Question

Solve the equation
\[x\left( x-1 \right)\left( x+2 \right)\left( x-3 \right)+8=0\]

Answer 1

Hint:First of all simplify the equation and form a quadratic equation. Now by hit and trial method, find the value of x which satisfies the equation. Now divide that factor by the given equation to find other factors. Keep repeating this method until you factorize the whole equation.

Complete step-by-step answer:
In this question, we have to solve the equation \[x\left( x-1 \right)\left( x+2 \right)\left( x-3 \right)+8=0\] and find the value of x. Let us consider the equation given in the question.
\[f\left( x \right)=x\left( x-1 \right)\left( x+2 \right)\left( x-3 \right)+8=0\]
By simplifying it, we get,
\[f\left( x \right)=\left( {{x}^{2}}-x \right)\left( {{x}^{2}}-x-6 \right)+8=0\]
\[f\left( x \right)={{x}^{4}}-2{{x}^{3}}-5{{x}^{2}}+6x+8=0\]
Now, start substituting the value of x = 0, 1, – 1, 2, – 2, etc. at which the value of the equation is 0 to find the factor. By hit and trial method, we substitute the value of x = 2, we get,
\[f\left( x \right)={{\left( 2 \right)}^{4}}-2{{\left( 2 \right)}^{3}}-5{{\left( 2 \right)}^{2}}+6\left( 2 \right)+8\]
\[f\left( x \right)=16-16-20+12+8\]
\[f\left( x \right)=0\]
We have found that for x = 2, f (x) = 0. So, we get, (x – 2) as a factor of f(x). So by dividing f (x) by (x – 2), we get,
\[x-2\overset{{{x}^{3}}-5x-4}{\overline{\left){\begin{align}
  & {{x}^{4}}-2{{x}^{3}}-5{{x}^{2}}+6x+8 \\
 & \underline{-{{x}^{4}}-2{{x}^{3}}} \\
 & \text{ }0\text{ 0}-5{{x}^{2}}+6x+8 \\
 & \text{ }-5{{x}^{2}}+10x \\
 & \text{ }\underline{+\text{ }-\text{ }} \\
 & \text{ }0\text{ }-4x+8 \\
 & \text{ }-4x+8 \\
 & \text{ }\underline{+\text{ }-\text{ }} \\
 & \text{ 0} \\
\end{align}}\right.}}\]
So, we get, \[f\left( x \right)=\left( x-2 \right)\left( {{x}^{3}}-5x-4 \right)=0\] and \[h\left( x \right)={{x}^{3}}-5x-4\]
Now, to further factorize f(x), we will again start substituting different values of x like 0, 1, – 1, etc.
By hit and trial method, we have found that for x = – 1,
\[h\left( x \right)={{\left( -1 \right)}^{3}}-5\left( -1 \right)-4\]
\[h\left( x \right)=-1+5-4=0\]
So f (x) is also equal to zero at x = – 1. So, x + 1 is a factor of h(x) as well as f(x).
Now by dividing h(x) by (x+1), we get,
\[\left( x+1 \right)\overset{{{x}^{2}}-x-4}{\overline{\left){\begin{align}
  & {{x}^{3}}-5x-4 \\
 & \underline{-{{x}^{3}}} \\
 & 0\text{ }-{{x}^{2}}-5x-4 \\
 & \text{ }-{{x}^{2}}-x \\
 & \text{ }\underline{+\text{ }-} \\
 & \text{ }0\text{ }-4x-4 \\
 & \text{ }-4x-4 \\
 & \text{ }\underline{+\text{ }+} \\
 & \text{ }0\text{ }0 \\
\end{align}}\right.}}\]
So we get, \[h\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-x-4 \right)\] and \[f\left( x \right)=\left( x-2 \right)\left( x+1 \right)\left( {{x}^{2}}-x-4 \right)=0\]
So, we get x = 2 or x = – 1 or \[{{x}^{2}}-x-4=0\].
Now, to find x term of the equation \[{{x}^{2}}-x-4=0\], we use the quadratic formula where
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
So, we get,
\[x=\dfrac{-\left( -1 \right)\pm \sqrt{1+16}}{2}\]
\[x=\dfrac{1\pm \sqrt{17}}{2}\]
So, we finally get the values of x from the given equation as
\[x=2,-1,\dfrac{1+\sqrt{17}}{2},\dfrac{1-\sqrt{17}}{2}\]

Note: In this question, many students make this mistake of writing the given equation as x (x – 1) (x + 2) (x – 3) = – 8 and then substituting x – 1 = – 8 and (x + 2) = – 8 and (x – 3) = – 8 and x = – 8 which is absolutely wrong because this method is only used when there is 0 in place of 8. So this mistake must be avoided. Also, students must cross-check their answers by substituting the values of x in the initial equation and checking if it is satisfied or not.