Question

Find the value of \[m\]so that \[2x - 1\] be a factor of\[8{x^4} + 4{x^3} - 16{x^2} + 10x + m\].
\[
  {\text{A}}{\text{. - 1}} \\
  {\text{B}}{\text{. 2}} \\
  {\text{C}}{\text{. - 2}} \\
  {\text{D}}{\text{. 0}} \\
 \]

Answer 1

Hint: Use the Remainder theorem which states that when a polynomial \[g\left( x \right)\] is divided by linear polynomial \[\left( {x - a} \right)\] then we get the remainder as \[g(a)\] where the remainder is the part of the polynomial which is left behind on carrying out the division. This theorem is applicable only when the polynomial is divided by a linear polynomial. A linear polynomial is generally represented as \[f\left( x \right) = ax + b\] where \[b\] is the constant term whereas polynomial is the expression that consists of variables and coefficients having the operations of addition, multiplication, subtraction, and exponent.

Complete step by step answer:
Here, \[8{x^4} + 4{x^3} - 16{x^2} + 10x + m\] is the dividend to which the number is to be divided and \[2x - 1\] is the divisor by which number is divided to find the value of \[m\].
Now equate the divisor which is a linear polynomial with \[0\] hence we get,
\[
  2x - 1 = 0 \\
  2x = 1 \\
  x = \dfrac{1}{2} \\
 \]
So we conclude when a polynomial \[8{x^4} + 4{x^3} - 16{x^2} + 10x + m\] is divided by the linear polynomial \[x = \dfrac{1}{2}\] we get the remainder as \[g\left( {\dfrac{1}{2}} \right)\] in terms of \[m\].
\[
  g\left( x \right) = 8{x^4} + 4{x^3} - 16{x^2} + 10x + m \\
  g\left( {\dfrac{1}{2}} \right) = 8{\left( {\dfrac{1}{2}} \right)^4} + 4{\left( {\dfrac{1}{2}} \right)^3} - 16{\left( {\dfrac{1}{2}} \right)^2} + 10\left( {\dfrac{1}{2}} \right) + m \\
   = 8 \times \dfrac{1}{{16}} + 4 \times \dfrac{1}{8} - 16 \times \dfrac{1}{4} + 10 \times \dfrac{1}{2} + m \\
   = \dfrac{1}{2} + \dfrac{1}{2} - 4 + 5 + m \\
   = 1 - 4 + 5 + m \\
   = 2 + m \\
 \]
Now, as \[2x - 1\] is a factor of \[8{x^4} + 4{x^3} - 16{x^2} + 10x + m\], so equate \[g\left( x \right) = 0\]
Hence we get,
\[
  g\left( x \right) = 0 \\
  2 + m = 0 \\
  m = - 2 \\
 \]

Therefore the value of \[m = - 2\].

Note: Before finding the remainder, ensure that the divisor is a linear polynomial otherwise division will not be possible. A polynomial function of degree n can be written as \[g(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + {a_{n - 2}}{x^{n - 2}} + ........... + {a_2}{x^2} + {a_1}{x^1} + {a_0}\]. Another method to find the remainder is by long division method which includes, dividing the dividend by the divisor which is the general method to find the remainder is long division method, \[\dfrac{{dividend}}{{divisor}} = \dfrac{{{x^4} + {x^3} - 2{x^2} + x + 1}}{{x - 1}}\]. However, this method is very long and time taking.