Factoring a 4th degree trinomial

Question

I am trying to factor $3x^4-8x^3+16$, but I have no idea how to even start. I put into Wolfram Alpha, and it said that the answer was $(x-2)^2 (3 x^2+4 x+4)$. How would you factor something like this by hand? Is there any way without using the quartic formula in this case?

Answer 1

You can use the rational root theorem: if this polynomial has a rational root, say $p/q$ in irreducible form, $p$ is a divisor of the constant term and $q$ a divisor of the leading coefficient (this is because it has integer coefficients). This makes a finite number of possibilities: $p=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16$ and $q=1$ or $3$. Furthermore, it can't have a negative root, since if $x<0$, $f(x)\ge 16$.

One checks at once that $2$ is a root. Thus it is divisible by $x-2$. Dividing, you obtain $$3x^4-8x^3+16=(x-2)(3x^3-2x^2-4x-8).$$ The second factor also has $2$ as a root. Dividing again by $x-2$, one obtains the given factorisation.

Answer 2

You can factor it if you know its roots. In general if $r$ is a root of a polynomial $f(x)$, then $(x-r)$ divides it.

For example, if you see that $f(x)=3 x^4−8 x^3+16$ has the root $2$, that is, $3(2)^4-8(2)^3+16=48-64+16=0$, then you know that $(x-2)$ divides it. You can then use long division to show that $3 x^4−8 x^3+16= (x-2)(3x^3-2x^2-4x-8) $. Now noticing that $2$ is also a root of $3x^3-2x^2-4x-8$ we have that $(x-2)$ divides that also. By long division $3x^3-2x^2-4x-8 = (x-2)(3x^2+4x+4)$. Now $f(x)=(x-2)^2(3x^2+4x+4)$.

So what you can do is first look for some roots using various methods.

Answer 3

-1 to @lhf's comment-I think that instead of the formula that you said,my friend,the best thing is to use the formula $(a^2-b^2)$ for this problem where a=$(2x^2+4)$ and b=$(x^2+4x)$ to get the following equation which is given as: $4x^4+16x^2+16-x^4-8x^3-16x^2=3x^4-8x^3+16$