How to Find the Zeros Using the Rational Roots Theorem?

In summary: No, there are not.In summary, Elissa found her negative zeros and then ran into a problem with finding the positive zeros. She was able to find them using synthetic division and possible zeros, but from there she is stuck. Elissa found her positive zeros using the same method, but her homework was done using a different program.
  • #1
Elissa89
52
0
I can't find the zeros to

\(\displaystyle 4x^5-10x^4-14x^3+49x^2-28x+4\)

I found my positive zeros, 2, 1/2 using synthetic division and possible zeros. But from there I'm stuck.
 
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  • #2
Elissa89 said:
I can't find the zeros to

\(\displaystyle 4x^5-10x^4-14x^3+49x^2-28x+4\)

I found my positive zeros, 2, 1/2 using synthetic division and possible zeros. But from there I'm stuck.

Hi Elissa89! Welcome to MHB! ;)

Which polynomial did you find after dividing out (x-2) and (x-1/2)?
Does it still have rational roots?
We should find a 2nd order polynomial with no rational roots, which we can solve to find the remaining irrational roots.
 
  • #3
Hello and welcome to MHB, Elissa! (Wave)

You say 2 and 1/2 are zeroes of the given polynomial...so let's use synthetic division to see what we get...

First 2:

\(\displaystyle \begin{array}{c|rr}& 4 & -10 & -14 & 49 & -28 & 4 \\ 2 & & 8 & -4 & -36 & 26 & -4 \\ \hline & 4 & -2 & -18 & 13 & -2 & 0 \end{array}\)

Next 1/2:

\(\displaystyle \begin{array}{c|rr}& 4 & -2 & -18 & 13 & -2 \\ \frac{1}{2} & & 2 & 0 & -9 & 2 \\ \hline & 4 & 0 & -18 & 4 &0\end{array}\)

What is your factorization so far?
 
  • #4
Ok, so from what you did, I divided further and got 4x^2+8x-2, which I applied the quadratic formula and got -2 +/- sqrt{6} all over 2. That's the same answer I've been getting but my homework is done in a program and it keeps telling me it's wrong.

Sorry, I don't know how to use some of the symbol/commands.
 
  • #5
Okay, what I have after doing the divisions indicated in my post above is:

\(\displaystyle f(x)=4x^5-10x^4-14x^3+49x^2-28x+4=(x-2)\left(x-\frac{1}{2}\right)\left(4x^3-18x+4\right)\)

Now, let's factor a 2 from the cubic factor and multiply it with the second linear factor to get:

\(\displaystyle f(x)=(x-2)(2x-1)\left(2x^3-9x+2\right)\)

Now, we see that 2 is a zero of the cubic factor, so we apply synthetic division again:

\(\displaystyle \begin{array}{c|rr}& 2 & 0 & -9 & 2 \\ 2 & & 4 & 8 & -2 \\ \hline & 2 & 4 & -1 & 0 \end{array}\)

And now we have:

\(\displaystyle f(x)=(x-2)^2(2x-1)\left(2x^2+4x-1\right)\)

Can you post your work so we can see why our quadratic factors differ?
 
  • #6
I don't know how to show it on here but I'll do the best I can.

So we have 4x^3-18x+4/ (x-2)

In synthetic division I used 4 0 -18 4 to divide by 2. I got 4x^2+8x-2 with remainder of 0. I applied the quadratic formula to 4x^2+8x-2 and got the answer -2 +/- sqrt{6} all over 2
 
  • #7
Elissa89 said:
I don't know how to show it on here but I'll do the best I can.

That's fine, we don't expect our users to be $\LaTeX$ experts right away. :)

Elissa89 said:
So we have 4x^3-18x+4/ (x-2)

In synthetic division I used 4 0 -18 4 to divide by 2. I got 4x^2+8x-2 with remainder of 0. I applied the quadratic formula to 4x^2+8x-2 and got the answer -2 +/- sqrt{6} all over 2

Sorry, somehow I missed that our quadratic differed only by a constant factor...so you did perform the division correctly. Applying the quadratic formula to my quadratic factor, I ultimately get:

\(\displaystyle x=\frac{-2\pm\sqrt{6}}{2}\)

And this is the same as you found.

Sometimes these online homework apps can be pretty picky about how the answers are input. If I were going to list the 5 roots of the given quintic polynomial using plain text, I would give:

2
2
1/2
(-2 + sqrt(6))/2
(-2 - sqrt(6))/2
 
  • #8
Well at least I know I did it right, thanks!
 
  • #9
Elissa89 said:
Well at least I know I did it right, thanks!

It's possible the app may want the roots in this form:

2
2
1/2
-1 + sqrt(1.5)
-1 - sqrt(1.5)

Or it may want decimal approximations for the irrational roots:

2
2
0.5
0.22474
2.2247

Are there any instructions provided on how to input the roots?
 

FAQ: How to Find the Zeros Using the Rational Roots Theorem?

What is the Rational Roots Theorem?

The Rational Roots Theorem is a mathematical rule that helps to find possible rational roots (or solutions) of a polynomial equation with integer coefficients.

How does the Rational Roots Theorem work?

The theorem states that if a polynomial equation with integer coefficients has a rational root, then that root must be a factor of the constant term (the number without a variable) divided by the leading coefficient (the number with the highest degree variable).

Why is the Rational Roots Theorem useful?

The theorem can save time and effort when trying to find the roots of a polynomial equation. It narrows down the possible rational roots to a finite list, making it easier to test and find the actual roots.

Are there any limitations to the Rational Roots Theorem?

Yes, the theorem only applies to polynomial equations with integer coefficients. It also only helps to find rational roots, not irrational or complex roots.

Can the Rational Roots Theorem be used for all polynomial equations?

No, the theorem can only be used for polynomial equations of degree 2 or higher. It does not apply to linear equations (degree 1) or constant equations (degree 0).

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