find value of 'a' such that (3x-1) is a factor of g(x)=6x³ + ax²+x-2
can someone help me with this question asap plsa
Answers
Answer:
Factorising x² - 3x + 2 we get
[Splitting the middle term]
x² - 2x - x + 2 = 0
=> x (x - 2) -1( x - 2) = 0
=> (x - 1)(x - 2) = 0
This means x - 1 and x - 2 are factors of
x³ + ax² - bx + 10
at x - 1 = 0
we get x = 1
Putting the value we get
(1)³ + a(1)² - b(1) + 10 = 0 (since it is a factor)
=> 1 + a - b + 10 = 0
=> a - b = -10 -1
=> a - b = -11.............(i)
Now at x - 2
x - 2 = 0
=> x = 2
Putting the value we get
(2)³ + a(2)² - b(2) + 10 = 0
=> 8 + 4a - 2b + 10 = 0
=> 4a - 2b + 18 = 0
=> 4a - 2b = -18
=> 2(2a - b) = -18
=> 2a - b =-18/2
=> 2a - b = -9.......... (ii)
Subtracting (i) from (ii) we get
2a - b -(a - b) = -9 - (-11)
=> 2a- b - a + b = -9 + 11
=> a = 2
Now putting the value of a in (i)
2 - b = -11
=> -b = -11 - 2
=> -b = -13
=> b = 13
a = 2 and b = 13
Step-by-step explanation:
hope this helps you
Given :-
( 3x - 1 ) is the factor of g ( x ) = 6x³ + ax² + x - 2
Required to find :-
- Value of ' a '
Method used :-
- Factor theorem
Solution :-
Given information :-
( 3x - 1 ) is a factor of g ( x ) = 6x³ + ax² + x - 2
We need to find the value of ' a '
Consider,
g ( x ) = 6x³ + ax² + x - 2
( 3x - 1 ) is the factor of g ( x )
So,
Let;
3x - 1 = 0
3x = 1
x = 1/3
Substitute this value in place of x in g ( x )
g ( ⅓ ) =
6 ( ⅓ )³ + a ( ⅓ )² + ⅓ - 2 = 0
6 ( 1/27 ) + a ( 1/9 ) + 1/3 - 2 = 0
2/9 + a/9 + 1/3 - 2 = 0
2 + a + 3 - 2 / 9 = 0
a + 5 - 2/9 = 0
a + 3/9 = 0
a + 3 = 0 x 9
a + 3 = 0
a = - 3
Therefore,
Value of ' a ' is - 3
Additional Information :-
- what is a polynomial ?
>> A polynomial is considered to be a special kind of algebraic expressions . If a algebraic Identity wants to become a polynomial then it should satisfy some conditions. The conditions are ;
- Variables must not be in the denominator
- Only powers should be whole numbers . ( non - negative integral powers )
>> The polynomials can be represented in the form of diagrams , graphs , words etc .
>> Some theorem used in solving polynomials are ; Factor theorem , Remainder theorem etc .
>> we can also verify the Division Algorithm for these polynomials