Solve the following equations: $\left( x-7 \right)\left( x-3 \right)\left( x+5 \right)\left( x+1 \right)=1680$.

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Hint: Take the average of all brackets given in the LHS of the equation and take it as another variable and form a new biquadratic equation. Now, solve it further to get values of that variable and hence values of ‘x’ as well.

Complete step-by-step answer:
The given equation is,
$\left( x-7 \right)\left( x-3 \right)\left( x+5 \right)\left( x+1 \right)=1680$…………….. (i)
Let us change the given variables ‘x’ by taking average of all brackets as a variables ’y’ in following way:-
$\Rightarrow$ $y=\dfrac{\left( x-7 \right)+\left( x-3 \right)+\left( x+5 \right)+\left( x+1 \right)}{4}$
$\Rightarrow$ $y=\dfrac{4x-4}{4}=x-1$
Hence, we get the value ’y’ as ‘x-1’. So, now ‘x’ can be written in form ‘y’ in following way:-
Let us replace variable ‘x’ from equation (i) with the help of equation (ii) hence we get,
$\Rightarrow$ $\left( y+1-7 \right)\left( y+1-3 \right)\left( y+1+5 \right)\left( y+1+1 \right)=1680$
$\Rightarrow$ \[\left( y-6 \right)\left( y-2 \right)\left( y+6 \right)\left( y+2 \right)=1680\]

Rewriting the above equation we get,
$\Rightarrow$ $\left( y-6 \right)\left( y+6 \right)\left( y-2 \right)\left( y+2 \right)=1680$…………. (iii)
Now, we can use an algebraic identity given as
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
So, we can simplify the equation (iii) using the above relation, we get
$\Rightarrow$ $\left( {{y}^{2}}-36 \right)\left( {{y}^{2}}-4 \right)=1680$……………. (iv)
Let ${{y}^{2}}=t$, so that we can solve the above equation in a simpler way. Hence, we get
$\Rightarrow$ $\left( t-36 \right)\left( t-4 \right)=1680$
Now, multiplying the both brackets, we get
  & {{t}^{2}}-4t-36t+36\times 4=1680 \\
 & {{t}^{2}}-40t+144=1680 \\
${{t}^{2}}-40t-1536=0$………. (v)
Now we can get values of ‘t’ from the above equation by using quadratic formula. We know roots of quadratic $A{{x}^{2}}+Bx+C=0$ can be given as
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$
Hence, value of ‘t’ from equation (V) be
  & t=\dfrac{-(-40)\pm \sqrt{{{(-40)}^{2}}-4\times 1\times (-1536)}}{2\times 1} \\
 & t=\dfrac{40\pm \sqrt{1600+4\times 1536}}{2} \\
 & t=\dfrac{40\pm \sqrt{7744}}{2} \\
Now, we can calculate the square root of 7744. So, we get
$t=\dfrac{40\pm 88}{2}$
So, two values of ‘t’ can be given as
t=64 or t=-24
We had assumed $t={{y}^{2}}$; hence, we get
 As we know squares of any number cannot be negative. ${{y}^{2}}$can never be -24.
So, we can ignore it. Hence,
Taking square root on both sides, we get
$y=\pm 8$……………(vi)
Now, we can use relation x=y+1 from equation (ii) to get values of ‘x’. Hence,
x =8+1 and x=-8+1
x =9 and x=-7
So, values of ’x’ satisfying the given equation are 9, -7.

Note: One can think that a given polynomial is of degree’4’ so why is there only two roots 9 and -7. We can get two values of x from the relation ${{x}^{2}}=-24$which will be imaginary and will be in terms ‘i’. So imaginary roots can be given as
So, there are two real and two imaginary roots of the given equation.
One can multiply the given brackets in the equation and can form a 4 degree polynomial. But guessing the roots 9 and -7 is very complex and difficult. So, take the average of all brackets and replace the given variable with another in these kinds of questions.
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