Question

solve for xe ________​

Answer 1

Answer:

$\frac{\left(x+1\right)\left(x^2+x+1\right)\cdot \left(x^2-4x+5\right)}{\left(x^2-4x+3\right)} \leq 0$

$\left(x+1\right)\left(x^2+x+1\right)\cdot \left(x^2-4x+5\right)\leq 0$

$x^5-2x^4-x^3+3x^2+6x+5\leq 0$

Best way to find the value of x is by Trial and Error Method since the equation is too complex

Let's take x as 1

$1^5-2\left(1\right)^4-1^3+3\left(1\right)^2+6\left(1\right)+5\\1-2-1+3+6+5\\12 \neq 0$

$x\neq 1$

Let's take x as (-1)

$-1^5-2\left(-1\right)^4-\left(-1\right)^3+3\left(-1\right)^2+6\left(-1\right)+5\\-1-2+1+3-6+5\\0=0$

∴, $x\leq -1$