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# Finding zeros of polynomials (example 2)

Sal finds all the zeros (which is the same as the roots) of p(x)=(3x⁴-8x³+15x-40)(3x-8)².

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• at , why did he not take into account the ^3 on 3x-8 when he was trying to figure out why it was equal to zero
• It is because (3x-8)³=0 is the same as (3x-8)(3x-8)(3x-8)=0. You can set all 3 of them equal to 0 if you want, but you'll end up getting the same answer for all 3.
• I tried doing the "Practice:Zeros of polynomials challenge problems", but I could not understand anything whatsoever. Can someone explain to me how to solve those problems?
• Take any polynomial first of all, factorize all the polynomial to a form like this, say [2x-x][10x-8x], and now equate this = 0 to earn zeroes of the polynomial. To know the zero of the polynomial either any one of the brackets should be equal to zero. Now in the first bracket, it turns out to be 2x-x=x so x = 0... In the second bracket 10x-8x=2x and if 2x = 0 then x= 0/2=0 so it turned out to be that 0 and 0 are the "zeros of the polynomial". This means 0 is the "zero" of this polynomial [2x-x][10x-8x]...

Always 0 may not be the "zero of the polynomial". It can be any number.
• At why is the answer not -1.71 i? I thought if you take the the root of a negative number you have to put i after the answer to specify that it is imaginary.
• He is taking the cube root of -5. If you take an odd powered root like 3, of a negative number your solution is real. For example the cube root of -8 is -2 because -2*-2*-2=-8.
• Just to make sure I understand this:
To find the zero of a polynomial, you have to factor the polynomial until you vet the last expressions. In this case, the polynomial simplified to (3x-8)^3 x (x^3+5). So we just solve (3x-8)^3 for 0 and then (x^3+5) for 0. To get 8/3 and sqrt3. -5 right?. If you enter these x values in the function, it will intercept the x axis, right? Also how would this be useful in real life?
• Yes that is correct, (3x-8)^3 * (x^3 +5) has the zeros x=8/3 and x= -cuberoot(5).
How is this useful in real life? Well, imagine you're throwing a ball and you want to know how far away from you it will hit the ground. If we think of the ground as the x-axis and yourself as the y-axis (your position is at x=0) then using your equation (and there are ways to find the equation of the ball's trajectory) you can find the zeros of the parabolic function and use them to see how far away from you the ball lands.
Hope this helps!
• In the Practice Problems - Zeros of polynomials challenge problems, I can't figure out how to work any of those problems. Below is one example, all of the problems are similar.

The position, p, of a particle is expressed in terms of time as:

p=(a-t)(t-b)(b-t) where a>b

I can't figure out how to work any of those challenge problems. Any help would be greatly appreciated. This is the only section that totally makes no sense to me.

• I understand finding the roots of a polynomial given that at least one of the roots are rational, but how would I find the EXACT roots of a polynomial (mainly a trinominal with four terms) given that all the roots are either irrational or imaginary?

Example: x^3 -2x^2 -4x -6
Grouping doesn't work here
I've tried using the rational roots theorem, then plugging each possible root into the equation, none of them are factors
I don't understand how to find the exact roots of the polynomial given that there are no rational roots.
(1 vote)
• Finding exact roots of a polynomial is, in general, a genuinely hard problem. After polynomials came into popular use, it was about 300 years before we had a decent set of tools for finding roots, with a branch of math called Galois theory.

As it happens, there is an explicit formula for finding the roots of degree-3 polynomials, and another, even more complicated one for degree-4. But there is not, and cannot be, such a formula for equations of degree 5 and up.

The way to find roots of 3rd-degree equations is described here:
https://qr.ae/TUhUjK
• how would we find the four remaining non-real complex roots?
• This polynomial has no complex roots.
Yes, the polynomial should have 6 roots. However, there reason we have only 2 roots is because there are duplicates.
(3x-8)^3=0 means there are 3 roots that are exactly the same.
If you write this without the exponent, you see the 3 factors
(3x-8)(3x-8)(3x-8)=0
Use the zero produce rule, separate & solve and you get 3 identical values:
3x-8 = 0 creates 8/3
3x-8 = 0 creates 8/3
3x-8 = 0 creates 8/3
This is called multiplicity-- Multiple roots that are the same.
The same thing is happening with x^3 = -5. This represents 3 matching roots.
So, all 6 roots are accounted for, there are just 4 values that are duplicates.
Hope this helps.
(1 vote)
• How are the other 2 complex roots found in this equation?
• how is 8/3 equal to zero?