Another Pre-Cal question

[Shaded]

Hey everyone,
Thanks again to the people that helped me solve my last question a couple weeks ago. I took my test on friday and I feel confident, however there was a bonus question that stumped me a tad. I wanted to ask it here now to see how it was properly done. I have it partly figured out, but couldn't get the last part.

Find all the real zeros of P(x)=x^5-4x^4-x^3+10x^2+2x-4

It says find the real zeros and I presume it's only the zeros that are real numbers, however I am a tad confused as I know there are 4 zeros to this as it is a fifth degree polynomial. Two of the zeros I have found, but I am having trouble finding the other two.

I know that because of the Rational zeros Theorem (p/q where P is a factor of the constant coefficient and q is a factor of the leading coefficient) that the possible zeros are ± 1, ± 2, and ± 4. But substituting these values in I try to see which ones give me zero.

After doing that I determined that two of the zeros are -1 and 2.

Now is where I get confused.

Finding the other two zeros I tried to factor the polynomial with synthetic division with the two known zeros in an attempt to factor it down to a 2nd degree polynomial than use the quadratic formula. (If this gets sloppy please forgive me I'll try my best though)

1 │ 1 -4 -1 10 2 -4
1 -3 -4 6 4
_________________
1 -3 -4 6 4 0
So after factoring out the first zero of -1 I now have (x-1)(x^4-3x^3-4x^2+6x+4)
Now I try to factor out the zero at 2

-2│1 -3 -4 6 4
-2 10 -12 12
______________
1 -5 6 -6 14
This is where I'm stuck. After doing synthetic division I am left with a remainder which shouldn't happen when dividing by a zero. Am I doing something wrong? Or did I go wrong somewhere else. The last two zeros of this polynomial completely elude me. Any help with this would be highly appreciated. Thanks again.

*Edit*, I'm having a very hard time trying to get the numbers for the synthetic division I did to line up properly. Trying to work on it though.

 

[Deleted member 4993]

Hey everyone,
Thanks again to the people that helped me solve my last question a couple weeks ago. I took my test on friday and I feel confident, however there was a bonus question that stumped me a tad. I wanted to ask it here now to see how it was properly done. I have it partly figured out, but couldn't get the last part.

Find all the real zeros of P(x)=x^5-4x^4-x^3+10x^2+2x-4

It says find the real zeros and I presume it's only the zeros that are real numbers, however I am a tad confused as I know there are 4 zeros to this as it is a fifth degree polynomial. Two of the zeros I have found, but I am having trouble finding the other two.

I know that because of the Rational zeros Theorem (p/q where P is a factor of the constant coefficient and q is a factor of the leading coefficient) that the possible zeros are ± 1, ± 2, and ± 4. But substituting these values in I try to see which ones give me zero.

After doing that I determined that two of the zeros are -1 and 2.

Now is where I get confused.

Finding the other two zeros I tried to factor the polynomial with synthetic division with the two known zeros in an attempt to factor it down to a 2nd degree polynomial than use the quadratic formula. (If this gets sloppy please forgive me I'll try my best though)

1 │ 1 -4 -1 10 2 -4
1 -3 -4 6 4
_________________
1 -3 -4 6 4 0
So after factoring out the first zero of -1 I now have (x-1)(x^4 - 5x^3 + 4x^2 + 6x - 4) ..... correction
Now I try to factor out the zero at 2

-2│1 -3 -4 6 4
-2 10 -12 12
______________
1 -5 6 -6 14
This is where I'm stuck. After doing synthetic division I am left with a remainder which shouldn't happen when dividing by a zero. Am I doing something wrong? Or did I go wrong somewhere else. The last two zeros of this polynomial completely elude me. Any help with this would be highly appreciated. Thanks again.

*Edit*, I'm having a very hard time trying to get the numbers for the synthetic division I did to line up properly. Trying to work on it though.

x⁵ -4x⁴ -x³ + 10x² + 2x - 4 = (x+1)(x-2)(x³ - 3x² - 2x + 2) = (x + 1)(x - 2)(x + 1)(x² - 4x + 2)

The other two roots are real but irrational.

 

[Shaded]

That much I know. I'm having trouble finding those irrational zeros. I already know that the roots are 2+√2 and 2-√2


However finding out those answers is what I'm having trouble with.

*Edit*

I didn't see your correction. Lemme work it out now

 

[Shaded]

x⁵ -4x⁴ -x³ + 10x² + 2x - 4 = (x+1)(x-2)(x³ - 3x² - 2x + 2) = (x + 1)(x - 2)(x + 1)(x² - 4x + 2)

The other two roots are real but irrational.

Ok after applying your correction I figured out where I went wrong initially. I divided by a positive 1 with instead of a negative which threw my division off. After making that adjustment the answer came out right and than with the 2 I got the first answer you got which was:
(x+1)(x-2)(x³-3x²-2x+2)

Now this might sound just a bit like I should know this (I feel I should), however I'm having a hard time understanding why we were able to factor another (x+1) out. Once we factor it out I can easily apply the quadratic formula and get the remaining zeros, however for some reason I'm hung up the reasoning for factoring another (x+1) out. Beyond that your answer made sense and I was easily able to figure out where my mistake was.

 

[stapel]

I'm having a hard time understanding why we were able to factor another (x+1) out.

Did you cheat and look at the graph? If you did (or do now), I'll bet the graph touches the x-axis at x = -1, and then reverses back the way it came. Whenever the graph just touches like that, you've got two (or four, or six, or...) same-value roots (and thus same-expression factors).

 

[Shaded]

Did you cheat and look at the graph? If you did (or do now), I'll bet the graph touches the x-axis at x = -1, and then reverses back the way it came. Whenever the graph just touches like that, you've got two (or four, or six, or...) same-value roots (and thus same-expression factors).

haha, well I submitted what I knew, unfortunately on the test even though I could peak at the graph on my calculator, my instructor isn't going to give me credit unless I can justify my answer, which is how it should be. So since what I did know is already turned in, I just wanted to know where I went wrong and the reasoning behind the irrational roots. Much appreciated to you guys that are posting here. Everyone here is extremely helpful and friendly.
I appreciate your explanation!

 

[stapel]

...on the test even though I could peak at the graph on my calculator, my instructor isn't going to give me credit unless I can justify my answer....

You'll peek at the calculator, see that you need to divide again by one of the roots you've already done, and then you'll justify your answer by showing that repeated division. You just won't mention anything about the peeking you did.