SOLUTION: The nature of the roots of the equation 3x^4 + 4x^3 + x - 1 = 0 is _____. A. Three positive real roots and one negative real root. B. Three negative real roots and one positive

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The nature of the roots of the equation 3x^4 + 4x^3 + x - 1 = 0 is _____. A. Three positive real roots and one negative real root. B. Three negative real roots and one positive      Log On


   

Question 1182855: The nature of the roots of the equation 3x^4 + 4x^3 + x - 1 = 0 is _____.
A. Three positive real roots and one negative real root.
B. Three negative real roots and one positive real root.
C. One negative real root and three complex roots.
D. One positive real root, one negative real root and two complex roots.

--> Please help me on answering the correct nature of the roots on the equation and correct steps of doing it. Thank you so much!
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The nature of the roots of the equation
3x%5E4+%2B+4x%5E3+%2B+x+-+1+=+0+
observe sign changes
signs are:
3x%5E4+%2B+4x%5E3+%2B+x+-+1+
+ + + -
=> number of sign changes is 1
it means => number of positive roots is 1
now replace x with -x and observe sign changes
3%28-x%29%5E4+%2B+4%28-x%29%5E3+%2B+%28-x%29+-+1+
3x%5E4+-4x%5E3+-x+-+1
+ - - -
=> number of sign changes is 1
=> number of negative roots of original polynomial is+1

so,
=> number of positive roots is 1
=> number of negative roots of original polynomial is 1 and
=> number of imaginary roots of original polynomial is:
n-%28p%2Bq%29 .......where n is number of degree,+p is number of positive roots, q is number of negative roots
n=4,p=1,q=1
n-%28p%2Bq%29+=4-%281%2B1%29=2
=> number of imaginary roots is 2

then answer is
D. One positive real root, one negative real root and two complex roots


Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Use Descartes' Rule of Signs.

(1) Put the polynomial in standard form -- that is, with decreasing powers of x.
(2) The possible numbers of positive real roots are the number of sign changes in the terms of the polynomial, minus any multiple of 2.
(3) The possible numbers of negative real roots are the number of sign changes in the terms of the polynomial when x is replaced by -x, minus any multiple of 2.

For the polynomial in your problem....

3x^4+4x^3+x-1
The signs of the terms are +, +, +, -
There is 1 sign change
The only possible number of positive real roots is 1

Replace x with -x in the polynomial; note that the signs of the coefficients of the even degree terms stay the same while the signs of the coefficients of the odd degree terms change.
3x^4-4x^3-x-1
The signs are +, -, -, -
There is 1 sign change
The only possible number of negative real roots is 1

The polynomial is degree 4, so it has 4 roots. There is only 1 positive real root and only 1 negative real root, so

ANSWER: D. The polynomial has one positive real root, one negative real root, and one pair of complex roots.

For a more interesting example of applying Descartes' Rule of Signs, consider this random polynomial:

4x^8+3x^7-5x^6-3x^5-2x^3+x^2+3x-1

The signs of the terms are +, +, -, -, -, +, +, -
There are 3 sign changes
The number of positive real roots is either 3 or 1

Replace x with -x:
4x^8-3x^7-5x^6+3x^5+2x^3+x^2-3x-1
The signs of the terms are +, -, -, +, +, +, -, -
There are 3 sign changes
The number of negative real roots is either 3 or 1

The nature of the roots is...
either 1 or 3 positive real roots
either 1 or 3 negative real roots

The total number of real roots is then either 2, or 4, or 6; Since there are 8 roots, there are...
either 6, or 4, or 2 complex roots