Question

What are the possible number of positive, negative, and complex zeros of f(x) = 2x3 – 5x2 – 6x + 4 ?

Answer 1

Apply Descartes' Rule of Signs.

Answer 2

i Know that, but I'm confused on whether I got it right or not? @ranga

Answer 3

And what did you get?

Answer 4

I got 1 positive 1 negative and I'm not sure about the complex zeros

Answer 5

f(x) = 2x^3 – 5x^2 – 6x + 4
Coefficients are: +2 -5 -6 +4
How many times does the coefficient change signs?

Answer 6

twice

Answer 7

That would be 2 positives?

Answer 8

It changes twice is correct.
That means the number of possible positive roots are: 2 or 0.
(you have to decrease by 2 each time until it reaches 0 or 1. If it changes sign 6 times then possible positive roots are: 6, 4, 2, 0.)

Answer 9

Ok that's understanding, so negative would be 1 or 0?

Answer 10

For negative roots, first replace x by -x in f(x):
f(x) = 2x^3 – 5x^2 – 6x + 4
f(-x) = 2(-x)^3 - 5(-x)^2 - 6(-x) + 4
f(-x) = -2x^3 - 5x^2 + 6x + 4
Coefficients: -2 -5 +6 +4
How many times does the coefficient change signs?

Answer 11

Once , 1 or 0 negative roots?

Answer 12

It changes sign once is correct.
The number of possible negative roots is 1.
(you cannot decrease the number of roots by 2 anymore because you have to stop when it reaches 0 or 1. here it is already 1 and so it stops.)

So there is no ambiguity here. This function definitely has ONE negative root.
But with positive it can have 2 or 0 roots.

Answer 13

wait when you substituted x by -x doesn't two negatives change to a positive?

Answer 14

f(x) = 2x^3 – 5x^2 – 6x + 4
f(-x) = 2(-x)^3 - 5(-x)^2 - 6(-x) + 4
f(-x) = -2x^3 - 5x^2 + 6x + 4
Coefficients: -2 -5 +6 +4

Which step you have a question?

Answer 15

For the 2nd and 3rd, if it's -5(-x)^2 wouldn't the coefficient be a positive 5

Answer 16

The -5 is not squared. Only the x is replaced by -x. So x^2 becomes (-x)^2 = +x^2
-5(x^2)
replace x by -x
-5(-x)^2 = -5 * x^2 = -5x^2

Answer 17

so it makes no difference that's why 5 is still a negative because -6(-x) became a positive 6

Answer 18

f(x) = 2x^3 – 5x^2 – 6x + 4
Do it term by term.
First term is 2x^3. Replace x by -x: 2(-x)^3 = -2x^3
Second term: -5x^2. Replace x by -x: -5(-x)^2 = -5x^2
Third term: -6x. Replace x by -x: -6(-x) = +6x
Fourth term: +4

Answer 19

Okay I got it. Every other term? and what about complex zeros?

Answer 20

When you replace x by -x, only the ODD powered x will change the sign of the coefficient.
The EVEN power will retain the same sign.

Answer 21

This is a third degree polynomial and so it should have a total of 3 roots.
We already know that it definitely has one negative root.
Positive root can be 2 or 0.
But in any case the total number of roots should always add up to 3.

List the possibilities:

Negative Positive Complex Total
1 2 - 3
1 0 2 3

Those are the two possibilities: a) 1 negative, 2 positive or b) 1 negative, 2 complex.

Answer 22

Okay so 2 or 0 complex. correct?

Answer 23

correct.
1 negative, 2 or 0 positive, 0 or 2 complex.

Answer 24

Thank you so much ( :

Answer 25

You are welcome.