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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:-
x=y+1………………(ii)
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
$\begin{align}
  & {{t}^{2}}-4t-36t+36\times 4=1680 \\
 & {{t}^{2}}-40t+144=1680 \\
\end{align}$
${{t}^{2}}-40t-1536=0$………. (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
$\begin{align}
  & t=\dfrac{-(-40)\pm \sqrt{{{(-40)}^{2}}-4\times 1\times (-1536)}}{2\times 1} \\
 & t=\dfrac{40\pm \sqrt{1600+4\times 1536}}{2} \\
 & t=\dfrac{40\pm \sqrt{7744}}{2} \\
\end{align}$
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
$t=\dfrac{40+88}{2}$or$t=\dfrac{40-88}{2}$
Or
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,
${{y}^{2}}=64$
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
$x=2\sqrt{6i}$and$x=-2\sqrt{6i}$
Where,
\[i=\sqrt{-1}\]
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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