Question

if 1 is a zero of x3-2x²-9x+18 find the other two zeroes ​

Answer 1

sorry 1 will never be zeroes of this polynomial your questions is wrong man

Answer 2

Given :

$x^3-2x^2-9x+18$ is the required polynomial.

2 is a zero of the above polynomial.

To find:

The other two zeroes = ?

Solution :

Since 2 is a zero of polynomial,

so, $x=2\implies x-2=0$

So, Divide the given polynomial by x-1 :

$\dfrac{x^3-2x^2-9x+18}{x-1}\\\\=\dfrac{x^2(x-2)-9(x-2)}{x-2}\\\\=\dfrac{(x-2)(x^2-9)}{x-2}\\\\=x^2-9$

So, the other two zeroes would be

$x^22-9=x^2-3^2=(x-3)(x+3)$

Hence, the other two zeroes would be $(x-3)(x+3)$