What is (4x^2-4x-8)/(10-2(2x+1))<= 0 ?

The solution to (4x^2-4x-8)/(10-2(2x+1)) =-1

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(4x^2-4x-8)/(10-2(2x+1))<= 0

\frac{4 x ^{2} - 4 x - 8}{10 - 2 ( 2 x + 1 )} \leq 0

x \geq - 1

Interval Notation

[- 1, \infty)

Solution steps

\frac{4 x ^{2} - 4 x - 8}{10 - 2 ( 2 x + 1 )} \leq 0

Factor

\frac{4 x ^{2} - 4 x - 8}{10 - 2 ( 2 x + 1 )} : \frac{4 ( x + 1 ) ( x - 2 )}{- 4 ( x - 2 )}

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

x + 1 \geq 0

Identify the intervals

x \geq - 1

\frac{m}{2} = 10

i^{17}

∣ x ∣ > 8

9 + 6

∣ 8 - 4 x ∣ < - 13

What is (4x^2-4x-8)/(10-2(2x+1))<= 0 ?
The solution to (4x^2-4x-8)/(10-2(2x+1))<= 0 is x>=-1