Question

Without actual division, prove that $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$ is exactly divisible by $\left( {{x^2} + 2x - 3} \right)$.

Answer 1

Hint: Here, we will proceed by factoring the polynomial $\left( {{x^2} + 2x - 3} \right)$ into its two linear factors and then equating both of these factors to zero. After this, the corresponding values of x when substituted in $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$ should bring it equal to zero.

Complete Step-by-Step solution:
Consider $\left( {{x^2} + 2x - 3} \right)$ and factorize this polynomial into linear factors by using splitting the middle term.
\[
  \left( {{x^2} + 2x - 3} \right) = {x^2} + 3x - x - 3 \\
   \Rightarrow \left( {{x^2} + 2x - 3} \right) = x\left( {x + 3} \right) - 1\left( {x + 3} \right) \\
   \Rightarrow \left( {{x^2} + 2x - 3} \right) = \left( {x + 3} \right)\left( {x - 1} \right) \\
 \]
$\left( {{x^2} + 2x - 3} \right)$ is factored into two linear factors \[\left( {x + 3} \right)\] and \[\left( {x - 1} \right)\]
In order to prove that $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$ is exactly divisible by $\left( {{x^2} + 2x - 3} \right)$, $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$ should be divisible by both the linear factors which are \[\left( {x + 3} \right)\] and \[\left( {x - 1} \right)\] i.e., x=-3 and x=1 should be the solution of ... This means that when we substitute x=-3 and x=1 in $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$ the value of this function should come equal to zero.
By putting x=-3 in $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$, we get
$
  {x^4} - 4{x^2} + 12x - 9 = {\left( { - 3} \right)^4} - 4{\left( { - 3} \right)^2} + 12\left( { - 3} \right) - 9 \\
   \Rightarrow {x^4} - 4{x^2} + 12x - 9 = 81 - \left( {4 \times 9} \right) - 36 - 9 \\
   \Rightarrow {x^4} - 4{x^2} + 12x - 9 = 81 - 36 - 36 - 9 \\
   \Rightarrow {x^4} - 4{x^2} + 12x - 9 = 81 - 81 \\
   \Rightarrow {x^4} - 4{x^2} + 12x - 9 = 0 \\
 $
Clearly, by putting x=-3 in $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$, the value of this polynomial reduces to zero.
By putting x=1 in $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$, we get
$
  {x^4} - 4{x^2} + 12x - 9 = {\left( 1 \right)^4} - 4{\left( 1 \right)^2} + 12\left( 1 \right) - 9 \\
   \Rightarrow {x^4} - 4{x^2} + 12x - 9 = 1 - 4 + 12 - 9 \\
   \Rightarrow {x^4} - 4{x^2} + 12x - 9 = 13 - 13 \\
   \Rightarrow {x^4} - 4{x^2} + 12x - 9 = 0 \\
 $
Clearly, by putting x=1 in $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$, the value of this polynomial reduces to zero.
Therefore, $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$ is exactly divisible by $\left( {{x^2} + 2x - 3} \right)$.

Note: In this particular problem, in order to make $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$ to be exactly divisible by $\left( {{x^2} + 2x - 3} \right)$ the solutions corresponding to this quadratic equation ${x^2} + 2x - 3 = 0$ should make the remainder zero when these solutions or values of x are substituted in $\left( {{x^4} - 4{x^2} + 12x - 9} \right)$. Here, we are prohibited to use the long division method.