1. ## Please solve using Descarte's Rule of Signs

Given f(x) = 2x^4-3x^3-10x^2+12x+8
1. Use Descarte's Rule of Signs to determine the possible number of negative and positive real roots and the number of complex roots. ( I do not know how to do this)
2. Use the rational roots theorem to find all possible rational roots (zeros)
+1, -1, +2, -2, +4, -4, +8, -8, +1/2, -1/2 (Is this correct? I found by ratio p/q)
3. Find all real and complex roots. (Real roots found by graphing -2, 2, -1/2. How to find complex roots?I factored the function: f(x)= (x-2) (x+2) (2x^2-3x-2). Then I solved the quadratic part by using quadratic formula: x=2 and x=-1/2. The initial function can be factored: (x-2)^2 (x+2) (x+1/2). Is this correct? How to find complex zeros? Does it have complex zeros?

Thank you very much for your help.

2. Originally Posted by oceanmd
Given f(x) = 2x^4-3x^3-10x^2+12x+8
1. Use Descarte's Rule of Signs to determine the possible number of negative and positive real roots and the number of complex roots. ( I do not know how to do this)
2. Use the rational roots theorem to find all possible rational roots (zeros)
+1, -1, +2, -2, +4, -4, +8, -8, +1/2, -1/2 (Is this correct? I found by ratio p/q)
3. Find all real and complex roots. (Real roots found by graphing -2, 2, -1/2. How to find complex roots?I factored the function: f(x)= (x-2) (x+2) (2x^2-3x-2). Then I solved the quadratic part by using quadratic formula: x=2 and x=-1/2. The initial function can be factored: (x-2)^2 (x+2) (x+1/2). Is this correct? How to find complex zeros? Does it have complex zeros?

Thank you very much for your help.
Descartes' Rule of sign..

$f(x) = 2x^4-3x^3-10x^2+12x+8$

since the sign changed from negative to positive and positive to negative twice, then there are 2 or zero positive real roots.

notice that if x is replaced with -x, the function becomes
$f(-x) = 2x^4+3x^3-10x^2-12x+8$

and notice also that there are two changes of sign, which implies that there 2 or no negative real roots..

the complex roots depends on the exixtence of positive and negative real roots. if there is no positive real root but 2 negatve real roots, then there must be 2 complex roots. also if there is no negative real root but 2 positive real roots, then there must be 2 complex roots. however, if there are no negative and positive real roots, then there must be 4 complex roots.

for your last question, there is no complex roots (assuming you got the right real roots) Ü note that if a function is of degree n (n is positive integer), then the function have exactly n roots in the system of Complex Numbers (of the form of a + bi). (remark: for real roots, the b there is 0). (Fundamental Theorem of Algebra)

3. ## Complex Zeros

Kalagota,
thank you very much for the explanation.
Did I find the rational possible roots correctly?
May I answer that complex zeros are 2, 2, -2, -1/2 according to a + bi and that b=0?
Did I find the real zeros correctly?
Thank you again.

4. Originally Posted by oceanmd
Kalagota,
thank you very much for the explanation.
Did I find the rational possible roots correctly?
May I answer that complex zeros are 2, 2, -2, -1/2 according to a + bi and that b=0?
Did I find the real zeros correctly?
Thank you again.
technically, they are complex zeros since the set of Real Numbers is just a subset of the set of Complex numbers. but since we always have the remark that complex roots must be in the form of a + bi, where b is not 0. so, if you got no "bi" part, just call it real roots. Ü