Question

If \[x + 2k\] is a factor of \[p(x) = {x^5} - 4{k^2}{x^3} + 2x + 2k + 3\]. Find k.

Answer 1

Hint:
In this problem we are given a polynomial with one of its factors. We have to find the value of k. Since \[x + 2k\] is a factor of the polynomial it should definitely satisfy the equation. So we will put x=-2k in \[p(x)\] and find the value of k. Let’s solve it!

Complete step by step solution:
Given that
\[p(x) = {x^5} - 4{k^2}{x^3} + 2x + 2k + 3\]
And \[x + 2k\] is one of the factors of \[p(x)\].
So \[
  x + 2k = 0 \\
   \Rightarrow x = - 2k \\
 \]
Putting this value of x in \[p(x)\] we get,
\[
  p( - 2k) = 0 \\
   \Rightarrow {\left( { - 2k} \right)^5} - 4{k^2}{\left( { - 2k} \right)^3} + 2\left( { - 2k} \right) + 2k + 3 = 0 \\
 \]
Now finding the respective powers,
\[ \Rightarrow - 32{k^5} + 4{k^2} \times 8{k^3} - 4k + 2k + 3 = 0.........\left( {{2^5} = 35,{2^3} = 8} \right)\]
Now using law of indices for second term
\[ \Rightarrow - 32{k^5} + 4{k^2} \times 8{k^3} - 4k + 2k + 3 = 0\]
\[ \Rightarrow - 32{k^5} + 32{k^5} - 2k + 3 = 0.......\left( {{a^m}.{a^n} = {a^{m + n}}} \right)\]
Now we can see the first and second term are the same with different signs. So they will be cancelled.
\[ \Rightarrow - 2k + 3 = 0\]
Now to find the value of k,
\[
   \Rightarrow 2k = 3 \\
   \Rightarrow k = \dfrac{3}{2} \\
\]
Here we get the answer!

Value of k is \[\dfrac{3}{2}\].

Note:
This is very easy to solve problems. Only students can make mistakes while writing the powers of 2. Always remember that the power of a negative number is negative and even the power of a negative number is positive. It means,
\[{( - )^{odd}} = - \] and \[{( - )^{even}} = + \].