How do you find the zeroes of $f (x) = x^{3} - 17 x^{2} + 81 x - 65$ $f(x)= x^3 - 17x^2 + 81x -65$ ?

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Answer 1 · 2018-03-03T12:43:45

Zeros are $x = 1, x = 8 - 1 i and x = 8 + 1 i$ $x=1 , x= 8-1i and x= 8+1i$

$f (x) = x^{3} - 17 x^{2} + 81 x - 65$ $f(x)=x^3-17x^2+81x-65$ or

$f (x) = x^{3} - x^{2} - 16 x^{2} + 16 x + 65 x - 65$ $f(x)=x^3-x^2-16x^2+16x+65x-65$ or

$f (x) = x^{2} (x - 1) - 16 x (x - 1) + 65 (x - 1)$ $f(x)=x^2(x-1)-16x(x-1)+65(x-1)$ or

$f(x)=(x-1)(x^2-16x+65) :. x=1$ is one zero

$f(x)= x^2-16x+65$ or

$f(x)= x^2-16x+64 +1$ or

$f(x)= (x-8)^2 - (i^2) [i^2=-1]$ or

$f(x)= (x-8+ i)(x-8-i) :.$ . Other zeros are

$x= 8-1i and x= 8+1i$

Zeros are $x=1 , x= 8-1i and x= 8+1i$ [Ans]

How do you find the zeroes of f(x)= x^3 - 17x^2 + 81x -65f(x)=x3−17x2+81x−65 f(x)= x^3 - 17x^2 + 81x -65?