Question

# Solve the following equations:$\left( {x + 9} \right)\left( {x - 3} \right)\left( {x - 7} \right)\left( {x + 5} \right) = 385$

Hint: In this question first multiply two factors together and the remaining two together so that it converts into a quadratic equation then substitute the same part to any other variable and multiply later on to apply the quadratic formula, so use these concepts to get the solution of the question.

Given equation is
$\left( {x + 9} \right)\left( {x - 3} \right)\left( {x - 7} \right)\left( {x + 5} \right) = 385$
Now multiply (x+9) and (x-7) together and remaining two together
$\left[ {\left( {x + 9} \right)\left( {x - 7} \right)} \right]\left[ {\left( {x - 3} \right)\left( {x + 5} \right)} \right] = 385 \\ \left[ {{x^2} + 9x - 7x - 63} \right]\left[ {{x^2} - 3x + 5x - 15} \right] = 385 \\ \left[ {{x^2} + 2x - 63} \right]\left[ {{x^2} + 2x - 15} \right] = 385 \\$
Let, $\left( {{x^2} + 2x} \right) = t...........\left( 1 \right)$
So, substitute this value in the above equation we have
$\left[ {t - 63} \right]\left[ {t - 15} \right] = 385 \\ \Rightarrow {t^2} - 15t - 63t + 945 = 385 \\ \Rightarrow {t^2} - 78t + 560 = 0 \\$
Now factorize the above equation we have,
$\Rightarrow {t^2} - 8t - 70t + 560 = 0 \\ \Rightarrow t\left( {t - 8} \right) - 70\left( {t - 8} \right) = 0 \\ \Rightarrow \left( {t - 8} \right)\left( {t - 70} \right) = 0 \\ \Rightarrow \left( {t - 8} \right) = 0{\text{ & }}\left( {t - 70} \right) = 0 \\ \Rightarrow t = 8{\text{ & }}t = 70 \\$
Now from equation (1)
$\left( {{x^2} + 2x} \right) = t \\ \Rightarrow \left( {{x^2} + 2x} \right) = 8{\text{ & }}\left( {{x^2} + 2x} \right) = 70 \\ \Rightarrow {x^2} + 2x - 8 = 0..........\left( 2 \right){\text{ & }}{x^2} + 2x - 70 = 0........\left( 3 \right) \\$
Now first solve equation (2) by factorization method we have,
$\Rightarrow {x^2} + 2x - 8 = 0 \\ \Rightarrow {x^2} + 4x - 2x - 8 = 0 \\ \Rightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \\ \Rightarrow \left( {x + 4} \right) = 0{\text{ & }}\left( {x - 2} \right) = 0 \\ \Rightarrow x = - 4{\text{ & }}x = 2 \\$
Now solve equation (3)
${x^2} + 2x - 70 = 0$
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4\left( 1 \right)\left( { - 70} \right)} }}{2} = \dfrac{{ - 2 \pm \sqrt {284} }}{2} = \dfrac{{ - 2 \pm 2\sqrt {71} }}{2} = - 1 \pm \sqrt {71}$
$x = \left( {2, - 4,\left( { - 1 \pm \sqrt {71} } \right)} \right)$