If x =4/3 is a root of the polynomial f(x) = 6x3-11x2+kx-20 find the value of k
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If 4/3 is a root of f (x)
f (4/3)=0
6 (4/3)^3-11 (4/3)^2+k (4/3)-20=0
6 (64/27)-11 (16/9)+4k/3-20=0
128/9-176/9+4k/3-20=0
4k/3=20+16/3=76/3
k=19
f (4/3)=0
6 (4/3)^3-11 (4/3)^2+k (4/3)-20=0
6 (64/27)-11 (16/9)+4k/3-20=0
128/9-176/9+4k/3-20=0
4k/3=20+16/3=76/3
k=19
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