Question

f(x) = x3 + 5x2 - 28x - 32

1 imaginary; 2 real

2 imaginary; 1 real

0 imaginary; 3 real

3 imaginary; 0 real

Answer 1

I got d

Answer 2

hint : Any cubic polynomial will have atleast 1 real root

Answer 3

Ive already found my answer is that right

Answer 4

hint2 : imaginary roots always come in pairs

Answer 5

did u take time to read my first hint

Answer 6

yes

Answer 7

ohhhh

Answer 8

:)

Answer 9

Having knowledge of those 2 hints, you can strike off few options. wat wud be they ?

Answer 10

a and d

Answer 11

yes answer is either b or c

Answer 12

you familiar wid descarte's rule of signs ?

Answer 13

um I dont think I know it by name, but if you told it to me I might
idk.. hahaha

Answer 14

f(x) = x3 + 5x2 - 28x - 32

Answer 15

count the sign changes

Answer 16

two

Answer 17

try again

Answer 18

one

Answer 19

yes, so 1 positive real root exists.

Answer 20

Next, take f(-x)

Answer 21

AYYYY thanks bro! its gotta be b

Answer 22

hey we're not finished yet, one more step is there to conlcude fully

Answer 23

oh ok

Answer 24

you need to count sign changes in f(-x) also

Answer 25

it 2

Answer 26

f(x) = x3 + 5x2 - 28x - 32

f(-x) = -x3 + 5x2 + 28x - 32

Answer 27

thank you so much

Answer 28

it has 2 sign changes. so it can have 2 or 0 negative real roots

Answer 29

answer is C