Question

Given $a{x}^2+bx+c$ is a quadratic polynomial in x and leaves remainders 6, 11 and 18 respectively when divided by $(x+1),(x+2)and(x+3)$. Find the value of $a+b+c.$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

Let $f\left(x\right)=a{x}^2+bx+c$.

Given, $f(-1)=6,f(-2)=11,f(-3)=18$

$a-b+c=6$ .....(1)

$4a-2b+c=11$ ......(2)

$9a-3b+c=18$ ..........(3)

Solving these equatiuons, we get,

$a=1,b=-2,c=3$

Therefore,

$a+b+c=2$