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}^{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{aligned}}\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{aligned} & {{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{aligned}}\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.