How to Factor 4x^4-x^2-18: Step-by-Step Solution and Tips

  • Thread starter Thread starter mindauggas
  • Start date Start date
  • Tags Tags
    Factoring
AI Thread Summary
The discussion focuses on factoring the polynomial 4x^4 - x^2 - 18, with participants sharing methods and insights. One approach involves substituting x^2 with t, transforming the equation into a quadratic form, and using factorization techniques. Another method discussed is completing the square, which leads to a different factorization. Participants emphasize the importance of understanding the reasoning behind the methods, particularly for educational purposes. The conversation highlights effective strategies for factoring quadratic trinomials in algebra.
mindauggas
Messages
127
Reaction score
0
Hello,

Homework Statement



Factor: 4x^{4}-x^{2}-18

The Attempt at a Solution



I solved a similar problem x^{4}-6x^{2}+9 by equating x^{2} to t and then reverse-FOIL'ing... this one just wouldn't give in...
Completing the square also does not help to get the answer (presuming of course that the answer is correct, which I wouldn't dare not to do before consulting in this forum)...

I have the answear: (x^2+2)(2x-3)(2x+3), so I need your help on the reasoning process guys.
 
Last edited:
Physics news on Phys.org
Replace x2 by t as you did before and complete the square. Then factorize further if it is possible.

ehild
 
There's also another method for factoring a quadratic in the form ax2 + bx + c. Let u = x2 so that we now have 4u2 - u - 18.

  1. Calculate a*c, which is -72 for this problem.
  2. Find two factors of -72 that add up to -1.
    For this problem, 8 and -9 are factors of -72, and they add to -1.
  3. Rewrite the quadratic with the middle term expanded using the factors found in step 2.
    4u2 - u - 18 = 4u2 + 8u - 9u - 18.
  4. Factor by grouping to get the two binomial factors.
    4u2 + 8u - 9u - 18 = 4u(u + 2) - 9(u + 2) = (4u - 9)(u + 2).

Don't forget to undo the substitution...
 
Thank you
 
Can't you directly complete the square & then factor ?

4x^4 - x^2 - 18 = \left(2x^2 -\frac{1}{4}\right)^2 - \left(\frac{17}{4}\right)^2 = (2x^2 + 4)(2x^2 - 4.5)
 
dextercioby said:
Can't you directly complete the square & then factor ?

4x^4 - x^2 - 18 = \left(2x^2 -\frac{1}{4}\right)^2 - \left(\frac{17}{4}\right)^2 = (2x^2 + 4)(2x^2 - 4.5)

Or (x^2+2)(4x^2-9)=(x^2+2)(2x+3)(2x-3)

ehild
 
Mark44 said:
  1. Calculate a*c, which is -72 for this problem.
  2. Find two factors of -72 that add up to -1.
    For this problem, 8 and -9 are factors of -72, and they add to -1.
  3. Rewrite the quadratic with the middle term expanded using the factors found in step 2.
    4u2 - u - 18 = 4u2 + 8u - 9u - 18.
  4. Factor by grouping to get the two binomial factors.
    4u2 + 8u - 9u - 18 = 4u(u + 2) - 9(u + 2) = (4u - 9)(u + 2).
This is a great method in factoring quadratic trinomials. I first learned of it in reading Lial's http://www.pearsonhighered.com/educator/product/Introductory-Algebra/9780321557131.page" book. It's interesting that when I learned factoring in school we were taught to just guess-and-check. I now teach this method to my freshmen Algebra I classes, even though their books use the guess-and-check method.
 
Last edited by a moderator:
Back
Top