For what value of a, 2y3 + ay2 + 11y + a +3 is exactly divisible by (2y -1)?
Answers
Step-by-step explanation:
Given :-
2y^3 + ay^2 + 11y + a +3
To find:-
For what value of a, 2y^3 + ay^2 + 11y + a +3 is exactly divisible by (2y -1)?
Solution:-
Given cubic Polynomial is
P(x) =2y^3+ay^2+11y+a+3
Given divisor = 2y-1
We know that
Factor theorem:-
P(x) be a Polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if P(x) is divisible by (x-a) . i.e. P(a) = 0 then (x-a) is a factor of P(x) vice -versa.
So given Polynomial is exactly divisible by (2y-1) then P(1/2) = 0
(Since 2y-1 = 0=> 2y= 1=> y=1/2)
=>P(1/2) = 0
=> 2(1/2)^3+a(1/2)^2+11(1/2)+a+3 = 0
=> 2(1/8)+a(1/4)+(11/2)+a+3 = 0
=> (2/8)+(a/4)+(11/2)+a+3 = 0
=> (1/4)+(a/4)+(11/2)+a+3 = 0
=> (1+a+22+4a+12)/4 = 0
=> (5a+35)/4 = 0
=> 5a+35 = 0×4
=> 5a+35 = 0
=> 5a = -35
=> a = -35/5
=> a = -7
Therefore, a= -7
Answer:-
The value of a for the given problem is -7
Used Concept:-
Factor Theorem:-
P(x) be a Polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if P(x) is divisible by (x-a) . i.e. P(a) = 0 then (x-a) is a factor of P(x) vice -versa.
Answer:
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