Question

if x=4/3 is root of the polynomial
${6x}^{3} - {11}^{2} + kx - 20$
find the value of k.
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Answer 1

6 x^3 - 11 x^2 + kx -20 = 0

put x= 4/3

6 ( 4/3)^ 3 - 11( 4/3)^2 + k(4/3) -20=0

6× 64/27 - 11× 16/9 + 4 k/3 - 20 = 0

128/9 - 176/9 + 4 k/3 - 20= 0

128 - 176 + 12k -180 = 0

-48 + 12k -180 = 0

12k = 228

k = 228/12.

= 19