What is (x+7)(8+2x-x^2)+(x^2+3x-28)>0 ?

The solution to (x+7)(8+2x-x^2)+(x^2+3x-28)>0 is x<-7\lor-1<x<4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x+7)(8+2x-x^2)+(x^2+3x-28)>0

(x + 7) (8 + 2 x - x^{2}) + (x^{2} + 3 x - 28) > 0

x < - 7 or - 1 < x < 4

Interval Notation

(- \infty, - 7) \cup (- 1, 4)

Solution steps

(x + 7) (8 + 2 x - x^{2}) + x^{2} + 3 x - 28 > 0

Factor

(x + 7) (8 + 2 x - x^{2}) + x^{2} + 3 x - 28 : - (x + 7) (x - 4) (x + 1)

- (x + 7) (x - 4) (x + 1) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x + 7) (x - 4) (x + 1)) (- 1) < 0 \cdot (- 1)

Simplify

(x + 7) (x - 4) (x + 1) < 0

Identify the intervals

x < - 7 or - 1 < x < 4

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

(x - 8) (x - 7) < 0

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

1 - x^{3} < 0

2 > \frac{( 6 x + 28 )}{( x + 2 )}

What is (x+7)(8+2x-x^2)+(x^2+3x-28)>0 ?
The solution to (x+7)(8+2x-x^2)+(x^2+3x-28)>0 is x<-7\lor-1<x<4