Question

Find the number of possible positive and negative real zeros for the function (2x^3)-(x^2)-+5x+2=P(x)

Answer 1

its -5x or +5x

Answer 2

i am actually asking about your question its not the answer btw

Answer 3

THANKS! immmm how'd you get that??

Answer 4

because u have used both the signs

Answer 5

oh sorry it's +5x

Answer 6

first find a root by hit and trial method.

Answer 7

Apply Descarte's Rule of Signs.

Answer 8

I am sorry I didn't understand.... can you please explain it again Sir/Ma'am

Answer 9

This is literally the question it gives me: Find the number of possible positive and negative real zeros for the function (2x^3)-(x^2)+5x+2=P(x)

And I'm a girl.... >:(

Answer 10

P(x) = 2x^3 - x^2 + 5x + 2
Descartes's Rule of Signs:
How many times does the coefficient change signs? +2 to -1; -1 to +5. Twice.
That means there are either 2 positive real roots or zero positive real roots (always goes down in steps of 2).

Answer 11

okay what about the negative roots??

Answer 12

i mean zeroes

Answer 13

P(-x) = 2(-x)^3 - (-x)^2 + 5(-x) + 2
p(-x) = -2x^3 - x^2 - 5x + 2
How many sign changes? Just once.
Therefore, there is definitely one negative real root.

Possibilities:
1 negative real root, 2 positive real root, no complex roots OR
1 negative real root, 0 positive real root, 2 complex roots.

Answer 14

"roots" "zeros" mean the same thing.

Answer 15

thanks so much!

Answer 16

You are welcome.
Since it was a third degree polynomial there will be a total of 3 roots as shown in the answer above.

Answer 17

Thanks so much from My Side too