Find the value of p for which the function f(x)=$\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^{5}-3x+ln5$ decreases $\forall$x $\in\mathbb{R}$

Question Find the value of p for which the function f(x)=$\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^{5}-3x+ln5$ decreases $\forall$x $\in\mathbb{R}$

MY Approach$\Longrightarrow$f'$\left(x\right)$= $\left(\frac{\sqrt{p+4}}{1-p}-1\right)5x^{4}-3$$\leq$0$\Longrightarrow$$\left(\frac{\sqrt{p+4}}{1-p}-1\right)\leq$$\frac{3}{5x^{4}}$$\Longrightarrow$$\left(\frac{\sqrt{p+4}}{1-p}-1\right)<0$$\Longrightarrow$x $\in$$\left(-\infty,\frac{3-\sqrt{21}}{2}\right)$$\bigcup$ $\left(\frac{3+\sqrt{21}}{2},\infty\right)$

But Book Says Answer is $\left[-4,\frac{3-\sqrt{21}}{2}\right]$$\bigcup$$\left(1,\infty\right)$

$$\frac{\sqrt{p+4}}{1-p}\le 1$$ if i square this equation i do not get the condition x belongs in (1,0) but $$\frac{\sqrt{p+4}}{1}\le1-p $$ this eqaution gives the condition x belongs in (1,0).How to know the correct approach