In general, it is very difficult to find zeros of polynomials higher than degree 2. So, you have to look for some low-hanging fruit using synthetic division, or try grouping.
Graphical methods may also be useful, if it's obvious that the roots are easy ones.
I usually start by trying ±1,±2
Doing that, I find
#1: the only rational roots are ±1,±1/2,±41,±41/2
It's fairly quick to find that 1/2 is a root, so
q(x) = (2x-1)(x^2-8x+41)
now use the quadratic formula.
#2: (x+3)(x^2-8x+17)
#3: (x+1)(x-3)(x^2-6x+10)
Luckily, these all had relatively easy-to-find rational roots.
Find all of the zeros of each function
1. q(x)=2x^3-17x^2+90x-41
2.x^3-5x^2-7x+51
3.x^4-9x^3+24x^2-6x-40
Please show work!!! I want to learn how to do it.
3 answers
I will do the one which potentially could have the most factors.
(the highest exponent matches the maximum number of linear factors, so this one could have a maximum of 4, but could be less )
look at your first term, it is 1x^4, so any linear factor must look like (x ± .... )
now look at the last term of -40.
It can only come from the multiplication of all the last terms of all the factors.
so any factor can only end in one of the factors of -40
so let's look for possible zeros of the function.
They could only be ±1, ±2, ±4, ±5, ±8, ±10, ±20 or finally ± 40
(This looks sort of hopeless, but let's see .... )
let f(x) = x^4-9x^3+24x^2-6x-40
f(1) = 1 - 9 + 24 - 6 - 40 ≠ 0 (all I care about is whether it is 0 or not)
f(-1) = 1 + 9 + 24 + 6 -40 = 0
Yeahhhh, lucky on the 2nd try.
So (x + 1) is one of the factors
now we do a long division of the original function by x+1
I usually use synthetic division for this stuff and I got
(x+1)(x^3 - 10x^2 + 34x - 40)
so let g(x) = x^3 - 10x^2 + 34x - 40
and the same argument about zeros applies
I tried ±1, ±2, with no luck, but
g(4) = 64 - 160 + 136 -40 = 0 YEAH again
so (x-4) is another factor.
again dividing g(x) by x-4 , using synthetic division, I got
g(x) = (x-4)((x^2 - 6x + 10)
now a quick inspection of x^2 - 6x + 10 shows that it does not factor
(there are no 2 numbers with a sum of -6 and a product of 10)
so x^4-9x^3+24x^2-6x-40
= (x+1)(x-4)(x^2 - 6x +10)
hint for the first:
only possible values are : ±1 , ±1/2 , ± 51/2
hint: x = 1/2 worked, so (2x - 1) is a factor
leaving you with a quadratic for which I again found no factors.
BTW, you can check your answers by simply expanding it again. Of course you must get the original.
(the highest exponent matches the maximum number of linear factors, so this one could have a maximum of 4, but could be less )
look at your first term, it is 1x^4, so any linear factor must look like (x ± .... )
now look at the last term of -40.
It can only come from the multiplication of all the last terms of all the factors.
so any factor can only end in one of the factors of -40
so let's look for possible zeros of the function.
They could only be ±1, ±2, ±4, ±5, ±8, ±10, ±20 or finally ± 40
(This looks sort of hopeless, but let's see .... )
let f(x) = x^4-9x^3+24x^2-6x-40
f(1) = 1 - 9 + 24 - 6 - 40 ≠ 0 (all I care about is whether it is 0 or not)
f(-1) = 1 + 9 + 24 + 6 -40 = 0
Yeahhhh, lucky on the 2nd try.
So (x + 1) is one of the factors
now we do a long division of the original function by x+1
I usually use synthetic division for this stuff and I got
(x+1)(x^3 - 10x^2 + 34x - 40)
so let g(x) = x^3 - 10x^2 + 34x - 40
and the same argument about zeros applies
I tried ±1, ±2, with no luck, but
g(4) = 64 - 160 + 136 -40 = 0 YEAH again
so (x-4) is another factor.
again dividing g(x) by x-4 , using synthetic division, I got
g(x) = (x-4)((x^2 - 6x + 10)
now a quick inspection of x^2 - 6x + 10 shows that it does not factor
(there are no 2 numbers with a sum of -6 and a product of 10)
so x^4-9x^3+24x^2-6x-40
= (x+1)(x-4)(x^2 - 6x +10)
hint for the first:
only possible values are : ±1 , ±1/2 , ± 51/2
hint: x = 1/2 worked, so (2x - 1) is a factor
leaving you with a quadratic for which I again found no factors.
BTW, you can check your answers by simply expanding it again. Of course you must get the original.
Go with Reiny on #3. I made a typo in my (x-3) factor.
1 answer