Question

16x^4-1

4x^4+39x^2-10


find the real number solutions of these equations.

3x^4+15x^2-72=0

x^3+2x^2-x=0


Thanks!

Answer 1

1. 16x^4 - 1 =
(4x^2+1)(4x^2-1) =
(4x^2+1)(2x+1)(2x-1).

2.

3. 3x^4 + 15x^2 - 72 = 0,
X = +- 2.19.
EXCEL Spreadsheets were used.

4. x^3 + 2x^2 - x = 0,
X= 0.

Answer 2

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^2-10) =
x^2(4x^2-1) + 10(4x^2-1) =
4x^2-1(x^2+10) =
(2x+1)(2x-1)(x^2+10).

Answer 3

Unless I am mistaken, you may want to check the EXCEL spreadsheet results.
#3
3x^4+15x^2-72=0 factors well to
3(x^2-3)(x^2+8)=0
which gives
x=±sqrt(3) or
x=±(2√2)i

and #4
x^3+2x^2-x=0
factors into
x(x^2+2x-1)
which gives x=0, or
x=(-2±√(2^2+4))/2
=-1±√8
=-1±2√2

Answer 4

#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!

Answer 5

You're welcome!

Answer 6

5p^2+14p+8