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? 
 A: You must have $bd= -3$ so $b= -3$ and $d=-1$ can not be. You must change the signs. 
Write $$x^3:\;\;\;\;a+c= 5$$ $$x^2:\;\;\;\;ac+b+d =4$$ $$x:\;\;\;\;ad+cb =-7$$
and solve this system...
A: As said in comments, there must be a typo in the question since it does not factorize.
If you write
$$(x^2 + ax + b) (x^2 + cx + d)=x^4 + 5x^3 + 4x^2 + 2x - 3$$ expanding and grouping terms, we end with
$$(b d+3)+x (a d+b c-2)+x^2 (a c+b+d-4)+x^3 (a+c-5)=0$$ and each coefficient must be zero.
So, trying, we have $d=-\frac 3b$, $c=5-a$ but the remaining equations are nightmares since we get
$$a=\frac{b (5 b-2)}{b^2+3}$$ and what is left to solve  for $b$ is
$$-27-36 b-39 b^2+47 b^3+13 b^4-4 b^5+b^6=0$$ which does not show any obvious roots (in fact, there are two real non-rational solutions).
If you had $b$, then all the other are available. 
A: 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. 
A: Note that you can split the four roots of a quartic into two pairs in three different ways. If you want to factor into two quadratics there are three different possibilities therefore (because the roots of the quadratics will be pairs of the original roots). Your equations, therefore, won't have a unique solution except in special cases. They may have a unique solution in integers, but this will not be readily obtainable by algebraic methods - though it may be identified by using the properties of integers and the fact that the search space is small.
Therefore, when you do a trial factorisation (the difference of two squares method illustrated by Michael Rosenberg is one of the ways to go), you expect to find a cubic which gives three different possibilities.
The factorisation of the quartic is "easy" only when the cubic has at least one "nice" root. Your method seems to depend on having an easy factorisation - most quartics don't have this.
A: $(x^2+ax+b)^2=x^4+2ax^3+a^2x^2+2bx^2+2abx+b^2$
we find:
$a=\frac{5}{2}$
$b=\frac{-9}{8}$
and we get following polynomial with these values:
$x^4+5x^3+4x^2-\frac{45}{8}x+\frac{81}{64}$
We compare this with given polynomial; if: 
$x^4+5x^3+4x^2+2x-3=0$
then:
$(x^2+\frac{5}{4}x-\frac{9}{8})^2=\frac{273}{64}-\frac{61}{8}x$
This equation has two real roots ; $x_1≈-4.202$ and $x_2≈0.53$ and two complex roots ; $x_3≈ -0.66-0.95 i$ and $x_3≈ -0.66+0.95 i$, so the factorized form can be:
$(x-0.52)(x+4.2)(x+0.66+095 i)(x-0.66+095 i)$
You can get the same result from Wolfram.
