Write the quartic equation $4x^4+12x^3-35x^2-300x+625$ as the product of two quadratic expressions. Write the quartic expression 
$4x^4+12x^3-35x^2-300x+625$ 
as the product of two quadratic expressions with real coefficients.
I would like to know the way of solving it other than solving with the simultaneous equation. 
 A: From the rational root test we get $x=5/2$ as a double root.
Thus we have a factor of $(x-2.5)^2 = x^2-5x+6.25$
Upon dividing by $x^2-5x+6.25$ and bringing $4$ into the first factor we can factor it as
$$4x^4+12x^3-35x^2-300x+625=(4x^2-20x+25)(x^2+8x+25)$$
but obviously you do not have real roots for the second one.
I do not know if it is even possible to factor it as required   
A: By the rational root theorem, if there is a rational root $p/q$ with $\gcd(p,q)=1$ then $p$ divides $625=5^4$ and $q$ divides $4=2^2$.
Note that if you find such root then $(qx-p)$ divides your polynomial.
Here we are quite lucky because it turns out $p/q=5/2$ is a double root.
So it suffices to divide your polynomial by $(2x-5)^2$ and you will find the second quadratic factor.
A: By the Ruffini's theorem, I have that $x=\frac{5}{2}$ is a rational zero of the polynomial. So, dividing by $x-\frac{5}{2}$, I have: $P(x)=(x-\frac{5}{2})(4x^3+22x^2+20x-250)=(x-\frac{5}{2})$. Using another time Ruffini's theorem, I have $\frac{5}{2}$ a zero, so $P(x)$ becomes: $$P(x)=2(x-\frac{5}{2})^2(x^2+8x+25)$$ $x^2+8x+25$ is irriducibile in $R$, so the factorization stop.
A: Let
$$f(x)=4x^4+12x^3-35x^2-300x+625$$
Use Rational Root Theorem to find at least 1 root, using the theorem, we can see that:
$a_0=625$, $a_4=4$
The dividers of $a_0=625$ are: $1, 5, 25, 125 , 625$
The dividers of $a_4$=4 are: $1,2 , 4$
So, we check the following numbers: +/- $\frac{1,5,25,125,625}{1,2,4}$.
Out of the above step, we discover that $f(\frac{5}{2})=0$.
Notice that $f'(x)=16x^3+36x^2-70x-300=0$ has a real root at $\frac{5}{2}$ also.
This hints to the fact that $\frac{5}{2}$ is a double root for the main equation $f(x)$.
Since $\frac{5}{2}$ is a root then $(2x-5)$ is a factor.
and since its a repeated root, then $(2x-5)(2x-5)$ are factors.
You could now divide f(x) by its factors, to get the other a second degree equation with complex roots
$$\frac{4x^4+12x^3-35x^2-300x+625}{\left(2x-5\right)\left(2x-5\right)}=x^2+8x+25$$
Now we can write:
$$4x^4+12x^3-35x^2-300x+625=(2x-5)^{2}(x^2+8x+25)$$
Note that maybe a numerical solution could get you the first root faster.
