factoring polynomials: 16x^3 + 18x^2 - 9x, 1 - 81x^4

[van]

i need some help here... i am pretty lost....

1. 16x^3 + 18x^2 - 9x

where do i start?

another one:

2. 1 - 81x^4

 

[van]

ok, so far i have this:
16x^3 + 18x^2 - 9x
16x^3 - 24x^2 + 6x - 9x
(16x^3 - 24x^2) + (6x - 9x)

not sure what to do next...

 

[jwpaine]

ok, so far i have this:
16x^3 + 18x^2 - 9x
16x^3 - 24x^2 + 6x - 9x
(16x^3 - 24x^2) + (6x - 9x)

not sure what to do next...

[16x^3 - 24x^2 + 6x - 9x] = [16x^2 - 24x^2 - 3x] which does NOT equal [16x^3 + 18x^2 - 9x]

If you're going to use the grouping method, make sure you do: (a)(c) = (16)(9) = 144, find two numbers divisible by 144 and add to 18, and then group.

You could always use the rational roots test: http://www.purplemath.com/modules/rtnlroot.htm

find some plausible factors and then divide off using synthetic division down to an easily factorable trinomial.

Best of luck.

 

[van]

find some plausible factors and then divide off using synthetic division down to an easily factorable trinomial.

i thought that i did that... ugh, now i am really lost..............................

 

[stapel]

1. 16x^3 + 18x^2 - 9x

where do i start?

A good start might be to take the common factor out front. Then you'll be left with a quadratic, which you can factor by whatever method you've studied.

2. 1 - 81x^4

Apply the difference-of-squares formula you memorized. Then apply it again to one of the factors you get. You should end up with two linear factors and one (prime) quadratic factor.

Eliz.