Write the quartic equation $4x^4+12x^3-35x^2-300x+625$ as the product of two quadratic expressions.

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

I like the following way.

For any real $k$ we have: $$4x^4+12x^3-35x^2-300x+625=$$ $$=(2x^2+3x+k)^2-4kx^2-6kx-9x^2-k^2-35x^2-300x+625=$$ $$=(2x^2+3x+k)^2-((4k+44)x^2+6(k+50)x+k^2-625).$$ Now, we'll choose a value of $k$ such that

$4k+44>0$ and $(4k+44)x^2+6(k+50)x+k^2-625$ is a perfect square, for which we need $$9(k+50)^2-(4k+44)(k^2-625)=0$$ or $$4k^3+35k^2-3400k-50000=0$$ or $$4k^3+160k^2+1600k-125k^2-5000k-50000=0$$ or $$(k+20)^2(4k-125)=0$$ and we see that $k=\frac{125}{4}$ is valid.

Id est, $$4x^4+12x^3-35x^2-300x+625=\left(2x^2+3x+\frac{125}{4}\right)^2-\left(169x^2+\frac{975}{2}x+\frac{5625}{16}\right)=$$ $$=\left(2x^2+3x+\frac{125}{4}\right)^2-\left(13x+\frac{75}{4}\right)^2=\left(2x^2-10x+\frac{25}{2}\right)(2x^2+16x+50)=$$ $$=(4x^2-20x+25)(x^2+8x+25)=(2x-5)^2(x^2+8x+25).$$