Question

The zeros of the polynomial x square - 2x -3​

Answer 1

3 & -1

Step-by-step explanation:

Zeroes of polynomial

x² - 2x - 3

=> x² - 3x + x - 3

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

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

So, zeroes

(i)

x - 3 = 0

=> x = 3

(ii)

x + 1 = 0

=> x = -1

So, zeroes of polynomial are 3 & (-1)

hope it helps.

Answer 2

ANSWER:

Given:

• x^2 - 2x - 3

To Find:

• Zeroes of polynomial.

Solution:

We are given that,

$\implies p(x)=x^2-2x-3$

METHOD 1: MIDDLE TERM SPLITTING

We have,

$\implies x^2-2x-3$

Splitting the middle term,

$\implies x^2-3x+x-3$

Taking x and 1 common,

$\implies x^2-3x+x-3$

$\implies x(x-3)+1(x-3)$

Taking x - 3 common,

$\implies (x-3)(x+1)$

So, zeroes are,

$\implies x-3=0\:\:\&\:\:x+1=0$

Hence,

$\implies\bf x=3\:\:\&\:\:x=-1$

METHOD 2: QUADRATIC FORMULA

We have,

$\implies x^2-2x-3$

We know, using Quadratic Formula,

$\implies x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a = 1, b = -2 and c = -3. So,

$\implies x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(-3)}}{2(1)}$

$\implies x=\dfrac{2\pm\sqrt{4+12}}{2}$

$\implies x=\dfrac{2\pm\sqrt{16}}{2}$

$\implies x=\dfrac{2\pm4}{2}$

So,

$\implies x=\dfrac{2\!\!\!/(1\pm2)}{2\!\!\!/}$

$\implies x=1\pm2$

So, zeroes are,

$\implies x=1+2\:\:\&\:\:x=1-2$

Hence,

$\implies\bf x=3\:\:\&\:\:x=-1$

Therefore, the zeroes are 3 and -1.