(x)$ = $x^{3}$ + a $x^{2}$ + bx + c

And given

$f(1)$ = 0 & $f(2)$ =0

$\Rightarrow$ $f(1)$ = 1 + a + b + c = 0 $\to$ (1)

$f(0)$ = c = 4

$\Rightarrow$ c = 4

By substituting c = 4 in (1), we get

1 + a + b + 4 = 0

$\Rightarrow$ a + b + 5 = 0

$\Rightarrow$ a + b = -5

Therefore a + b + c = -5 + 4 = $\boxed{-1}$