Inequality $(x^2 + 3x + 1)\cdot(x^2 + 3x - 3) \geq 5$

Question

This is the question I have in front of me. Clearly, I need to find the range of values of $x$ for the given inequality.

Having taken a cue from a similar question, here's how I approached it -

Let $ a = x^2 + 3x + 1$ $\implies ax^2 + 3ax - (3a + 5) \geq 0$

Now, for this quadratic expression to be greater than or equal to $0$, the coefficient of $x^2$ must be $>0$, with the discriminant being equal to $0$. But, the coefficient of $x^2$ i.e. $a = x^2 + 3x + 1$ can take negative values also, which does not ensure the requirement.

So, how shall I go about this one? Any help will be appreciated. Thanks.

Answer 1

not quite. Let $$ b = x^2 + 3 x - 1 $$ $$ (b+2)(b - 2) \geq 5 $$ $$ b^2 \geq 9 $$ $$ b \leq -3 \; \; \mbox{OR} \; \; b \geq 3 $$

Answer 2

hint

Observe that if

$$f (x)=(x^2+3x+1)(x^2+3x-3) $$

then $$f (1)=f (-1)=f (-2)=f (-4)=5$$ thus

$$f (x)-5=$$ $$(x-1)(x+1)(x+2)(x+4) $$