What is ((6-2x))/((x^2+5))>= 0 ?

The solution to ((6-2x))/((x^2+5))>= 0 is x<= 3

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((6-2x))/((x^2+5))>= 0

\frac{( 6 - 2 x )}{( x ^{2} + 5 )} \geq 0

x \leq 3

Interval Notation

(- \infty, 3]

Solution steps

\frac{6 - 2 x}{x ^{2} + 5} \geq 0

Factor

\frac{6 - 2 x}{x ^{2} + 5} : \frac{- 2 ( x - 3 )}{x ^{2} + 5}

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

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 2 ( x - 3 ) ) ( - 1 )}{x ^{2} + 5} \leq 0 \cdot (- 1)

Simplify

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

Divide both sides by

2

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

Simplify

\frac{x - 3}{x ^{2} + 5} \leq 0

Identify the intervals

x \leq 3

\frac{( x ^{3} + 1 )}{( x ^{4} + x ^{3} - 10 x ^{2} - 13 x + 21 )} > 0

(x + 1) (x^{2} + x - 1) < (x + 1)

(x - 1)^{2} \cdot (x + 1)^{3} \cdot (x - 4) \geq 0

4 x^{3} - 3 x^{2} - 16 x + 12 < 0

\frac{( m + 1 )}{m} \leq 0

What is ((6-2x))/((x^2+5))>= 0 ?
The solution to ((6-2x))/((x^2+5))>= 0 is x<= 3