Question

2 Find
the
zeroes
of
x² - 2x -3​

Answer 1

Step-by-step explanation:

${x}^{2} - 2x - 3 \\ {x}^{2} - 3x + 1x - 3 \\ x(x - 3) + 1(x - 3) \\ (x + 1)(x - 3) \\ zeroes \: \\ x + 1 = 0 \\ x = - 1 \\ x - 3 = 0 \\ x = - 3$

Answer 2

Answer:

-1 and +3

Step-by-step explanation:

given polynomial ;

$x {}^{2} - 2x - 3$

By splitting the middle term,

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

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

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

$if \: ab = 0 \: then \: either \: a = 0 \: or \: b = 0$

$(x + 1) = 0 \\ x = - 1$

Now

$(x - 3) = 0 \\ x = + 3$

Thus zeroes are -1 and +3