SOLUTION: Im having trouble factoring polynomials:
please show the steps and explain (simply) what changes were made and why they were made between steps:
a few examples of what i am h
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please show the steps and explain (simply) what changes were made and why they were made between steps:
a few examples of what i am h
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Question 634771: Im having trouble factoring polynomials:
please show the steps and explain (simply) what changes were made and why they were made between steps:
a few examples of what i am having trouble with:
(2x^3) + (3x^2) - 8x - 12 = 0
(x/(x+2) + (1/4) = 5
x^(5/4) - 32 = 0
(6x^4) - (23x^2) + 20 = 0 Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! (2x^3) + (3x^2) - 8x - 12 = 0
factor by grouping
(2x^3 - 8x)+ (3x^2 - 12) = 0
2x(x^2-4)+3(x^2-4)=0
factor out common term
(x^2-4)(2x+3)=0
solve for x
x^2-4=0
x^2=4
x=±√4=±2
or
2x+3=0
2x=-3
x=-3/2
x=-2, -3/2, 2
..
(x/(x+2) + (1/4) = 5
add(-1/4) to each side
(x/(x+2) = 20/4-1/4=19/4
(x/(x+2)=19/4
cross-multiply
4x=19(x+2)=19x+38
15x=-38
x=-38/15
x≈2.5333..
..
x^(5/4) - 32 = 0
x^(5/4) = 32
raise both sides to the 4/5 power
x=32^(4/5)=16 (take fifth root of 32=2, then raise it to the 4th power=2^4=16
..
(6x^4) - (23x^2) + 20 = 0
6x^4-23x^2+20 = 0
let u=x^2
u^2=x^4
equation:
6u^2-23u+20=0
solve for u, using following quadratic formula:
a=6, b=-23, c=20
ans:
u=2.5
or
u=4/3
solve for x:
x^2=u=2.5
x=±√2.5
or
x^2=4/3
x=±√(4/3)
..
x=-√(4/3), -√(2.5), √2.5, and √(4/3)
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