What is (x^2-7x+6)/(x+3)<= 0 ?

The solution to (x^2-7x+6)/(x+3)<= 0 is x<-3\lor 1<= x<= 6

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(x^2-7x+6)/(x+3)<= 0

\frac{x ^{2} - 7 x + 6}{x + 3} \leq 0

x < - 3 or 1 \leq x \leq 6

Interval Notation

(- \infty, - 3) \cup [1, 6]

Solution steps

\frac{x ^{2} - 7 x + 6}{x + 3} \leq 0

Factor

\frac{x ^{2} - 7 x + 6}{x + 3} : \frac{( x - 1 ) ( x - 6 )}{x + 3}

\frac{( x - 1 ) ( x - 6 )}{x + 3} \leq 0

Identify the intervals

x < - 3 or 1 \leq x \leq 6

2 (x - 3) (2 x + 1)^{4} + 4 (x - 3)^{2} (2 x + 1)^{3}

\frac{x ^{4}}{x ^{5}}

\frac{4}{\frac{5}{6}}

6 r^{2} t - 3 r t^{2}

0 \leq x \leq 8

What is (x^2-7x+6)/(x+3)<= 0 ?
The solution to (x^2-7x+6)/(x+3)<= 0 is x<-3\lor 1<= x<= 6