Question

find all the zeroes of x3-3x2-x+3 if one of its zero is 1

Answer 1

x^3 - 3x^2 - x + 3 = 0
x^2(x - 3) - 1(x - 3) = 0
x - 3(x^2 - 1) = 0
(x - 3)(x+1)(x - 1) = 0
Zeroes of the Polynomial are 3 , 1 , -1

Answer 2

The remaining zeroes are -1 and 3.

Step-by-step explanation:

The given polynomial is

$P(x)=x^3-3x^2-x+3$

We need to find the zeroes of the given polynomial.

Equate the polynomial equal to 0.

$P(x)=0$

$x^3-3x^2-x+3=0$

$(x^3-3x^2)+(-x+3)=0$

Taking common factors from each parenthesis.

$x^2(x-3)-(x-3)=0$

$(x^2-1)(x-3)=0$

$(x-1)(x+1)(x-3)=0$              $\because a^2-b^2=(a-b)(a+b)$

Using zero product property we get

$x-1=0\Rightarrow x=1$

$x+1=0\Rightarrow x=-1$

$x-3=0\Rightarrow x=3$

It is given that one of its zero is 1.

Therefore, the remaining zeroes are -1 and 3.

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