Question

it being given that 2 is one of the zeros of the polynomial (x)=x^3-4x^2+x+6. find it's other zeros​

Answer 1

Step-by-step explanation:

take the help from the pic

hope it helps

Answer 2

Answer:

$x =3 \: and \: x = - 1 \: \: are \: the \: other \: two \: roots$

Step-by-step explanation:

Let,

$f(x) = {x}^{3} - 4 {x}^{2} + x + 6$

$x = 2$

is one of the zeros.

Therefore, (x-2) is a factor of f(x)

$q(x) = \frac{f(x)}{x - 2}$

$q(x) = \frac{ {x}^{3} - 4 {x}^{2} + x + 6 }{x - 2}$

$\frac{ {x}^{3} - 4 {x}^{2} + x + 6 }{x - 2} = {x}^{2} - 2x - 3$

Now we factorise the quotient....

${x}^{2} - 2x - 3 = {x}^{2} - 3x + x - 3$

$= x(x - 3) + 1(x - 3) = (x - 3)(x + 1)$

Therefore,

x=3 and x=-1

are the other two roots....