Using Descartes' Rule of signs, what are the roots of 6x^3 + 5x^2 - 6x + 1 = 0

Question

I managed to find that the real positive roots: 2 or 0 & real negative roots: exactly 1.

 

I can't figure out what number you substitute for x to get it to equal 0. 

 

Algebra 2.

 

Thanks.

Answer 1

possible + roots are 1/2, 1/3, 1/6

x=1/2 works, it's a root

divide x-1/2 into the cubic to get a quadratic which you can easily solve for the other 2 roots

1/2 | 6 5 -6 1

3 4 -1

6 8 -2 0

quotient = 6x^2+8x-2

3x^2+4x-1=0

x= -4/6 +/-(1/6)sqr(16+12)=0

x=-2/3+/-1/3sqr7= about -1.55 or 0.22= the other two roots

another method is use a graphing calculator and see where the cubic curve intersects the x axis

such as desmos

or use an on line equation solver