Question

x=4/3 is zero of polynomial f(x) = 6x^3 -11x^2+kx -20 then find the value of k.

Answer 1

$f(x) = 6x {}^{3} - 11x {}^{2} + kx - 20$
If x=4/3 is the zero of polynomial.
Then f(4/3)=0
So,
$6 \times ( \frac{4}{3} ) {}^{3} - 11 \times ( \frac{4}{3} ) {}^{2} + k \times \frac{4}{3} - 20 = 0 \\ 6 \times \frac{64}{27} - 11 \times \frac{16}{9} + k \times \frac{4}{3} = 20 \\ \frac{128}{9} - \frac{176}{9} + \frac{4k}{3} = 20 \\ \frac{128 - 176}{9} + \frac{4k}{3} = 20 \\ \frac{ - 48}{9} + \frac{4k}{3} = 20 \\ \frac{ - 16}{3} + \frac{4k}{3} = 20 \\ \frac{4k}{3} = 20 + \frac{16}{3} \\ \frac{4k}{3} = \frac{60 + 16}{3} \\ \frac{4k}{3} = \frac{76}{3} \\ k = \frac{76 \times 3}{3 \times 4} \\ k = 19$

Answer 2

hi friend

Step-by-step explanation:

here is your answer