<s>[INST] Factoring $x^4 + 5x^3 + 4x^2 + 2x - 3$ 
So I have to factor this polynomial
  $$x^4 + 5x^3 + 4x^2 + 2x - 3$$

I got $(x^2 + 2x -3)(x^2 + 3x + 1)$
but when I multiplied it, I got a different equation:
$$x^4 + 5x^3 + 4x^2 - 7x - 3$$
I don’t really understand how to find the factors. I only found numbers that could add up to 5 and multiply to -3 but they’re wrong. What did I do wrong and what can I do to fix this?
In the format of: 
($x^2$+ __ x +__ ) ($x^2$ + __ x + __ )
or
$$(x^2 + ax + b) (x^2 + cx + d)$$
Also, how does comparing coefficients help factor this polynomial? 
 [/INST] The following way always works:
$$x^2+5x^3+4x^2-7x-3=\left(x^2+\frac{5}{2}x-1\right)^2-\left(\frac{x}{2}+2\right)^2=(x^2+2x-3)(x^2+3x+1)=...$$
We can get this factoring by the following reasoning.
For any $k$ we obtain:
$$x^2+5x^3+4x^2-7x-3=\left(x^2+\frac{5}{2}x+k\right)^2-2kx^2-5kx-\frac{25}{4}x^2-k^2+4x^2-7x-3=$$
$$=\left(x^2+\frac{5}{2}x+k\right)^2-\left(\left(2k+\frac{9}{4}\right)x^2+(5k+7)x+k^2+3\right).$$
Now, we need to choose a value of $k$ for which $2k+\frac{9}{4}\geq0$ and $\left(2k+\frac{9}{4}\right)x^2+(5k+7)x+k^2+3$ is a perfect square.
We immediately see that $k=-1$ is valid, but if we don't see it we can write a condition for a perfect square:
$$(5k+7)^2-4\left(2k+\frac{9}{4}\right)(k^2+3)=0$$ or
$$4k^3-8k^2-23k-11=0$$ or
$$4k^3+4k^2-12k^2-12k-11k-11=0$$ or
$$(k+1)(4k^2-12k-11)=0$$ and since $2\cdot(-1)+\frac{9}{4}>0$, we obtain the difference of squares and the needed factoring. 
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