Question

If the expression $a{x}^2+bx+c$ is equal to $4$, when $x=0$, leave the remainder $4$ when divided by $\left(x+1\right)$ and leaves a remainder $6$ when divided by $\left(x+2\right)$, then the values of $a$, $b$ and $c$ are respectively,

• A. $1,1,4$
• B. $2,2,4$
• C. $3,3,4$
• D. $4,4,4$

Answer 1

The correct option is A

$1,1,4$

If $\mathit{f}\left(x\right)$ is divisible by $\left(x-a\right)$, then the remainder is $\mathit{f}\left(a\right)$.

Step 1. Find the value of $\mathit{c}$.

Here, the given polynomial is $a{x}^2+bx+c$,

Substitute, $x=0$ in the given expression.

$$\begin{gathered} a{\left(0\right)}^2+b\left(0\right)+c=4 \\ \Rightarrow c=4 \end{gathered}$$

Step 2. Find the remainder.

Here, the given polynomial is divisible by $\left(x+1\right)$ and by $\left(x+2\right)$,

So, the remainder are $f\left(-1\right)$ and $f\left(-2\right)$

$f\left(-1\right)=4$ and $f\left(-2\right)=6$

$\begin{array}{rcl}f\left(-1\right)& =& 4\\ a{\left(-1\right)}^2+b\left(-1\right)+c& =& 4\\ a-b+4& =& 4\left[c=4\right]\\ a-b& =& 0\to \left(1\right)\end{array}$

Now,

$\begin{array}{rcl}f\left(-2\right)& =& 6\\ a{\left(-2\right)}^2+b\left(-2\right)+c& =& 6\\ 4a-2b+4& =& 6\left[c=4\right]\\ 4a-2b& =& 2\\ 2a-b& =& 1\to \left(2\right)\end{array}$

Step 3. Solve the equation $\left(1\right)$ and $\left(2\right)$.

Subtract the equation $\mathrm{\left(}\mathrm{1}\mathrm{\right)}$ and $\mathrm{\left(}\mathrm{2}\mathrm{\right)}$ to obtain the value of $\mathrm{a}$.

$\begin{array}{rcl}-a& =& -1\\ & \Rightarrow & a=1\end{array}$

Substitute the value of $\mathrm{a}$ into equation $\mathrm{\left(}\mathrm{1}\mathrm{\right)}$ to obtain the value of $\mathrm{b}$.

$\begin{array}{rcl}a-b& =& 0\\ -b& =& -a\\ & \Rightarrow & b=1\end{array}$

Therefore, the value of $a$, $b$ and $c$ are $\mathrm{1}$,$\mathrm{1}$,$\mathrm{4}$ respectivley.

Hence, the correct option is $\left(A\right)$.