Factorization of polynomials with degree higher than 2

I need help to factorize $x^4-x^2+16$. I have tried to take $x^4$ as $(x^2)^2$ and factorize it in the typical way of factorizing a quadratic expression but that did not help. Can someone help me to factor this and also introduce me to the procedure that i need to follow to factorize expressions with degree higher than two?

• Google "StackExchange LaTeX guide" to learn how to format equations. Apr 3, 2016 at 15:12
• Just a hint: if you are given something to factorize at classes, the first step is to try guessing the root, using factorization of the last term. It usually works with easy tasks :) Apr 3, 2016 at 21:08

$$x^4-x^2+16$$

$$=[(x^2)^2+2\cdot x^2\cdot 4+4^2]-9x^2$$

$$=(x^2+4)^2-(3x)^2$$

$$=(x^2+4-3x)(x^2+4+3x)$$

by using $a^2-b^2=(a+b)(a-b)$.

• So is the middle term in the brackets 2*x^2*4 because the exponents in the original equation are 2 and 4? Or for different reason? Apr 9, 2016 at 14:41
• Nevermind. It's 2*x^2*4 to make it dissapear when simplified to (a+c)^2. Apr 9, 2016 at 14:43

$$x^4+8x^2+16-9x^2=(x^2+4)^2-(3x)^2=$$ $$=(x^2-3x+4)(x^2+3x+4)$$

• How did you arrive into this??? Can you please explain the procedure? Apr 3, 2016 at 15:13
• Yes, I'd like to know what is the procedure. Please explain in detail. Apr 3, 2016 at 15:16
• For example $x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt2 x)^2=(x^2-\sqrt2 x+1)(x^2+\sqrt2 x+1)$ Apr 3, 2016 at 15:16
• @KKZiomek I have added a detailed answer below... Apr 3, 2016 at 15:16
• @AyanShah if you have any question pointed directly towards someone,use @ symbol and then add the user name... Apr 3, 2016 at 15:23

Can be done by making perfect squares $$Let\ ax^2 +bx + c=0 \\Try\ to\ make\ the\ equation\ look\ in\ the\ form \\ax^2 +2\sqrt {ca}x +c-(2\sqrt{ca}-b)x=0 \\You\ will\ see\ that\ ax^2 +2\sqrt {ca}x +c\ makes\ a \ perfect\ square \\\therefore \quad ax^2 +2\sqrt {ca}x +c=(\sqrt ax + \sqrt c )^2 \\Thus\ the\ equation\ will\ convert\ to\ (\sqrt ax + \sqrt c )^2-(2\sqrt{ca}-b)x=0 \\Consider \ the\ term\ (2\sqrt{ca}-b)x \\We\ can\ write\ it\ as\ {[\sqrt{(2\sqrt{ca}-b)x}]}^2 \\Thus\ the\ equation\ is\ converted\ to\ form\ p^2-q^2=0 \\where\ p=\sqrt ax + \sqrt c \quad and\quad q=\sqrt{(2\sqrt{ca}-b)x} \\Equation\implies [\sqrt ax + \sqrt c]^2 - {[\sqrt{(2\sqrt{ca}-b)x}]}^2=0$$ Now you can apply $p^2-q^2=(p+q)(p-q)$ to get the required factorization.

$$a= 1,b=-1 ,c=16 \ and\ usual \ substitution\ x=x^2\ (both\ x\ are\ different) \\ \therefore p=x^2+4 \quad q=3\sqrt {x^2}=3x$$
• @JimmyKudo Consider using \text{} in math mode for more readable text Apr 3, 2016 at 18:54
Can be done by a simple completion of squares. The presence of $x^4$ and 16 hints towards a possibility of a $(x^2+4)^2$ so just calculate that and compare it with your question function. $(x^2+4)^2 = x^4+8x^2+16$ which leaves us with a difference only in the $x^2$ term. This difference turns out to be $9x^2$. So our function reduces to $$(x^2+4)^2-(3x)^2$$ It's easy to see after this, a simple use of $a^2-b^2 = (a+b)(a-b)$ gives the final answer to be $$(x^2-3x+4)(x^2+3x+4)$$