College Algebra

Question

Use the rational zeros theorem to list all possible zeros of the function f(x)=2x3+4x2+x+7f(x)=2x3+4x2+x+7.

Enter the possible zeros separated by commas. You do not need to factor the polynomial.

Answer 1

P = +or- {1,7}

Q = +or- {1,2}

P/Q = +or- {1,1/2,7,7/2}

Courtesy of MS Excel

f(1) = 14

f(-1) = 8

f(1/2) = 8.75

f(-1/2) = 7.25

f(7) = 896

f(-7) = -490

f(7/2) = 145.25

f(-7/2) = -33.25

So there are NO RATIONAL SOLUTIONS.

Note that all of the coefficients are positive,

which means that any solutions MUST be negative!

Moreover, DeCartes rule of signs says there are no

positive solutions. Therefore, there are only

negative real solutions and complex solutions.

The real solution must be found numerically if

accuracy is important. However, by examining the

graph, the solution is ABOUT -2.4

Bisection Method says the approximate solution

is about -2.39951085035778794 correct to 15 digits

The other two solutions are complex,imaginary