P(x)=6x³-11x²+kx-20
X=4by3
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On putting the value of root of a polynomial into that polynomial we get 0
so, 6(4/3)^3-11(4/3)^2+k4/3-20=0
→128/9-176/9+k4/3 -20=0
→-48/9+k4/3 -20=0
→-16/3+k4/3 -20=0
→(4k-16)/3=20
→4k-16=60
→4k=76
→k=19
Answered by
1
On putting the value of root of a polynomial into that polynomial we get 0
so, 6(4/3)^3-11(4/3)^2+k4/3-20=0
→128/9-176/9+k4/3 -20=0
→-48/9+k4/3 -20=0
→-16/3+k4/3 -20=0
→(4k-16)/3=20
→4k-16=60
→4k=76
→k=19
so, 6(4/3)^3-11(4/3)^2+k4/3-20=0
→128/9-176/9+k4/3 -20=0
→-48/9+k4/3 -20=0
→-16/3+k4/3 -20=0
→(4k-16)/3=20
→4k-16=60
→4k=76
→k=19
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