Question

please give me the correct answer ​

Answer 1

Step-by-step explanation:

first put x=1

(1)³+3(1)²-1-3

1+3-1-3

4-4

=0

so 1 is zeros of cubic polynomial x³+3x²-x-3

put x=2

(2)³+3(2)²-2-3

8+12-5

20-5

15≠0 so 2 is not a zeros of the given cubic polynomial

$put \: x = \frac{1}{2}$

(1/2)³+3(1/2)²-1/2-3

1/8+3/4 -7/2

$\frac{1 + 6}{8} - \frac{7}{2}$

7/8 -7/2≠0

so 1/2 is not a zeros of the given cubic polynomial

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