find the value of M and n so that polynomial f(x)=x^3-6x^2+mx-n is exactly divisible by (x-1),(x-2)
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(x-1) & (x-2) are the factors of f (x)
so we can substitute the the value of x as x=1 or x=2.
firstly x=1
x^3-6x^2+mx-n=0
(1)^3-6 (1)^2+m (1)-n=0
1-6+m-n=0
m-n=5. ...(1)
Now x=2
x^3-6x^2+mx-n=0
(2)^3-6 (2)^2+m (2)-n=0
8-24+2m-n=0
2m-n=16. ... (2)
By subtracting equation (1) from (2) we get
m=11
n=6
value of m&n are 11&6 respectively
so we can substitute the the value of x as x=1 or x=2.
firstly x=1
x^3-6x^2+mx-n=0
(1)^3-6 (1)^2+m (1)-n=0
1-6+m-n=0
m-n=5. ...(1)
Now x=2
x^3-6x^2+mx-n=0
(2)^3-6 (2)^2+m (2)-n=0
8-24+2m-n=0
2m-n=16. ... (2)
By subtracting equation (1) from (2) we get
m=11
n=6
value of m&n are 11&6 respectively
kamakshisharma:
How the value of m is 11....and of n is 6......plzzzzz xplain
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