Question

If ax2+bx+c is exactly divisible by (x-1) (x-2) and leaves a remainder 6 when divided by (x+1) then the values of a,b and c are respectively​

Answer 1

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Answer 2

a, b and c  values  are 1,-3 and 2 respectively.

Step-by-step explanation:

Given:

$ax^{2} +bx+c$ is exactly divisible by (x-1) (x-2).

$ax^{2} +bx+c$  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- $ax^{2} +bx+c$  is exactly divisible by (x-1) (x-2).

Considering $ax^{2} +bx+c$ is  exactly divisible by (x-1)

It means  x=1 ,satisfy the equation $ax^{2} +bx+c=0$                       $a(1)^{2} +b(1)+c=0$      

$a +b+c=0$                    ---------------------- equation no.01

Considering $ax^{2} +bx+c$ is exactly divisible by (x-2).

It means  x=2, satisfy the equation $ax^{2} +bx+c=0$

 $a(2)^{2} + b(2)+c=0$

 $4a +2b+c=0$             ---------------------------------equation no.02

As  given- $ax^{2} +bx+c=0$ leaves a remainder 6 when divided by (x+1).

It means x= -1 , satisfy the equation $ax^{2} +bx+c=6$

$a(-1)^{2} +(-1)b+c=6$  

 $a -b+c=6$                   ---------------- 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.