Question

The value of p for which the polynomial ${x^3} + 4{x^2} - px + 8$ is exactly divisible by $(x - 2)$is
A) 0
B) 3
C) 5
D) 16

Answer 1

Hint: We should know the theorem that whenever we put the value of one of the roots of the equation in the equation itself the value of the function would be 0.

Complete step-by-step answer:
We know that if ${x^3} + 4{x^2} - px + 8$ is exactly divisible by $(x - 2)$ then we know for sure that $(x - 2)$is a factor of ${x^3} + 4{x^2} - px + 8$.
By remainder theorem for polynomials we know that if $(x - a)$is a factor of $f(x)$ then the value of $f(a)$ will be 0.
Applying the same property to our question given above we get that the value of $a$ will be equal to 2 and the function $f(x)$ is equivalent to ${x^3} + 4{x^2} - px + 8$. Therefore, by the remainder theorem we get that the value of $f(2)$would be 0.
Putting the value of $x$ as 2 we get,
$
   = (2 \times 2 \times 2) + (4 \times 2 \times 2) - (p \times 2) + 8 \\
   = 8 + 16 - 2p + 8 \\
   = 32 - 2p \\
$
Now by the theorem we know that its value should be equal to 0.
$
  32 - 2p = 0 \\
  32 = 2p \\
  p = 16 \\
$
Therefore, the value of $p$ is equal to 16 and so the correct option is Option D.


Note: Prior knowledge of the remainder theorem is necessary to solve the problem correctly. If we know the above theorem then the question can be solved easily. If we cannot remember this particular theorem, we can always put the different values of $p$ and then check with the long division process whether the function is divisible by the given factor. But this method is time consuming and should not be always adopted as this is a tedious process.