Question

please help asapp !!!

Answer 1

Solution:

${x}^{2} - 5 = 0 \\ \\ {x}^{2} = 5 \\ \\ x = + - \sqrt{5} \\ \\ \alpha = - \sqrt{5} \\ \\ \beta = + \sqrt{5}$
So

$1 + \alpha = 1 - \sqrt{5} \\ \\ 1 + \beta = 1 + \sqrt{5}$
sum of zeros

$1 + \alpha + 1 + \beta = \frac{ - b}{a} \\ \\ 1 - \sqrt{5} + 1 + \sqrt{5} = \frac{ - b}{a} \\ \\ 2 = \frac{ - b}{a}$
Multiplication of zeros

$(1 - \sqrt{5} )(1 + \sqrt{5} ) = \frac{c}{a} \\ \\ 1 - 5 = \frac{c}{a} \\ \\ \frac{c}{a} = - 4$
New polynomial

$a {x}^{2} + bx + c \\ \\ {x}^{2} - ( \frac{ - b}{a} )x + \frac{c}{a} \\ \\ {x}^{2} - (2)x - 4 \\ \\ {x}^{2} - 2x - 4$
is the required polynomial