9 x ^ 3 + 6 x ^ 2 - 29 x - 10
Since the constant in the given equation is a 10 , we know that the integer root must be a factor of 10.
The possible values are :
+ OR - 1
+ OR -2
+ OR - 5
and
+ OR - 10
You can use the factor theorem to test the possible values by trial and error :
f ( 1 ) = 9 + 6 - 29 - 10 = -24 ¡Ù 0
f ( ¨C 1 ) = ¨C 9 + 6 + 29 ¨C 10 = 16 ¡Ù 0
f ( 2 ) = 72 + 24 ¨C 58 ¨C 10 = 28 ¡Ù 0
f ( - 2 ) = - 72 + 24 + 58 ¨C 10 = 0
f ( 5 ) = 1125 + 150 ¨C 145 ¨C 10 = 1120 ¡Ù 0
f ( - 5 ) = - 1125 + 150 + 145 ¨C 10 = - 840 ¡Ù 0
f ( 10 ) = 9000 + 600 ¨C 290 ¨C 10 = 9300 ¡Ù 0
f ( - 10 ) = - 9000 + 600 + 290 ¨C 10 = - 8120 ¡Ù 0
The integer root is - 2
Divide 9 x ^ 3 + 6 x ^ 2 - 29 x - 10 with [ x - ( - 2 ) ]
x - ( - 2 ) = x + 2 so :
( 9 x ^ 3 + 6 x ^ 2 - 29 x - 10 ) / ( x + 2 ) =
9 x ^ 2 - 12 x - 5
Now :
9 x ^ 3 + 6 x ^ 2 - 29 x - 10 =
( x + 2 ) ( 9 x ^ 2 - 12 x - 5 )
Now in google type:
quadratic equation online
When you see list of results click on:
Free Online Quadratic Equation Solver:Solve by Quadratic Formula
When page be open click option :
Solve by Factoring (includes Factoring by Grouping)
then in rectangle type :
9 x ^ 2 - 12 x - 5 = 0
and click optin: solve it!
You will see solution step by step
So solution of your equations are :
x = - 2
x = - 1 / 3
and
x = 5 / 3
________________________________________
P.S.
If you don't know how to divide 9 x ^ 3 + 6 x ^ 2 - 29 x - 10 with x + 2
go on :
calc101 dot com
When page be open click option :
long division
Then in up rectangle type :
9 x ^ 3 + 6 x ^ 2 - 29 x - 10
In down rectangle type :
x + 2
and click option :
DO IT
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List all potential rational zeros of 9x^3+6x^2-29x-10
2 answers
¡Ù
mean :
different of
f ( ¨C 1 ) = ¨C 9 + 6 + 29 ¨C 10 = 16 ¡Ù 0
mean :
f ( - 1 = 16 differen of 0
mean :
different of
f ( ¨C 1 ) = ¨C 9 + 6 + 29 ¨C 10 = 16 ¡Ù 0
mean :
f ( - 1 = 16 differen of 0