Question

zeros of p(X)=x²-2x-3 are​

Answer 1

Answer:

$x^{2} - 2x - 3\\x^{2} -3x + x - 3\\x(x-3) +1(x-3)\\(x+1)(x-3)\\x = -1,3$

Answer 2

Step-by-step explanation:

Given: The expression $p(x)=x^{2} -2x-3$

To Find: The zeroes of the given expression

We know, to find the zeroes of the expression, we equate the expression with 0.

⇒ $x^{2} -2x-3 = 0$

Solving the expression by splitting the middle term (finding factors of 3 such that they add up to 2)

⇒  $x^{2} +x-3x-3=0$

⇒ $x(x+1) -3(x+1) = 0$

⇒$(x+1)(x-3) = 0$

Equating each term with 0 separately,

⇒ $x+1= 0$

⇒ $x=-1$

⇒ $x-3= 0$

⇒ $x=3$

Hence, the zeroes of the given expression $p(x)$ are $x=-1,3$.