what are the real zeros of x^4-5x^3+6^2-x-2

Question

i need help with this problem canyou explain please

Answer 1

Hi, Mikia!

 

I assume that the question is asking for the real zeros of x⁴ - 5x³ + 6x² - x - 2 = 0.  Based on that assumption, I would find the real solution values by graphing the equation on your graphing calculator and looking at where the function crosses the x-axis.  To find the exact values of where the function crosses the x-axis, you can either:

1. Look in TABLE for all of the places y=0 and hope all of the real solutions show up there

or

2. Use your calculator's "zero" function

- Push 2nd, CALC [above TRACE], 2:zero

- Answer some questions:

- left bound? [hit ENTER on some part of the function to the left of where it crosses the x-axis - you can move the tracer by using the left or right arrow keys, it will automatically stay on the function until you hit enter after "guess?"]

- right bound? [hit ENTER on some part of the function to the right of where it crosses the x-axis]

- guess? [hit ENTER somewhere in between left bound and right bound]

- Repeat for each zero that you can see on the graph.  You may need to zoom in to be able to isolate each zero.

 

I prefer the "zero" function method, which shows me that there are 2 real zeros for this function.  One is at x=-.4331071, and the other is at x=3.3522244

 

Good luck!

 

Patty

Answer 2

There are two real zeros of this polynomial.   Neither one of them is a rational number (which can be shown with aid of the rational zeros theorem).  A graphing calculator shows the two zeros to be

 

x = -.4331071

 

x = 3.3522244

 

If the last term were -1 instead of -2, there would be a double zero at x = 1.   Long division, followed by factoring, could be used to find the other two zeros.