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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?

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  • $\begingroup$ Google "StackExchange LaTeX guide" to learn how to format equations. $\endgroup$
    – user21820
    Apr 3, 2016 at 15:12
  • $\begingroup$ 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 :) $\endgroup$
    – sooobus
    Apr 3, 2016 at 21:08

4 Answers 4

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

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

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

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

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

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  • $\begingroup$ 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? $\endgroup$
    – KKZiomek
    Apr 9, 2016 at 14:41
  • $\begingroup$ Nevermind. It's 2*x^2*4 to make it dissapear when simplified to (a+c)^2. $\endgroup$
    – KKZiomek
    Apr 9, 2016 at 14:43
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$$x^4+8x^2+16-9x^2=(x^2+4)^2-(3x)^2=$$ $$=(x^2-3x+4)(x^2+3x+4)$$

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    $\begingroup$ How did you arrive into this??? Can you please explain the procedure? $\endgroup$
    – Ayan Shah
    Apr 3, 2016 at 15:13
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    $\begingroup$ Yes, I'd like to know what is the procedure. Please explain in detail. $\endgroup$
    – KKZiomek
    Apr 3, 2016 at 15:16
  • $\begingroup$ For example $x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt2 x)^2=(x^2-\sqrt2 x+1)(x^2+\sqrt2 x+1)$ $\endgroup$
    – Roman83
    Apr 3, 2016 at 15:16
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    $\begingroup$ @KKZiomek I have added a detailed answer below... $\endgroup$
    – Soham
    Apr 3, 2016 at 15:16
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    $\begingroup$ @AyanShah if you have any question pointed directly towards someone,use @ symbol and then add the user name... $\endgroup$
    – Soham
    Apr 3, 2016 at 15:23
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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.


Regarding your question
Consider
$$ a= 1,b=-1 ,c=16 \ and\ usual \ substitution\ x=x^2\ (both\ x\ are\ different) \\ \therefore p=x^2+4 \quad q=3\sqrt {x^2}=3x $$
After this its easy mathematics,you have to factorize 2 more easy equations thus giving you 4 roots.I hope , I did well to explain you :)

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  • $\begingroup$ ....thank you for your concern $\endgroup$
    – Ayan Shah
    Apr 3, 2016 at 16:00
  • $\begingroup$ No problem :) . Up vote if u found it useful :P $\endgroup$
    – Jimmy Kudo
    Apr 3, 2016 at 17:18
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    $\begingroup$ @JimmyKudo Consider using \text{} in math mode for more readable text $\endgroup$ Apr 3, 2016 at 18:54
  • $\begingroup$ @michael Thanx for the suggestion, I was unaware of it.I will implement it from next answers :) $\endgroup$
    – Jimmy Kudo
    Apr 3, 2016 at 18:58
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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 $$(x^2-3x+4)(x^2+3x+4)$$

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  • $\begingroup$ By far the best answer here in my opinion, at least for the level of knowledge the OP seems to have on this material. Not too simple in that you just show completing the square without pointing out why, but not too complicated beyond the OP's level either. $\endgroup$ Apr 6, 2016 at 1:30

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