Find all zeros of the function. Show all of your work - synthetic division, factoring, quadratic formula, etc. f(x)=3x^3-5x^2+5x-14

Question

The leading coefficient of 3 has an exponent of 3 so shouldn't there be 3 roots? Yet the only root I found was x=2. I am confused.

Answer 1

f(x)=3x³-5x²+5x-14

0=3x³-5x²+5x-14

0=3x³+x²+7x-6x²-2x-14

0=x(3x²+x+7)-2(3x²+x+7)

0=(x-2)(3x²+x+7)

First root is x=2, second root can be found with quadratic formula:

x = [-1±√(1²-4(3)(7))]/[2(3)]

x = [-1±√(1-84)]/6

x = [-1±√-83]/6

x = [-1±i√83]/6 <-- These are the two remaining roots

Only one root is real while the other two are complex. This is because by Descartes' Rules of Signs, there are 3 sign changes for the positive-root case. This means there are either 3 or 1 positive real roots. If we look at the negative root case, there are 2 sign changes, which means there are either 2 or 0 real negative roots.

If we assumed there were 3 real positive roots and 0 real negative roots, then there would be no imaginary roots. If there were 1 real positive root and 2 real negative roots, then there would be no imaginary roots either. There's also the case where there is 1 real positive roots and 0 real negative roots, which would leave 2 imaginary numbers. Since we found x=2 as a positive root, and the discriminant of 3x²+x-7 is negative, then there has to be 2 complex roots in addition to the positive root.

Hope this helped!

Answer 2

Divide the polynomial by x - 2 the suppresed quadratic is 3x² + x + 7.

Use the Qudratic Formula to determine the othe two roots.

Answer 3

3x^3 -5x^2 +5x -14 = 0

3(8) -5(4) +5(2) -14

= 24-20+10-14

= 4-4=0

2 is one zero

x=2

x-2 = 0

divide x-2 into the cubic to get a quadratic, then factor or use the quadratic formula to find 2 other zeros

by des cartes' sign rule there are no negative real zeros and either 1 or 3 positive real zeros

if only 1 real, then the other two solutions are imaginary

3x^2 +x + 7

x = -1/6 +/-(1/6)sqr(1 -4(3)(7))

the other two zeros are imaginary since the discriminant <0, b^2-4ac<0, 1-4(3)(7)<0

there are 3 roots, but only 1 is real, 2 are imaginary

if you have a graphing calculator, graph the cubic and it only intersects the x axis once

the real roots or zeros are the same as the x intercepts

and when you graph the quadratic, it never intersects the x axis