Question

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

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.

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.