What is (x^2+3x+1)(x^2+3x-3)>= 5 ?

The solution to (x^2+3x+1)(x^2+3x-3)>= 5 is x = 1

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(x^2+3x+1)(x^2+3x-3)>= 5

(x^{2} + 3 x + 1) (x^{2} + 3 x - 3) \geq 5

x \leq - 4 or - 2 \leq x \leq - 1 or x \geq 1

Interval Notation

(- \infty, - 4] \cup [- 2, - 1] \cup [1, \infty)

Solution steps

(x^{2} + 3 x + 1) (x^{2} + 3 x - 3) \geq 5

Rewrite in standard form

x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8 \geq 0

Factor

x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8 : (x + 1) (x - 1) (x + 2) (x + 4)

(x + 1) (x - 1) (x + 2) (x + 4) \geq 0

Identify the intervals

x \leq - 4 or - 2 \leq x \leq - 1 or x \geq 1

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

x^{6} - 9 x^{3} + 8 \leq 0

\frac{( - x ^{2} - 1 )}{( 9 - 2 x )} \leq 0

\frac{( 1 + 3 x ^{2} )}{( 2 x ^{2} - 21 x + 40 )} < 0

\frac{x ^{2} + 2 x - 3}{x ^{2} + 1} < 0

What is (x^2+3x+1)(x^2+3x-3)>= 5 ?
The solution to (x^2+3x+1)(x^2+3x-3)>= 5 is x<=-4\lor-2<= x<=-1\lor x>= 1