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

[mindauggas]

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.

 

[ehild]

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

ehild

 

[Mark44]

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...

 

[mindauggas]

Thank you

 

[dextercioby]

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]

 

[ehild]

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

 

[eumyang]

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.