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Finding zeros of polynomials -HELP!

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I have an attachment showing the question - #10. Please hep me solve this. 

What I know is that p(x) = 0, and that I have to try and factor the polynomial - but I'm stuck on that part. Please help?

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[For convenience, I'm going to write everything with x instead of z]

Ew this problem is pretty gross and requires some obscure knowledge from Algebra II. It's called the rational root theorem. Let's say you had a polynomial P(x)=qx^n+...+p, so q is the coefficient of the leading term and is the constant term. The rational root theorem states that all of the positive and negative factors of divided by all of the positive and negative factors of are possible zeroes, and you need to check them by synthetic division (hooray!). In #10, q=2 and p=-6. The factors of are +/- 1,2 and the factors of are +/- 1,2,3,6. Dividing each of these gives the possible zeroes to be: +/- 1/1,1/2,2/1,2/2,3/1,3/2,6/1,6/2 => +/- 1/2,1,3/2,2,3. 

Now you need to start using synthetic division. I usually start with +/- 1, just because it's easy to work with. 

1| 2 -5  13  -4 -6

      2  -3   10  6 

   2 -3  10   6  0

Since the synthetic division gives a remainder of 0 (the last value below the line), x-1 is a factor. You might be looking at the synthetic division and saying: "What on Earth is going on?" Basically, you're trying to divide P(x) by x-a, and show that P(a)=0 (if the remainder is 0, then this is true). So you set x-a=0 and solve for x, so x=a, and write this in the little box at the top left.. Then you write out all of the coefficients (including zeroes if a term is not included, so if I had 2x^2+1, I would write 2 0 1) at the top. Then you drop the first coefficient below the line. Next, multiply that number by (here a=1) and write it below the next term (-5). Then add the two, and write the result below the line (-5+2=-3). Repeat the process until you reach the end.

What comes out is another polynomial, but with (x-a) factored out. This new one is 2x^3-3x^2+10x+6, and P(x)=(x-1)(2x^3-3x^2+10x+6) Now you'd have to do the whole and thing again if they changed; In this case they didn't! :D You still have to keep doing synthetic division again, though.

1| 2 -3  10  6 

       2  -1   9  

   2 -1   9  15 

The remainder isn't zero, so x-1 is not a factor.

-1| 2 -3  10   6 

       -2   5   -15  

   2  -5   15  -9

The remainder isn't zero, so x+1 is not a factor. At this point, you have to start using the ugly ones :( I'll start with +/- 1/2

.5| 2 -3  10  6 

        1   -1  4.5  

    2  -2   9  10.5

-.5| 2 -3  10  6 

         -1  2   -6  

   2   -4   12  0

The remainder is finally 0! Sometimes, you might have to go through all the factors, and that's a real pain. The new polynomial becomes 2x^2-4x+12, and P(x)=(x-1)(x+1/2)(2x^2-4x+12)

Now we have a quadratic, and you can use the quadratic formula to solve for the zeroes. It turns out that there are no real zeroes, so P(x) in factored form remains (x-1)(x+1/2)(2x^2-4x+12). Thus, the only factors are x=1 and x=-1/2.

 

 

 

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Guest

Thanks! However, I don't recall learning the process of synthetic division, only long division. So, is it possible if this question could be answered using the long division method?

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6 minutes ago, frank!e said:

Thanks! However, I don't recall learning the process of synthetic division, only long division. So, is it possible if this question could be answered using the long division method?

Yeah it's just called long division for a reason, because it's long :P
You get the exact same answer as with synthetic division, because really only the coefficients of the poly and a matter in long division. Synthetic just allows you to avoid dealing with all of that.

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2 minutes ago, TheNintendoChip said:

Yeah it's just called long division for a reason, because it's long :P
You get the exact same answer as with synthetic division, because really only the coefficients of the poly and a matter in long division. Synthetic just allows you to avoid dealing with all of that.

Alright - but I don't have a divisor. So - how would I start?

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10 minutes ago, frank!e said:

Alright - but I don't have a divisor. So - how would I start?

Look at the and things that I did before doing synthetic division. All of the factors of divided by the factors of (in your example,  +/- 1/2,1,3/2,2,3 ) are potential factors. Set equal to each of them and then make it equal to zero. All of these are the divisors now (i.e., x+1/2,x-1/2,x+1,x-1,x+3/2,x-3/2,x+2,x-2,x+3,x-3). 

Edited by TheNintendoChip

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