If (x+2) is a factor of p(x)=ax^3+bx^2+x-6 and p(x) when divided by(x-2) leaves the remainder 4. Prove that a=0 and b=2 irrelevant answers will be reported
Answers
sorry dont know the answer
Step-by-step explanation:
Given :-
(x+2) is a factor of p(x)=ax^3+bx^2+x-6 and p(x) when divided by(x-2) leaves the remainder 4.
To find :-
Prove that a=0 and b=2
Solution :-
Given Polynomial is ax³+bx²+x-6
Let P(x) = ax³+bx²+x-6
Given factor of P(x) = (x+2)
we know that ,by Factor Theorem
If (x-a) is a factor of P(x) then P(a) = 0
Now,
If (x+2) is a factor of P(x) then P(-2) = 0
=> a(-2)³+b(-2)²+(-2)-6 = 0
=> a(-8)+b(4)-2-6 = 0
=> -8a+4b-8 = 0
=> 4(-2a+b-2) = 0
=> -2a+b-2 = 0/4
=> -2a+b-2= 0
=> b = 2a+2 -----------(1)
and
Given that
P(x) is divided by(x-2) leaves the remainder 4.
By Remainder Theorem
If P(x) is divided by x-a then the remainder is P(a).
Now,
P(x) is divided by(x-2) then the remainder = P(2).
According to the given problem.
P(2) = 4
=> a(2)³+b(2)²+2-6 = 4
=> a(8)+b(4)-4 = 4
=> 8a+4b-4 = 4
=> 8a+4b = 4+4
=> 8a+4b = 8
=> 4(2a+b) = 8
=> 2a+b = 8/4
=>2a+b = 2
=> 2a+2a+2 = 2
=> 4a +2 = 2
=> 4a = 2-2
=> 4a = 0
=> a = 0/4
=> a = 0
On substituting the value of a in (1) then
=> b = 2a+2
=> b = 2(0)+2
=> b = 0+2
=> b = 2
Therefore, a = 0 and b = 2
Hence, Proved.
Answer:-
The value of a = 0 and b = 2
Used formulae:-
Remainder Theorem:-
Let 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 divided by (x-a) then the remainder is P(a).
Factor Theorem:-
Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if (x-a) is a factor of P (x) then P(a) = 0 vice-versa.