Question

Find the value of \[a\]and \[b\] such that ${x^2} - 2x - 3$ is a factor of the polynomial ${x^3} - 3{x^2} + ax - b$.

Answer 1

Hint: To solve this problem ,use remainder theorem and on substituting factors in the given equation remainder will always be equal to zero.


Complete step by step solution:
1. We know that factor of any polynomial must satisfy it.
It is given that ${x^2} - 2x - 3$ is a factor of polynomial ${x^3} - 3{x^2} + ax - b$
So, roots of the polynomial ${x^2} - 2x - 3$ will satisfy ${x^3} - 3{x^2} + ax - b$.
2. First we calculate the factor of the equation ${x^2} - 2x - 3$ by splitting the middle term.
i.e. ${x^2} - 2x - 3$
$
   = {x^2} - 3x + 1x - 3 \\
   = x\left( {x - 3} \right) + 1\left( {x - 3} \right) \\
   = \left( {x - 3} \right)\left( {x + 1} \right) \\
 $
3. Hence, from above we see that factors of ${x^2} - 2x - 3$are$\left( {x - 3} \right)\left( {x + 1} \right)$.
4. As discussed above in step \[1\] that ${x^2} - 2x - 3$ is a factor of ${x^3} - 3{x^2} + ax - b$.
So, $\left( {x - 3} \right)$ and $\left( {x + 1} \right)$ will also be the factors of ${x^3} - 3{x^2} + ax - b$
5. Therefore, by remainder theorem we have
$x - 3 = 0$
$ \Rightarrow x = 3$ will satisfy the equation ${x^3} - 3{x^2} + ax - b$
Substituting value of \[x = 3\]and equating to zero as $x - 3$ is a factor of the ${x^3} - 3{x^2} + ax - b$
$
   \Rightarrow {\left( 3 \right)^3} - 3{\left( 3 \right)^2} + a\left( 3 \right) - b = 0 \\
   \Rightarrow 27 - 27 + 3a - b = 0 \\
   \Rightarrow 3a - b = 0 \\
   \Rightarrow b = 3a......(i) \\
 $
6. Repeating step \[5\] now for $x + 1$ we have
$x + 1 = 0$
$x = - 1$
$x = - 1$ will satisfy the equation ${x^3} - 3{x^2} + ax - b$
Substituting value of \[x = - 1\]and equating to zero as$x + 1$ is a factor of the ${x^3} - 3{x^2} + ax - b$
$ \Rightarrow {( - 1)^3} - 3{\left( { - 1} \right)^2} + a\left( { - 1} \right) - b$
$
   \Rightarrow - 1 - 3 - a - b = 0 \\
   \Rightarrow - 4 - a - b = 0 \\
   \Rightarrow a + b + 4 = 0.......(ii) \\
 $
7. In step \[5\] and \[6\] we got two linear equations in variable ‘\[a\]’ and ‘\[b\]’. Now solving these two equations by method of substitution we got respective values of ‘\[a\]’ and ‘\[b\]’.
Substituting value of $b = 3a$ from (i) in (ii) we have
$a + 3a = 4$
$
   \Rightarrow 4a = 4 \\
   \Rightarrow a = 1 \\
 $
Now from (i) we have
$
  b = 3(a) \\
  b = 3 \times 1 \\
  b = 3 \\
 $
Hence, required values of ‘\[a\]’ and ‘\[b\]’ are \[1\] and \[3\] respectively.

Note: While finding the values by factors method, students should take care that when we put the value of factor in the equation it must satisfy the given condition and $LHS = RHS = 0$.