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by shashank.ism » Sat Feb 06, 2010 1:41 pm
Let P(x) = kx3 + 2k2x2 + k3. Find the sum of all possible real numbers k for which (x - 2) is a factor of P(x).

a) - 8
b) 16
c) 4
d) 32
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by rohan_vus » Sat Feb 06, 2010 1:51 pm
P(x) is of form (x-2)F(x)...as as (x-2) is factor of P(x)..Take F(x)(another polynomial with 1 degree lesser than P(x))...

P(x) = (x-2)F(x)--(1)

This means for x = 2 , P(x) = 0

Above gives k^3 + 8k^2 + 8K = 0
(k)(k^2 + 8K + 8) = 0 ---(2)

Sum of all values of k = - 8

A it is
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by shashank.ism » Sat Feb 06, 2010 2:00 pm
rohan_vus wrote:P(x) is of form (x-2)F(x)...as as (x-2) is factor of P(x)..Take F(x)(another polynomial with 1 degree lesser than P(x))...

P(x) = (x-2)F(x)--(1)

This means for x = 2 , P(x) = 0

Above gives k^3 + 8k^2 + 8K = 0
(k)(k^2 + 8K + 8) = 0 ---(2)

Sum of all values of k = - 8

A it is
Good Rohan you have solved it pretty fast.... keep up the spirit....
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by ajith » Sat Feb 06, 2010 9:09 pm
shashank.ism wrote:Let P(x) = kx3 + 2k2x2 + k3. Find the sum of all possible real numbers k for which (x - 2) is a factor of P(x).

a) - 8
b) 16
c) 4
d) 32
2 is a factor of kx3 + 2k2x2 + k3
=> 8k+8k^2+k^3 =0
k(k^2+8k+8)

now k =0 and roots of (k^2+8k+8) =0 are the possible real numbers

sum of roots of equation (k^2+8k+8) =0 is -8 (-b/a = -8/1)

Sum of all possible roots = -8+0 = -8

A for me
Last edited by ajith on Sun Feb 07, 2010 2:47 am, edited 1 time in total.
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by shashank.ism » Sun Feb 07, 2010 1:40 am
rohan_vus wrote:P(x) is of form (x-2)F(x)...as as (x-2) is factor of P(x)..Take F(x)(another polynomial with 1 degree lesser than P(x))...

P(x) = (x-2)F(x)--(1)

This means for x = 2 , P(x) = 0

Above gives k^3 + 8k^2 + 8K = 0
(k)(k^2 + 8K + 8) = 0 ---(2)

Sum of all values of k = - 8

A it is

Rohan I would like to know what if the equation does come out to be quadratic or the polynomial which is given is of higher degree,
Do plugin method would be the only possible way to solve the problem ..
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by rohan_vus » Sun Feb 07, 2010 2:11 am
Rohan I would like to know what if the equation does come out to be quadratic or the polynomial which is given is of higher degree,
Do plugin method would be the only possible way to solve the problem ..
Not sure if you are asking in general about (x-2) being factor of any polynomial or are you asking about sum of roots of any polynomial ..?

Well either way , sum of roots of any polynomial you dont need to find out roots individually , as sum of roots = -(Coefficient of the term one degree less than first term )/(Coefficient of the first term of the polynomial)..I factorized it in this particular case , but you dont need to as you can directly derive sum of roots looking at the polynomial .

In general if any expression claims to be a factor of any polynomial then roots of that expression is one of the the solution of the given polynomial .

In this particular case x -2 is the expression given , and x=2 is the root of this expression , so x=2 must be solution to P(x). i.e P(2) = 0.

Yes substituting values of roots of factors is one of the best possible ways ( to my knowledge)
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by sunil_snath » Sun Feb 07, 2010 4:11 am
Hi Shashank,

what does this mean?

kx3 + 2k2x2 + k3

Is kx3 a 3 digit number with x as a digit? or is it K*3 in which case what is the difference between kx3 and k3.

I tried hard to figure what these terms mean. Even by looking at the other posts, i am not able to figure this out.

Please help.

thanks,
Sunil
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