Mastering Factoring Polynomials: Tips, Tricks, and Examples to Help You Succeed!

AI Thread Summary
Factoring polynomials involves techniques such as factoring out common terms, recognizing perfect squares, and using grouping methods. For the polynomial 7h^3 + 448, factoring out a 7 simplifies it to 7(h^3 - 64), which can be further factored using the formula for the difference of cubes. The expression y^4 - 81 can be factored as a difference of squares into (y^2 - 9)(y^2 + 9), where the first factor can be further factored. In grouping, the polynomial 3n^3 - 10n^2 - 48n + 160 can be rearranged and factored to reveal common factors, ultimately leading to a more manageable expression. Understanding these techniques is crucial for mastering polynomial factoring.
CamTheLamb
Messages
3
Reaction score
0
Alright, I'll be honest. I was extremely tired and slept all through the lesson in Algebra today lol.
And now I need help with factoring polynomials.

Example problems that I need help on:
7h3+448
Perfect square factoring - y4-81
Grouping - 3n3-10n2-48n+160

You don't have to answer those problems (Though it would help =P), I just need a quick lesson or the formula for factoring these.
Thanks =)
 
Physics news on Phys.org
CamTheLamb said:
Alright, I'll be honest. I was extremely tired and slept all through the lesson in Algebra today lol.
And now I need help with factoring polynomials.

Example problems that I need help on:
7h3+448
The first thing I would do is try to factor out a 7: 7(h^3- 64) which is wonderful because it is now 7(x^3- 4^3).
You need to know that a^2- b^3= (a- b)(a^2+ ab+ b^2)

Perfect square factoring - y4-81
Yes, those are perfect squares: (y^2)^2- 9^2 and, of course, a^2- b^2= (a-b)(a+b). After you have used that you will still have a "difference of squares" in one factor and can use that again. There is no way to factor [math]a^2+ b^2[/math] with real coefficients.

Grouping - 3n3-10n2-48n+160
Well, 10 isn't divisible by 3 but 48 is so I would try 3n^3- 48n= 3n(n^2- 16). Aha! Now it's easy to see that -10n^2+ 160= -10(n^2- 16)

3n^3- 10n- 48n+ 160= 3n(n^2- 16)- 10(n^2- 16)
That isn't the answer- you need to finish it.

You don't have to answer those problems (Though it would help =P), I just need a quick lesson or the formula for factoring these.
Thanks =)
 
Thanks for the help, I knew it was something similar to this, just didn't know any formulas.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top