finding zeros of a polynomial function.. Tecnical help about graphing calculator.. chose question.. if u want?

Question

If you can help me understand this.. i basically understand the examples in my book but when it comes to problems I run into factoring problems.
Find the zeros algebraically.
#1 f(x)=2x^4-2x^2-40
Then I also have to analyze it.. in these problems..
#2 f(x)=2x^4-6x^2+1
#3 f(x)=x^5+3x^3-x+6

these have to be done with a graphing utility to approximate the zeros and then check by calculating.. another problem my graphing calculator i s not very accurate.. it is not broken not old.. it just is not very exact.. it jumps when i have it on trace mode.. from like x=0.5 to next is x= -0.8 and i am wanting to find X=0 .. the only way to get to that is to not use the trace but still then it only gives me one decimal place when i should be getting 3..
if you know how to fix that then i will be happy.. but if you don't think it is much of a problem that is ok.. but i keep on getting the wrong answer for the 3 decimal points.

Answer 1

Hi,

Find the zeros algebraically.
#1 f(x)=2x^4-2x^2-40

2x^4-2x^2-40 = 0

Factor out the GCF.

2(x^4- x^2 - 20) = 0
Now factor the trinomial.

2(x² - 5)(x² + 4) = 0

Set each factor with an "x" equal to zero and solve.
x² - 5 = 0
x² = 5._
x = ±√5

x² + 4 = 0
x² = -4...__
√x² = ±√-4
x = ±2i
...._
±√5 and ±2i are the answers


#2 f(x)=2x^4-6x^2+1

On the TI83 I would enter Y1 = 2x^4 - 6x^2 + 1. With the appropriate window I could graph the equation and see a"W"-shaped graph that crosses the x axis 4 times. These 4 points are the zeros.

To find them I would press 2nd, TRACE, ans select 2 Zero, the zero command. For each point I need to pick a left bound , any point to the left of the x intercept, a right bound , any point to the right of the x intercept, and a guess for its value. If my left and right bound only enclose one x intercept, I can just hit Enter for the guess. If there are multiple x intercepts between my left and right bound, I need to have my guess be an "x" value closer to the zero I want to find. These steps will be repeated for each x intercept.

This equation has zeros at -1.68, -.421, .421, and 1.68.

#3 f(x)=x^5+3x^3-x+6
Enter this into the calculator as Y1 = x^5 + 3x^3 - x + 6
Use the zero command again with a left bound of -2, right bound of 0, and a guess of -1, the calculator finds a zero at x = -1.178. The other zeros are all complex numbers with "i" and can not be found from the calculator.

I hope that helps you!!

Answer 2

Once you know what the polynomial as a whole looks like, you can zoom in on each of the zeroes (I presume) replot and get increased accuracy that way.

For #1, the first thing to look for is a common factor, in this case, 2:

2x^4 - 2x^2 - 40
= 2(x^4 - x^2 - 20)

The part still in parentheses is a quadratic in x^2, as might be more readily seen by replacing x^2 with another variable like y:

x^4 - x^2 - 20
= y^2 - y - 20
= (y - 5)(y + 4)
= (x^2 - 5)(x^2 + 4)

After replacing with x again, the two factors can again be factored by the difference of squares. The tricky part is to recognize that the factor on the right is also the difference of squares and what the one on the left is the square of:

= (x^2 - (√5)^2)(x^2 - (2i)^2)
= (x + √5)(x - √5)(x + 2i)(x - 2i)

so f(x) has zeroes at ±√5 and ±2i.

#2, though not quite so tractable, is also a quadratic in x^2. After you have made the substitution y = x^2, you can solve for y with the quadratic formula, getting

y = { 6 ± √[(-6)^2 - 4(2)(1)] } / (2*2)
= 3 ± √6

so x = ±√(3 ± √6)
where the ± symbols are independent of each other, i.e., you are free to take the - side of one symbol and the + side of the other.

I haven't figured out what to do with #3 yet. That buggy +6 at the end bollixes what would otherwise be a polynomial that an x could be factored out of...

Answer 3

#1 f(x)=2x^4-2x^2-40
= 2(x^4-x^2-20) =2(x^2-5)(x^2+4)
So x = +/- sqrt(5) and +/- 2i
You have two real roots and two imaginary roots.

#2 f(x)=2x^4-6x^2+1
Let z = x^2
Then 2z^2 -6z +1
z=[6 +/- sqrt(36-8)]/4 =[3 +/- sqrt(7)]/2
So x^2 =[3 +/- sqrt(7)]/2
Now take sqrt of both sides and you will get 4 answers for x.
The four answers will all be real.
If you graph the curve it will look like a w and be symmetric to the y-axis. Two of the roots will be negative and the other two will be positive but equal in magnitude to the two negative roots. When you are in trace, find a place where the y value changes sign as you cross the x-axis. Then Press Zoom, then 2, then enter.Now press TRACE again and you will zero in ont value of x. Keep repeating this until you hav arrived at the desired accuracy.

#3 f(x)=x^5+3x^3-x+6
Graphing shows that this equation has 1 real root and 4 imaginary roots. The real root lies between -1.1783 and -1.1781. I had to ZOOM six times to get that accuracy.
So -1.178 is the answer to the required accuracy.
I don't think there is a way to get an exact answer algebraically.

Answer 4

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RE:
finding zeros of a polynomial function.. Tecnical help about graphing calculator.. chose question.. if u want?
If you can help me understand this.. i basically understand the examples in my book but when it comes to problems I run into factoring problems.
Find the zeros algebraically.
#1 f(x)=2x^4-2x^2-40
Then I also have to analyze it.. in these problems..
#2 f(x)=2x^4-6x^2+1
#3...

Answer 5

3x^4-5x^3-4x^2+5x-2