Question

Let p(x) be a quadratic polynomial such that p(0) = 1. If p(x) leaves remainder 4 when divided by x − 1 and it leaves remainder 6 when divided by x + 1; then which one is correct?
(a) p(2) = 11
(b) p(─2) = 19
(c) p(─2) = 11
(d) p(2) = 19

Answer 1

Hint: It is given that p(x) is a quadratic polynomial. The standard form of quadratic polynomial is given as $p\left( x \right)=a{{x}^{2}}+bx+c$. We now have to find the values of a, b and c. It is given p(0) = 1. It means, we have to substitute x = 0 in p(x). This will give us the value of c. Since, it is given that the when the polynomial is divided by another polynomial of lower order, there is a remainder. Thus, we will use these conditions to form equations comprising of the coefficients and we’ll solve those equations to get the coefficient. Once we get values for all the three coefficients, we will substitute x = 2 and x = −2 to find p(2) and p(−2).

Complete step by step answer:
Let p(x) be $p\left( x \right)=a{{x}^{2}}+bx+c$.
It is given that p(0) = 1.
So, we will substitute x = 0
$\begin{align}
  & \Rightarrow p\left( 0 \right)=a{{\left( 0 \right)}^{2}}+b\left( 0 \right)+c \\
 & \Rightarrow c=1 \\
\end{align}$
Now, it is given that if p(x) is divided by x – 1, the remainder is 4.
This means, if we put x = 1, p(1) = 4.
$\Rightarrow $ a + b + c = 4 …… (1)
Similarly, it is given that if p(x) is divided by x + 1, the remainder is 6.
This means, if we substitute x = –1 in p(x), p( –1) = 6.
$\Rightarrow $ a – b + c = 6 …… (2)
Subtract (2) from (1).
$\Rightarrow $ 2b = – 2
$\Rightarrow $ b = – 1
We know that c = 1 and b = –2. Substitute these in equation (1).
$\Rightarrow $ a – 1 + 1 = 4
$\Rightarrow $ a = 4
Therefore, the quadratic polynomial is $p\left( x \right)=4{{x}^{2}}-x+1$.
Now, substitute x = −2.
$\Rightarrow $ p(−2) = 16 + 2 + 1
$\Rightarrow $ p(−2) = 19
And we will see p(2) by substituting x = 2.
$\Rightarrow $ p(2) = 16 – 2 + 1
$\Rightarrow $ p(2) = 15

So, the correct answer is “Option B”.

Note: If it is given that (x – 1) can divide the polynomial and remainder is 4, this means $p\left( x \right)=\left( x-1 \right)g\left( x \right)+4$. Thus if x = 1, p(1) = 4, where $g\left( x \right)$ is the other factor of $p\left( x \right)$.