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Solve \[\left( x-5 \right)\left( x-7 \right)\left( x+4 \right)\left( x+6 \right)=504\].

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Hint: Find the value of x; take the value of \[\left( {{x}^{2}}-x \right)\]as t.
Find the roots of t and the substitute it in \[t={{x}^{2}}-x\].
Solve the quadratic equations formed and you will get 4 values of x.

Complete step-by-step answer:
Given \[\left( x-5 \right)\left( x-7 \right)\left( x+4 \right)\left( x+6 \right)=504\]
We can consider \[\left[ \left( x-5 \right)\left( x+4 \right) \right]\]together and \[\left[ \left( x-7 \right)\left( x+6 \right) \right]\].
Open the bracket and form a quadratic equation of the form\[a{{x}^{2}}+bx+c\].
\[\begin{align}
  & \left[ \left( x-5 \right)\left( x+4 \right) \right]\left[ \left( x-7 \right)\left( x+6 \right) \right]=504-(1) \\
 & \left( x-5 \right)\left( x+4 \right)={{x}^{2}}-5x+4x-20={{x}^{2}}-x-20 \\
 & \left( x-7 \right)\left( x+6 \right)={{x}^{2}}-7x-6x-42={{x}^{2}}-x-42 \\
\end{align}\]
Put \[t={{x}^{2}}-x-(2)\]
\[\begin{align}
  & \Rightarrow \left( t-20 \right)\left( t-42 \right)=504 \\
 & \Rightarrow {{t}^{2}}-20t-42t+840-504=0 \\
 & \Rightarrow {{t}^{2}}-62t+336=0-(3) \\
\end{align}\]
By substituting value of t in equation (1), we can simplify it to equation (3)
Now we get a quadratic equation \[{{t}^{2}}-62t+336=0\].
The general form \[a{{x}^{2}}+bx+c=0\], by comparing the general form and equation (3), we get
a=1, b=-62 and c=336
substituting these values in \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], quadratic formula we get the roots.


\[\begin{align}
  & =\dfrac{-\left( -62 \right)\pm \sqrt{{{\left( -62 \right)}^{2}}-4\times 1\times 336}}{2}=\dfrac{62\pm \sqrt{3844-1344}}{2} \\
 & =\dfrac{62\pm \sqrt{2500}}{2}=\dfrac{62\pm 50}{2} \\
\end{align}\]
The roots are \[\left( \dfrac{62+50}{2} \right)\]and \[\left( \dfrac{62-50}{2} \right)\]= 56 and 6
\[\therefore \]Roots of t = 56 and 6
We know, \[t={{x}^{2}}-x\].
Put the values of t = 56.
\[\Rightarrow {{x}^{2}}-x-56=0-(4)\]
Now find the roots of equation (4) by using quadratic equation
a=1, b = -1, c = -56
\[\begin{align}
  & =\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times \left( -56 \right)}}{2}=\dfrac{-1\pm \sqrt{1+224}}{2} \\
 & =\dfrac{1\pm \sqrt{225}}{2}=\dfrac{1\pm 15}{2} \\
\end{align}\]
The roots are \[\left( \dfrac{1+15}{2} \right)\]and \[\left( \dfrac{1-15}{2} \right)\]= 8 and -7
Similarly, \[t={{x}^{2}}-x\] , put value of t = 6
\[\Rightarrow {{x}^{2}}-x-6=0\]
a = 1, b = -1, c = -6
 \[\begin{align}
  & =\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times \left( -6 \right)}}{2}=\dfrac{-1\pm \sqrt{1+24}}{2} \\
 & =\dfrac{1\pm \sqrt{25}}{2}=\dfrac{1\pm 5}{2} \\
\end{align}\]
The roots are \[\left( \dfrac{1+5}{2} \right)\]and \[\left( \dfrac{1-5}{2} \right)\]= 3 and -2.
\[\therefore \]The values of x are 8, -7, 3 and -2.

Note: The pair to be multiplied should be chosen in a way that \[t={{x}^{2}}-x\]. Taking \[\left( x-5 \right)\left( x-7 \right)\]and \[\left( x+4 \right)\left( x+6 \right)\]won’t give the required answer. Therefore, we choose \[\left( x-5 \right)\left( x+4 \right)\]and \[\left( x-7 \right)\left( x+6 \right)\], while forming the equation to get the value of x. Solving the value of t to get the roots.
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