12x^5-4x^4+7x^3-2x^2-5x+2
Find the zeros of the polynomial as the product of prime binomial factors. Determine all the possible rational roots and the y-intercept. And determine if there are any prime quadratic factors over the real number field.
Find the zeros of the polynomial as the product of prime binomial factors. Determine all the possible rational roots and the y-intercept. And determine if there are any prime quadratic factors over the real number field.
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The y intercept is (0,2) ... Plug in zero for x.
By the rational root theorem:
The possible rational roots are (factors of 2)/( factors of 12)
= +/-{ 1,2}/ { 1,2,3,4,6,12}
Now make all possible ratipnal roots= +/-{ 1,1/2,1/3,1/4,1/6,1/12, 2, 2/3 }
12x^5-4x^4+7x^3-2x^2-5x+2
First try 1 and -1. (Add these points to your graph as you go)
f(1)= 12-4+7-2-5+2= 10
f(-1)= -12-4-7-2+5+2= -18
Since there is a sign change , the function crosses the x axis at least once between -1 and 1.
And since f(0)=2, then we can narrow it down between 0 and -1.
Now the best way to try the other fractions is by synthetic division ( you know how to do this, right?)
-1/2). 12... -4.... 7.... -2... -5...2
------ ----(-6)---(5)---(-6)--(4)---(1/2)
...........12...-10....12.....-8...-1.… 2.5
So f(-1/2)= 2.5.....add this point to your graph.
Notice it is higher than the y intercept.
-2/3) 12....-4.. ....7... ...-2.....-5.....2
-----------(-8)- --(8)---(-10)---(8)---(-2)
.........12....-12....15... -12........3. ...0..< remainder is (0) and this row of numbers is the other factor
F(-2/3)= 0
Add this point to your graph.
This means the f(x) factors as (x+2/3)( 12x^4-12x^3+15x^2-12x+3)
= 3(x+2/3)(4x^4-4x^3+5x^2-4x+1)
=(3x+2)(4x^4-4x^3+ 5x^2-4x+1)
Using rational root thm again, we can try +/- 1/2 or 1/4
1/2) 4..... -4.... 5.... -4.... 1
------------(2)--(-1)--(2)---(-1)-
........4..... -2......4... .(-2)...(0)<----- remainder is zero.
(x-1/2) is a factor , leaving (4x^3-2x^2+4x-2)
==(3x+2)(x-1/2)(4x^3-2x^2+4x-2)
Factor out 2 on the cubic:
By the rational root theorem:
The possible rational roots are (factors of 2)/( factors of 12)
= +/-{ 1,2}/ { 1,2,3,4,6,12}
Now make all possible ratipnal roots= +/-{ 1,1/2,1/3,1/4,1/6,1/12, 2, 2/3 }
12x^5-4x^4+7x^3-2x^2-5x+2
First try 1 and -1. (Add these points to your graph as you go)
f(1)= 12-4+7-2-5+2= 10
f(-1)= -12-4-7-2+5+2= -18
Since there is a sign change , the function crosses the x axis at least once between -1 and 1.
And since f(0)=2, then we can narrow it down between 0 and -1.
Now the best way to try the other fractions is by synthetic division ( you know how to do this, right?)
-1/2). 12... -4.... 7.... -2... -5...2
------ ----(-6)---(5)---(-6)--(4)---(1/2)
...........12...-10....12.....-8...-1.… 2.5
So f(-1/2)= 2.5.....add this point to your graph.
Notice it is higher than the y intercept.
-2/3) 12....-4.. ....7... ...-2.....-5.....2
-----------(-8)- --(8)---(-10)---(8)---(-2)
.........12....-12....15... -12........3. ...0..< remainder is (0) and this row of numbers is the other factor
F(-2/3)= 0
Add this point to your graph.
This means the f(x) factors as (x+2/3)( 12x^4-12x^3+15x^2-12x+3)
= 3(x+2/3)(4x^4-4x^3+5x^2-4x+1)
=(3x+2)(4x^4-4x^3+ 5x^2-4x+1)
Using rational root thm again, we can try +/- 1/2 or 1/4
1/2) 4..... -4.... 5.... -4.... 1
------------(2)--(-1)--(2)---(-1)-
........4..... -2......4... .(-2)...(0)<----- remainder is zero.
(x-1/2) is a factor , leaving (4x^3-2x^2+4x-2)
==(3x+2)(x-1/2)(4x^3-2x^2+4x-2)
Factor out 2 on the cubic:
12
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