Question

what are the possible numbers of positive negative and complex zeros of f(x)= -2x^3 - 5x^2 +6x +4

what method used?

Answer 1

Doesn't say rational does it?

Answer 2

nope :o

Answer 3

Well, its been a while since I've done these, and I'm having trouble finding the factor for it. But according to wolfram alpha there is a simplified version of the equation:

-(1+2 x) (-4+2 x+x^2)

from here you can set each part to 0 in order to solve.

http://www.wolframalpha.com/input/?i=-%281%2B2+x%29+%28-4%2B2+x%2Bx%5E2%29&lk=1&a=ClashPrefs_*Math-

Answer 4

First try the Rational roots theorem. By the theorem, the potential rational roots could be \(\bf x= \pm 4, \pm 2, \pm 1, \pm \frac{1}{2} \).

Checking for \(\bf x= -\frac{1}{2}\), we notice that it's a zero hence f(x) can be re-written as:\[\bf f(x)=\left( x+\frac{ 1 }{ 2 } \right)R(x)\]Where R(x) is some quadratic. To find this quadratic, we have to do long division. So divide our expanded form of f(x), which is given to us, by one of the factors x + 1/2 to get R(x):\[\bf R(x)=\frac{ -2x^3-5x^2+6x+4 }{ \left( x+\frac{ 1 }{ 2 } \right) }=-2x^2-4x+8=-2(x^2+2x-4)\]So now we can re-write f(x) as: \[\bf f(x)=\left( x+\frac{ 1 }{ 2 } \right)R(x)=-2\left( x+\frac{ 1 }{ 2 } \right)(x^2+2x-4)\]Notice here we can't factor the quadratic. We know that one of the roots of f(x) is x = -1/2, to find the other two roots of f(x) given by the quadratic, we will use the quadratic formula:\[\bf x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a } \rightarrow x = \frac{ -2 \pm \sqrt{4-4(1)(-4)} }{ 2 }=\frac{ -2 \pm \sqrt{20} }{ 2 }\]The quadratic formula gives no complex roots for the quadratic hence f(x) has no complex zeroes.

Therefore, the zeroes of f(x) are:\[\therefore \ \bf x= -\frac{ 1 }{ 2 }, -1+\sqrt5 ,-1- \sqrt5\]
@FlyinSolo_424

Answer 5

so are they all three negative zeros? @genius12

Answer 6

We know x = -1/2 is definitely negative. To figure out if the rest are negative or not, punch them in to your calculator. This is what I get when I punch in to my calculator:\[\bf -1+ \sqrt5 \approx 1.237\]\[\bf -1 - \sqrt5 \approx -3.236\]So 2 zeroes are negative and 1 is positive.

@FlyinSolo_424

Answer 7

THANK YOU SO MUCH

Answer 8

Thanks for the reminder genius, its been a while since I've had to use the rational roots theorem.