Question

Let p(x) = x2 – 3x – 18.
(a) Find the sum of the zeroes of p(x).
(b) Find the product of the zeroes of p(x).

Answer 1

Answer:

x2-3x-18

Step-by-step explanation:

x2+3x-6x-18

x(x+3) -6(x+3)

(x-6) (x+3)

x=6 , x= -3

sum of zeroes

6+(-3)

6-3

3

product of zeroes =6×(-3)

=. -18

Answer 2

Step-by-step explanation:

x^2-3x-18=0

${x}^{2} - 3x - 18 \\ x {}^{2} - 6x + 3x - 18 \\ x(x - 6)3(x - 6) \\ (x - 6)(x + 3) \\ x - 6 = 0 \: \: \: x = 6 \\ x + 3 = 0 \: \: \: x = - 3 \\ a = 1 \: \: \: \: b = - 3 \: \: \: c = - 18(from \: given \: equation) \\ sum \: of \: zeros \: = \alpha + \beta = - b \div a \\ 6 - 3 = - ( - 3) \div 1 \\ 3 = 3 \\ product \: of \: zeros \: = \alpha \beta = c \div a \\ 6( - 3) = - 18 \div 1 \\ - 18 = - 18$