List all potential rational zeros of 9x^3+6x^2-29x-10
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
________________________________________
¡Ù
mean :
different of
f ( ¨C 1 ) = ¨C 9 + 6 + 29 ¨C 10 = 16 ¡Ù 0
mean :
f ( - 1 = 16 differen of 0
To find the potential rational zeros of a polynomial, you can use the Rational Root Theorem. According to this theorem, if a rational number p/q is a zero of a polynomial with integer coefficients, then p must be a factor of the constant term (in this case, -10), and q must be a factor of the leading coefficient (in this case, 9).
Let's find all the factors of -10 and 9 to determine the potential rational zeros:
Factors of -10: ±1, ±2, ±5, ±10
Factors of 9: ±1, ±3, ±9
By testing all possible combinations of these factors, we can find the potential rational zeros:
p/q = ±1/±1, ±2/±1, ±5/±1, ±10/±1, ±1/±3, ±2/±3, ±5/±3, ±10/±3, ±1/±9, ±2/±9, ±5/±9, ±10/±9
After simplifying these fractions, we can conclude that the potential rational zeros of the polynomial 9x^3 + 6x^2 - 29x - 10 are:
±1, ±2, ±5, ±10, ±1/3, ±2/3, ±5/3, ±10/3, ±1/9, ±2/9, ±5/9, ±10/9
Please note that this list only includes potential zeros, and some of them may not actually be zeros of the polynomial. To determine the actual zeros, you would need to apply a method such as synthetic division, graphing, or factoring.