If ax^2+bx+c is exactly divisible by (x-1),(x-2) and leaves remainder 6 when divised by (x+1) find a,b and c
Answers
respectively.
Step-by-step explanation:
Given:
ax^{2} +bx+cax
2
+bx+c is exactly divisible by (x-1) (x-2).
ax^{2} +bx+cax
2
+bx+c leaves a remainder 6 when divided by (x+1).
To Find:
The values of a,b and c .
Formula Used:
Dividend = Divisor x Quotient + Remainder
Solution:
As given- ax^{2} +bx+cax
2
+bx+c is exactly divisible by (x-1) (x-2).
Considering ax^{2} +bx+cax
2
+bx+c is exactly divisible by (x-1)
It means x=1 ,satisfy the equation ax^{2} +bx+c=0ax
2
+bx+c=0 a(1)^{2} +b(1)+c=0a(1)
2
+b(1)+c=0
a +b+c=0a+b+c=0 ---------------------- equation no.01
Considering ax^{2} +bx+cax
2
+bx+c is exactly divisible by (x-2).
It means x=2, satisfy the equation ax^{2} +bx+c=0ax
2
+bx+c=0
a(2)^{2} + b(2)+c=0a(2)
2
+b(2)+c=0
4a +2b+c=04a+2b+c=0 ---------------------------------equation no.02
As given- ax^{2} +bx+c=0ax
2
+bx+c=0 leaves a remainder 6 when divided by (x+1).
It means x= -1 , satisfy the equation ax^{2} +bx+c=6ax
2
+bx+c=6
a(-1)^{2} +(-1)b+c=6a(−1)
2
+(−1)b+c=6
a -b+c=6a−b+c=6 ---------------- equation 03.
Subtracting equation no. 01 by equation no.03, we get.
b=- - 3
Putting b= -3 in equation no. 02 and equation no. 01 . After that Subtracting equation no. 01 by equation no.02, we get.
a=1
Putting b= -3 and a=1 in equation no. 01, we get.
c=2
Thus, The values of a, b and c are 1,-3 and 2 respectively.