Question

Given that the zeros of the cubic polynomial \[{x^3} - 6{x^2} + 3x + 10\] are of the form \[a,a + b\,\,{\text{and}}\,\,a + 2b\] for some real numbers \[a\,\,{\text{and}}\,\,b\], find the value of \[a\,\,{\text{and}}\,\,b\].

Answer 1

Hint: In this problem, we need to use the formula of the root of a cubic polynomial, to obtain the equation in \[a\,\,{\text{and}}\,\,b\]. Now, solve the obtained equations to find the value of \[a\,\,{\text{and}}\,\,b\]. Next, verify the obtained value of \[a\,\,{\text{and}}\,\,b\].

Complete step by step answer:
Consider the given polynomial \[{x^3} - 6{x^2} + 3x + 10\] as \[p\left( x \right)\].
\[p\left( x \right) = {x^3} - 6{x^2} + 3x + 10\]
Now, the zeroes of the polynomial \[p\left( x \right) = {x^3} - 6{x^2} + 3x + 10\] are \[a,a + b\,\,{\text{and}}\,\,a + 2b\].
Then,
\[
  \,\,\,\,\,\,\,a + a + b{\text{ + }}a + 2b = - \dfrac{{\left( { - 6} \right)}}{1} \\
   \Rightarrow 3a + 3b = 6 \\
   \Rightarrow a + b = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right) \\
\]
Again,
\[
  \,\,\,\,\,\,\,\left( a \right)\left( {a + b} \right) + \left( {a + b} \right)\left( {a + 2b} \right){\text{ + }}a\left( {a + 2b} \right) = \dfrac{3}{1} \\
   \Rightarrow {a^2} + ab + {a^2} + 2ab + ab + 2{b^2} + {a^2} + 2ab = 3 \\
   \Rightarrow 3{a^2} + 2{b^2} + 6ab = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \\
\]
Further,
\[
  \,\,\,\,\,\,\,a\left( {a + b} \right)\left( {a + 2b} \right) = - \dfrac{{\left( {10} \right)}}{1} \\
   \Rightarrow {a^3} + {a^2}b + 2{a^2}b + 2a{b^2} = - 10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 3 \right) \\
\]
From equation (1), substitute \[2 – a\] for \[b\] in equation (2).
\[
  \,\,\,\,\,\,3{a^2} + 2{\left( {2 - a} \right)^2} + 6a\left( {2 - a} \right) = 3 \\
   \Rightarrow 3{a^2} + 2\left( {4 - 4a + {a^2}} \right) + 12a - 6{a^2} = 3 \\
   \Rightarrow - {a^2} + 4a + 5 = 0 \\
   \Rightarrow {a^2} - 4a - 5 = 0 \\
   \Rightarrow {a^2} - 5a + a - 5 = 0 \\
   \Rightarrow \left( {a - 5} \right)\left( {a + 1} \right) = 0 \\
   \Rightarrow a = 5\,\,{\text{or}}\,\,a = - 1 \\
\]
For \[a = 5\], \[b = - 3\], and for \[a = - 1\], \[b = 3\],
Substitute, \[a = 5\], \[b = - 3\] in equation (3).
\[
  \,\,\,\,\,\,\,{\left( 5 \right)^3} + {\left( 5 \right)^2}\left( { - 3} \right) + 2{\left( 5 \right)^2}\left( { - 3} \right) + 2\left( 5 \right){\left( { - 3} \right)^2} = - 10 \\
   \Rightarrow 125 - 75 - 150 + 90 = - 10 \\
   \Rightarrow - 10 = - 10\left( {true} \right) \\
\]
Substitute, \[a = - 1\], \[b = 3\] in equation (3).
\[
  \,\,\,\,\,\,\,{\left( { - 1} \right)^3} + {\left( { - 1} \right)^2}\left( 3 \right) + 2{\left( { - 1} \right)^2}\left( 3 \right) + 2\left( { - 1} \right){\left( 3 \right)^2} = - 10 \\
   \Rightarrow - 1 + 3 - 6 - 18 = - 10 \\
   \Rightarrow - 22 \ne - 10\left( {false} \right) \\
\]

Thus, the value of \[a\] is 5 and \[b\] is -3.

Note: The formula for the roots \[\alpha ,\beta \,\,{\text{and}}\,\,\gamma \] of cubic polynomial \[a{x^3} + b{x^2} + cx + d = 0\] is shown below.
\[\begin{aligned}
  \alpha + \beta + \gamma = - \dfrac{b}{a} \\
  \alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{c}{a} \\
  \alpha \beta \gamma = - \dfrac{d}{a} \\
\end{aligned}\]