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I want to factorize the polynomial $9a^4-a^2+16$. $(3a^2+5a+4)(3a^2-5a+4)$ is the solution according to the book, but it doesn't provide a step-by-step solution. I tried the PhotoMath app, and the first step according to it was to rewrite the expression in the following way:

$9a^4-15a^3+15a^3+12a^2-25a^2+12a^2+20a-20a+16$

The problem is, unlike some other easier expressions, the application of a special product/algebraic formula doesn't seem self-evident to me in this case.

There could be many ways I could rewrite such polynomials, but I don't want to waste time in my exams. I just wanna figure out the correct way to rewrite the expression just by looking at it and applying an algorithm instead of brainstorming infinite ways. What's the algorithm for rewriting a polynomial for factorization in cases like these?

nonuser
  • 90,026
Divyansh Adwani
  • 51

2 Answers2

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You can try to write it as a product of two polinomials (with unknow coefficents) both of degree 2 or one of degree 1 and second of degree 3. $$9x^4-x^2+16 = (ax^2+bx+c)(dx^2+ex+f)$$ and try to find the coefficients. Multiply those terms in bracket and do a sutable comparisons, say $ad = 9$, $ae+bd=0$,...

If there is no slution to this system then try

$$9x^4-x^2+16 = (ax^3+bx^2+cx+d)(ex+f)$$ and do the same again.

In general it could happen that you won't get a solution to second also which means that given polynomial is irreducibile.

nonuser
  • 90,026
1

First and third terms remaind me on $x^2+2xy+y^2$ with $2xy$ left. So what do I have to add?

\begin{align} \color{blue}{9a^4}-a^2+\color{blue}{16} &= \color{blue}{9a^4}+\color{red}{24a^2}+\color{blue}{16}-a^2-\color{red}{24a^2} \\ &=(3a^2+4)^2-25a^2\\ &= \Big((3a^2+4)-5a\Big)\Big((3a^2+4)+5a\Big) \end{align}

nonuser
  • 90,026