Solve the following equation:
$(x - 1) (x - 2) (x - 3) (x - 4) = 15$

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Solution

$(x - 1) (x - 2) (x - 3) (x - 4) = 15$
$(x^{2} - 5 x + 4) (x^{2} - 5 x + 6) = 15$
Let $x^{2} - 5 x$ be $t$ .
$(t + 4) (t + 6) = 15$
$t^{2} + 10 t + 24 - 15 = 0$
$t^{2} + 10 t + 9 = 0$
$(t + 9) (t + 1) = 0$
$t = - 9, - 1$
When $t = - 1$
$x^{2} - 5 x = - 1$
$x^{2} - 5 x + 1 = 0$
$x = \frac{5 \pm \sqrt{21}}{2}$
When $t = - 9$
$x^{2} - 5 x = - 9$
$x^{2} - 5 x + 9 = 0$
$x = \frac{5 \pm 2 \sqrt{2}}{2}$

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Solve the following equation:(x−1)(x−2)(x−3)(x−4)=15

(x−1)(x−2)(x−3)(x−4)=15(x2−5x+4)(x2−5x+6)=15Let x2−5x be t.(t+4)(t+6)=15t2+10t+24−15=0t2+10t+9=0(t+9)(t+1)=0t=−9,−1When t=−1x2−5x=−1x2−5x+1=0x=5±√212When t=−9x2−5x=−9x2−5x+9=0x=5±2√22

Solve the following equation:
$(x - 1) (x - 2) (x - 3) (x - 4) = 15$