Answer:
c. (x + 3)
Step-by-step explanation:
using factor theorem
if x - 3 is a factor then p(a) = 0
p(a)= x^3 - 3x^2 - 4x + 12
a.(x-3)
p(3) = (3)^3 - 3(3)^2 - 4(3) + 12
= 27 - 27 - 12 + 12
= 0
therefore x-3 is a factor
b.(x + 2)
p(-2) = (-2)^3 - 3(-2)^2 - 4(-2) + 12
= -8 -12 + 8 + 12
,= 0
therefore x + 2 is a factor
c.(x + 3)
p(-3) = (-3)^3 - 3(-3)^2 - 4(-3) + 12
= -27 -27 + 12 + 12
= -30
therefore x + 3 is not a factor
d.(x-2)
p(2) = (2)^3 - 3(2)^2 - 4(2) + 12
= 8 -12 - 8 + 12
= 0
therefore x - 2 is a factor
Given that <span>454 out of 478 adults aged 18 - 29 answered yes, the </span><span>point estimate of the proportion of adults aged 18 – 29 who use the internet</span> is given by
Check the picture below.
now hmmm let's observe, hmmmm 11 has two digits, and low and behold, the 2 is right in the middle, then 1 and 1 on flanking it.
let's check 111 hmmmm has three digits and has a 3 right in the middle as well, and low and behold, is flanked by a regressing count, namely an "outwards" count.
let's check 1111 and 11111, same happens, one has a digit of 4 in the middle and the other 5, and yes, the "outwards" count on each, outwards count towards 1 each time btw.
so, if we were to take that pattern, we can say that eight 1's, namely 11111111 x 11111111, will have an 8 in the middle, and will look like 1234567 8 7654321.
C- is either a or c
Explanation -