What is (((x^2-5)(x^2-4)))/((x-1))<= 0 ?

The solution to (((x^2-5)(x^2-4)))/((x-1))<= 0 is x<=-sqrt(5)\lor-2<= x<1\lor 2<= x<= sqrt(5)

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(((x^2-5)(x^2-4)))/((x-1))<= 0

\frac{( ( x ^{2} - 5 ) ( x ^{2} - 4 ) )}{( x - 1 )} \leq 0

x \leq - 5 or - 2 \leq x < 1 or 2 \leq x \leq 5

Interval Notation

(- \infty, - 5] \cup [- 2, 1) \cup [2, 5]

Decimal Notation

x \leq - 2.23606 \dots or - 2 \leq x < 1 or 2 \leq x \leq 2.23606 \dots

Solution steps

\frac{( x ^{2} - 5 ) ( x ^{2} - 4 )}{x - 1} \leq 0

Factor

\frac{( x ^{2} - 5 ) ( x ^{2} - 4 )}{x - 1} : \frac{( x + 2 ) ( x - 2 ) ( x + 5 ) ( x - 5 )}{x - 1}

\frac{( x + 2 ) ( x - 2 ) ( x + 5 ) ( x - 5 )}{x - 1} \leq 0

Identify the intervals

x \leq - 5 or - 2 \leq x < 1 or 2 \leq x \leq 5

\frac{( x ^{2} - 5 x + 12 )}{( x ^{2} - 4 x + 5 )} > 3

\frac{( 2 x - x ^{2} )}{( ( 3 + x ) ( 5 - x ) )} < 0

\frac{1}{( x - 1 )} > \frac{1}{( x + 1 )}

\frac{( x - 1 ) ( x - 5 )}{( x - 3 )} > 0

\frac{1}{x} + \frac{1}{x ^{2}} > 0

What is (((x^2-5)(x^2-4)))/((x-1))<= 0 ?
The solution to (((x^2-5)(x^2-4)))/((x-1))<= 0 is x<=-sqrt(5)\lor-2<= x<1\lor 2<= x<= sqrt(5)