Question

Solve the equation: $ x(x - 1)(x + 2)(x - 3) + 8 = 0 $

Answer 1

Hint: The roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. The roots of an equation can be found out by factorizing the equation and also by a special formula called completing the square method. So, we have to first simplify the given equation and then find out its roots.

Complete step-by-step answer:
We are given that $ x(x - 1)(x + 2)(x - 3) + 8 = 0 $
Multiplying the terms on the left-hand side with each other, we get –
 $
  x(x - 1)(x + 2)(x - 3) + 8 = 0 \\
  ({x^2} - x)(x + 2)(x - 3) + 8 = 0 \\
  ({x^3} + 2{x^2} - {x^2} - 2x)(x - 3) + 8 = 0 \\
  ({x^3} + {x^2} - 2x)(x - 3) + 8 = 0 \\
  {x^4} - 3{x^3} + {x^3} - 3{x^2} - 2{x^2} + 6x + 8 = 0 \\
  {x^4} - 2{x^3} - 5{x^2} + 6x + 8 = 0 \;
  $
Now, the above equation has degree four, so it has four solutions. The equation can’t be solved by factorization easily, so we use the hit and trial method for finding one of the roots of the equation.
Let $ f(x) = x(x - 1)(x + 2)(x - 3) + 8 $
At x=1,
 $
  f(x) = 1(1 - 1)(1 + 2)(1 - 3) + 8 \\
  f(x) = 1(0)(3)( - 2) + 8 \\
  f(x) = 8 \;
  $
That is $ f(x) \ne 0 $ , thus x=1 is not the root of the equation.
At x=2
 $
  f(x) = 2(2 - 1)(2 + 2)(2 - 3) + 8 \\
  f(x) = 2(1)(4)( - 1) + 8 \\
  f(x) = - 8 + 8 = 0 \;
  $
That is $ f(x) = 0 $ , hence $ x - 2 = 0 $ is the factor of the above equation. So we divide the given equation by $ x - 2 $ .
On dividing, we get –
 $ f(x) = (x - 2)({x^3} - 5x - 4) $
We know that
 $
  {x^4} - 2{x^3} - 5{x^2} + 6x + 8 = 0 \\
  (x - 2)({x^3} - 5x - 4) = 0 \\
   \Rightarrow {x^3} - 5x - 4 = 0 \;
  $
For factorizing the above equation we have to add and subtract $ {x^2} $ and rewrite the terms of the equation as shown –
 $
  {x^3} + {x^2} - {x^2} - x - 4x - 4 = 0 \\
  {x^2}(x + 1) - x(x + 1) - 4(x + 1) = 0 \\
  (x + 1)({x^2} - x - 4) = 0 \\
   \Rightarrow x = - 1,\,{x^2} - x - 4 = 0 \;
  $
So, $ x = -1 $ is another root of the equation. To find the rest of the two roots, we factorize $ {x^2} - x - 4 = 0 $ , but is factorization is not possible so we solve this equation by completing the square method –
 $
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
  x = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4(1)( - 4)} }}{{2(1)}} = \dfrac{{1 \pm \sqrt {1 + 16} }}{2} \\
  x = \dfrac{{1 \pm \sqrt {17} }}{2} \\
   \Rightarrow x = \dfrac{{1 + \sqrt {17} }}{2},\,x = \dfrac{{1 - \sqrt {17} }}{2} \;
  $
Hence, the four roots of the given equation are –
 $ x = 2,\,x = - 1,\,x = \dfrac{{1 + \sqrt {17} }}{2},\,x = \dfrac{{1 - \sqrt {17} }}{2} $
So, the correct answer is “ $ x = 2,\,x = - 1,\,x = \dfrac{{1 + \sqrt {17} }}{2},\,x = \dfrac{{1 - \sqrt {17} }}{2} $ ”.

Note: The points on the x-axis at which the y-coordinate of the function is zero are called the roots of the equation. In a polynomial equation, the highest exponent of the polynomial is called its degree. And according to the Fundamental Theorem of Algebra, a polynomial equation has exactly as many roots as its degree.