If the two zeroes of the polynomial f(x) = x3 – 6x2 + mx + n are 2 + root5 and 2 - root 5, then find third zero and values of m and n.
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Answered by
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Step-by-step explanation:
x=1
f(1)=1^3-6(1)^2+m×1-n=0
=1-6+m-n=0
=-5+m-n=0
=m=n+5 ... (1)
If f(x) =x^3-6x^2+mx-n is exactly divisible by x-2
So x-2=0
x=2
f(2)=2^3-6(2)^2+m×2-n=0
=8-24+2m-n=0
=-16+2m-n=0
=2m-n=16 ... (2)
From eq 1&2
2(n+5)-n=16[m=n+5]
2n+10-n=16
n=16-10
n=6
Putting the value of n in eq 1
m=6+5
m=11
Answered by
0
Answer:
Step-by-step explanation:
x=1
f(1)=1^3-6(1)^2+m×1-n=0
=1-6+m-n=0
=-5+m-n=0
=m=n+5 ... (1)
If f(x) =x^3-6x^2+mx-n is exactly divisible by x-2
So x-2=0
x=2
f(2)=2^3-6(2)^2+m×2-n=0
=8-24+2m-n=0
=-16+2m-n=0
=2m-n=16 ... (2)
From eq 1&2
2(n+5)-n=16[m=n+5]
2n+10-n=16
n=16-10
n=6
Putting the value of n in eq 1
m=6+5
m=11
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