Question

Find the condition that the zeros of \[P(x) = {x^3} - p{x^2} + qx - r\] may be in A.P.

Answer 1

Hint: Since the degree of the given polynomial is three so the number of zero of the polynomial will be three, when three numbers are in AP they have the same common difference. We will assume three numbers which are in AP then we will find their sum, product and the sum of product of those zeros of the polynomial and by substituting the values we will get the required equations.

Complete step-by-step answer:
The given equation is \[P(x) = {x^3} - p{x^2} + qx - r\]
Now let the zeros of the given polynomial who are in AP be\[\left( {a - d} \right),a,\left( {a + d} \right)\]
Let the sum of the zeros of the polynomial be S, hence we can write
\[S = \left( {a - d} \right) + a + \left( {a + d} \right)\]
Hence by solving we get
\[
  S = 3a \\
  a = \dfrac{S}{3} - - (i) \\
 \]
Also let the product of zeros of the polynomial be T, so we get
\[
\Rightarrow T = \left( {a - d} \right) \times a \times \left( {a + d} \right) \\
   = a\left( {{a^2} - {d^2}} \right) \\
 \]
Now since \[a = \dfrac{S}{3}\]from equation (i), so we get
\[
\Rightarrow T = \dfrac{S}{3}\left( {\dfrac{{{S^2}}}{9} - {d^2}} \right) \\
  \dfrac{{3T}}{S} = \dfrac{{{S^2}}}{9} - {d^2} \\
  {d^2} = \dfrac{{{S^2}}}{9} - \dfrac{{3T}}{S} - - (ii) \\
 \]
Now we take the sum of the product of the zeros of the polynomial taken 2 at a time, so we can write
\[\Rightarrow U = a\left( {a - d} \right) + a\left( {a + d} \right) + \left( {a - d} \right)\left( {a + d} \right)\]
By further solving this we get
\[
\Rightarrow U = a\left( {a - d} \right) + a\left( {a + d} \right) + \left( {a - d} \right)\left( {a + d} \right) \\
   = {a^2} - ad + {a^2} + ad + {a^2} - ad + ad - {d^2} \\
   = 3{a^2} - {d^2} \\
 \]
Now substitute the value of a and d from equation (i) and (ii) in the above equation
\[
\Rightarrow U = 3{\left( {\dfrac{S}{3}} \right)^2} - \left( {\dfrac{{{S^2}}}{9} - \dfrac{{3T}}{S}} \right) \\
   = 3 \times \dfrac{{{S^2}}}{9} - \dfrac{{{S^2}}}{9} + \dfrac{{3T}}{S} \\
   = \dfrac{{{S^2}}}{3} - \dfrac{{{S^2}}}{9} + \dfrac{{3T}}{S} \\
   = \dfrac{{3{S^2} - {S^2}}}{9} + \dfrac{{3T}}{S} \\
   = \dfrac{{2{S^2}}}{9} + \dfrac{{3T}}{S} \\
 \]
Hence the condition at which the zeros of \[P(x) = {x^3} - p{x^2} + qx - r\] are in A.P is when
\[U = \dfrac{{2{S^2}}}{9} + \dfrac{{3T}}{S}\]

Note: Students must note that the number of roots of a polynomial depends on the degree of the given polynomial where if the polynomial is a trinomial then that polynomial has three roots.