How to Find the Zeros Using the Rational Roots Theorem?

[Elissa89]

I can't find the zeros to

$$4x^5-10x^4-14x^3+49x^2-28x+4$$

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

 

[I like Serena]

I can't find the zeros to

$$4x^5-10x^4-14x^3+49x^2-28x+4$$

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.

 

[MarkFL]

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:

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

Next 1/2:

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

What is your factorization so far?

 

[Elissa89]

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.

 

[MarkFL]

Okay, what I have after doing the divisions indicated in my post above is:

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

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

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

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

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

And now we have:

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

Can you post your work so we can see why our quadratic factors differ?

 

[Elissa89]

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

 

[MarkFL]

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. :)

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:

$$x=\frac{-2\pm\sqrt{6}}{2}$$

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

 

[Elissa89]

Well at least I know I did it right, thanks!

 

[MarkFL]

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?