# How to factor a fourth degree polynomial

I'm working on a math problem but I am having a hard time figuring out the method used by my textbook to make this factorization:

$$x^4 + 10x^3 + 39x^2 + 70x + 50 = (x^2 + 4x + 5)(x^2 + 6x + 10)$$

I've tried to see if this equation can be factored by grouping or by long division to no avail. Any help would be greatly appreciated.

• @EpsilonDelta Yes, it looks like the mistake was corrected. I meant to say 5 terms, not 5th degree.
– Loop
Jul 4 '19 at 7:38
• A related question. Jul 7 '19 at 16:22

The only really general way of which I am aware is to guess at the form of the factorization. Since it is monic (the highest term has coefficient 1), you know that the factors should also be so. Thus, there are really only 2 possible factorizations you need to think of, at least at start, which may then be further reducible through easier methods. If we denote the polynomial by $$P(x)$$, we produce the following candidate factorization equations:

1. one is factorization to a linear term and cubic term, i.e.

$$P(x) = (x + a)(x^3 + b_2 x^2 + b_1 x + b_0)$$

1. the other is factorization to two quadratic terms, i.e.

$$P(x) = (x^2 + a_1 x + a_0)(x^2 + b_1 x + b_0)$$

The "obvious" next case of this would simply result in now getting a third-degree polynomial on the left and first on the right, but that's just case 1 thanks to the commutative property, so this is exhaustive. The second case is what you have here. The first case is most easily tested and solved by a simple application of the rational root theorem which will, if it's possible, give the value for $$a$$ - followed by a polynomial long division to get the rest.

For the second case, there isn't really a much better method than to just multiply it all out:

$$(x^2 + a_1 x + a_0)(x^2 + b_1 x + b_0) = x^4 + c_3 x^3 + c_2 x^2 + c_1 x + c_0$$

where we've introduced for notational cleanliness (I had the computer multiply this out for me because it's there)

$$c_3 := a_1 + b_1$$ $$c_2 := a_0 + a_1 b_1 + b_0$$ $$c_1 := a_1 b_0 + a_0 b_1$$ $$c_0 := a_0 b_0$$

Then you just set the $$c_j$$ equal to the appropriate coefficient values read off from the terms of the given polynomial (i.e. $$c_0 = 50$$ in your given example), and try to find whole number values for $$a_j$$ and $$b_j$$ that work. You'd probably want to start with $$c_3$$ and $$c_0$$ first.

• A really good explaination!
– user482325
Jul 4 '19 at 7:47
• @Emilia314 : aww thanks :) #Mehhr. Jul 4 '19 at 7:47
• Awesome! I was able to use the value of a0b0 to help me solve for the constants. Thank you so much for your help. Is there a specific name for this method?
– Loop
Jul 4 '19 at 20:07
• @Loop : Thanks. a warm little #Mehhr to little Youzzle, too :) #mehhr. Jul 5 '19 at 1:49
• A further clue is to see that there are no sign changes in the coefficients, so there can be no positive roots. Jul 9 '19 at 19:36

Let

$$f (x) = x^4 + 10 x^3 + 39 x^2 + 70 x + 50$$

Converting to a depressed quartic see here, we see that the $$x$$ term drops out as well

$$f\left(x-\frac{5}{2}\right)=x^4+\frac{3 x^2}{2}+\frac{25}{16}=\left(x^2+\frac{5}{4}\right)^2-x^2=\left(x^2+\frac{5}{4}+x\right)\left(x^2+\frac{5}{4}-x\right)$$

The factorization of $$f$$ is obtained by replacing $$x$$ with $$x+\frac{5}{2}$$

• (+1) I think you should mention how you obtain that $-\frac{5}{2}$ term. Jul 4 '19 at 17:53

Hint: Make the ansatz $$x^4+10x^3+39x^2+70x+50=(x^2+ax+b)(x^2+cx+d)$$ Expanding the right-hand side $$x^4+x^3(a+c)+x^2(b+d+ac)+x(bc+ad)+bd$$ And you will get $$a+c=10,b+d+ac=39,bc+ad=70,bd=50$$

• Definitely the way to do this. Jul 4 '19 at 7:39
• Thank you! How would I determine the constants a and c? My initial instinct was to follow the quartic form and have ax^4+bx^3...=0
– Loop
Jul 4 '19 at 7:42
• Start with $bd$ is a factorization of $50$. Each such factorization gives you that $ac$ is a factorization of $ac = 30-b-d$ (the second equation among the coefficients) and you already have $a+c = 10$ to quickly reject candidates. Jul 4 '19 at 16:37
• I think you meant 39 there, not 30. Jul 4 '19 at 17:23
• This is true for any quartic, though, and they're generally messy to solve. How does this make solving it easier? Jul 5 '19 at 8:34

Sometimes the best way to proving that $$A=B$$ is to simply forget all about $$B,$$ and just have a lot of fun independently exploring the beauty of $$A$$ anew, with fresh eyes, for one's own intellectual pleasure, without having any needless worries concerning the much expected arrival at point $$B$$ menacingly hovering over one's mind, as a dark cloud, of sorts. This is basically the mathematical equivalent of the well-worn philosophical adage about a true traveler not knowing his destination.

Now, any road, no matter how long, begins with a simple step. So, what if I were to simply tell you that $$(x+a)^2=x^2+2ax+a^2~?$$ You'd probably say that, apart from being painfully obvious, it is also of no practical use to us, since we are dealing with a fourth degree $$($$or quartic$$)$$ expression, rather than a humble quadratic. But what if we'd replace $$x$$ by $$x^2$$ ? Then the polynomial expression would soon become $$(x^2+a)^2=x^4+2ax^2+a^2,$$ bringing it much closer to our intended form for $$A(x)$$. Just two “small” problems: $$39$$ is odd, and $$50$$ is not a perfect square. So let's put this on pause for a second, and take a look at the remaining two terms, $$10x^3+70x:$$ is there nothing that can be done here ? “Well, sure there is!”, you might retort. “Both share a common factor, $$10x.$$” So let's see where that takes us, shall we ? $$10x^3+70x=10x~(x^2+7).$$ But, wait a second here, doesn't the latter expression, $$x^2+7,$$ look suspiciously similar to our initial one, $$x^2+a$$ ? In which case, $$a^2=7^2=49,$$ which comes incredibly close to our original $$50=49+1,$$ and $$2a=2\cdot7=14,$$ whose difference until $$39$$ is $$39-14=25=5^2,$$ which $$($$remaining$$)$$ coefficient fits ever so nicely with the $$x^2$$ it multiplies. Wrapping it all up, the polynomial becomes $$A(x)=(x^2+7)^2+2\cdot5x~(x^2+7)+(5x)^2+1,$$ at which point the factoring $$A(x)=\Big[(x^2+7)+5x\Big]^2+1$$ should become rather transparent.
Further writing $$1=-i^2,$$ and using $$a^2-b^2=(a-b)(a+b),$$ we eventually obtain $$A(x)=(x^2+5x+7-i)~(x^2+5x+7+i).$$ Since $$(7-i)~(7+i)=7^2-i^2=49+1=50,$$ we are most likely looking for something like $$A(x)=\Big[x^2+(5-n)~x+p\Big]\cdot\Big[x^2+(5+n)~x+q\Big],$$ with $$pq=50$$. $$($$Would you like me to finish this for you, or do you, by any chance, already feel confident enough to take it from here $$?).$$

For real valued rational roots you can use the rational root theorem.

• The rational root theorem helps to find rational roots, not real roots. But I don't think that this polynomial has real roots at all. Jul 4 '19 at 7:48
• @MartinR yep you are right. It doesn't. But it's a good tool in the toolbox. Jul 4 '19 at 7:50

For all real $$k$$ we obtain: $$x^4+10x^3+39x^2+70x+50=$$ $$=(x^2+5x+k)^2-25x^2-k^2-10kx-2kx^2+39x^2+70x+50=$$ $$=(x^2+5x+k)^2-((2k-14)x^2+(10k-70)x+k^2-50).$$ Now, we'll choose $$k$$ such that we'll get a difference of squares.

For which we need $$25(k-7)^2-(2k-14)(k^2-50)=0$$ or $$(k-7)(2k^2-25k+75)=0$$ or $$(k-7)(k-5)(2k-15)=0.$$ We see that only $$k=7.5$$ is valid and we obtain: $$x^4+10x^3+39x^2+70x+50=(x^2+5x+7.5)^2-(x^2+5x+6.25)=$$ $$=(x^2+5x+7.5)^2-(x+2.5)^2=(x^2+4x+5)(x^2+6x+10).$$

For an investigative approach:

Since the coefficient of $$x^4$$ is 1, you know both your quadratics must start with $$x^2$$.

Now look at the factors of 50, which are 1, 50 or 2, 25 or 5, 10, so you have a basis from which to make an educated guess for the quadratic's constant terms.

Next, you will need two coefficients of $$x$$ for your quadratics that will sum to the coefficient of $$10x^3$$ in your original polynomial.

Expanding the ansatz approach will give you the exact conditions required.