precalculus

Question

Let p(x) = 2x3- 2x5+x2-2

(a) Use the upper and lower Bounds theorem to show that the real roots (zeros0 of p lie between -1 and 1

(b) Does p(x) have any rational root ? justitf your answer by using the result in (a) and the Rational Zeros Theorem

Answer 1

p(x) has no zero between -1and +1, but does have a root between -2 and -1 at -1.1229 approx. That -2 is a lower bound is shown by using synthetic division:

-2 | -2  0  2  1    0  -2

    |      4 -8 12 -26 52

      -2  4 -6 13 -26 50

The alternating signs show that -2 is a lower bound. At x=0 p(0)=-2 and p(1)=-1. p remains negative as x is more positive. When x<-1.1229 p is positive.

Rational Zeros Theorem has 1 and 2 as the factors of the highest power of x and the constant term. These create the quotients +/-1 +/-1/2 +/-2. We already know that there appears to be but one zero p(-1.1229)=0.