factoring polynomials help?
posted by Skye on .
16x^41
4x^4+39x^210
find the real number solutions of these equations.
3x^4+15x^272=0
x^3+2x^2x=0
Thanks!

1. 16x^4  1 =
(4x^2+1)(4x^21) =
(4x^2+1)(2x+1)(2x1).
2.
3. 3x^4 + 15x^2  72 = 0,
X = + 2.19.
EXCEL Spreadsheets were used.
4. x^3 + 2x^2  x = 0,
X= 0. 
2. 4x^4 + 39x^2  10 = 0.
AC Method:
A*C = 4*(10) = 40 = 1*40.
4x^4 + (x^2+40x^2) 10,
Group terms into factorable pairs:
(4x^4  x^2) + (40x^210) =
x^2(4x^21) + 10(4x^21) =
4x^21(x^2+10) =
(2x+1)(2x1)(x^2+10). 
Unless I am mistaken, you may want to check the EXCEL spreadsheet results.
#3
3x^4+15x^272=0 factors well to
3(x^23)(x^2+8)=0
which gives
x=±sqrt(3) or
x=±(2√2)i
and #4
x^3+2x^2x=0
factors into
x(x^2+2x1)
which gives x=0, or
x=(2±√(2^2+4))/2
=1±√8
=1±2√2 
#3. Your procedure was easy to follow:
I will use it on similar problems.
EXCEL doesn't work properly when imaginary numbers are involved.
Thanks for the INFO! 
You're welcome!

5p^2+14p+8