Factoring 4x^4-x^2-18

  • Thread starter mindauggas
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  • #1
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Hello,

Homework Statement



Factor: 4x[itex]^{4}[/itex]-x[itex]^{2}[/itex]-18

The Attempt at a Solution



I solved a similar problem x[itex]^{4}[/itex]-6x[itex]^{2}[/itex]+9 by equating x[itex]^{2}[/itex] 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.
 
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Answers and Replies

  • #2
ehild
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Replace x2 by t as you did before and complete the square. Then factorize further if it is possible.

ehild
 
  • #3
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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...
 
  • #5
dextercioby
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Can't you directly complete the square & then factor ?

[tex] 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) [/tex]
 
  • #6
ehild
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Can't you directly complete the square & then factor ?

[tex] 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) [/tex]
Or [tex](x^2+2)(4x^2-9)=(x^2+2)(2x+3)(2x-3)[/tex]

ehild
 
  • #7
eumyang
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  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" [Broken] 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.
 
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