Zeros of p(x) = x2 − 2x − 3 are
Question:

Zeros of $p(x)=x^{2}-2 x-3$ are

(a) 1, −3
(b) 3, −1
(c) −3, −1
(d) 1, 3

 

Solution:

(b) 3,-1

Here, $\mathrm{p}(\mathrm{x})=x^{2}-2 x-3$

Let $x^{2}-2 x-3=0$

$=>x^{2}-(3-1) x-3=0$

$=>x^{2}-3 x+x-3=0$

$=>x(x-3)+1(x-3)=0$

$=>(x-3)(x+1)=0$

$=>x=3,-1$

 

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