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Please help me brush up with my factoring.
- Thread starter Alpha6
- Start date
I haven't done this in a while...
I'm attempting to simplify X4-82X2+81 = 0
I took out X3 and reduced it to X3(X-81)(X-1) = 0
If I messed up, I'd appreciate any help. Thanks!
What do you get if you multiply that expression out? The second part of the equation,
(x-81)(x-1) = 0
looks almost correct.
Do you think it would help to substitute
u = x2
in the original equation to get
u2 - 82 u + 81 = 0
then, after you factor that, put back in the x2 instead of u for two equations which then become the difference of two squares.
TeacherSatya
New member
- Joined
- Oct 19, 2014
- Messages
- 10
Solution
X4-82X2+81 = 0
Alright here we go:
This is called AC, B Method/Factoring middle term.
AC = 81
B = -82
Now we have to find two numbers that give us the product of 81 and sum of -82 and the 2 numbers are (-81; -1).
Now we will substitute the middle term with those two factors there so we will have:
x^4 -81x^2 -1x^2 +81 =0 >> Now we factorize
x^2(x^2-81)-1(x^2-81) =0
(x^2 - 1)(x^2 - 81)=0
Hope it helps.
X4-82X2+81 = 0
Alright here we go:
This is called AC, B Method/Factoring middle term.
AC = 81
B = -82
Now we have to find two numbers that give us the product of 81 and sum of -82 and the 2 numbers are (-81; -1).
Now we will substitute the middle term with those two factors there so we will have:
x^4 -81x^2 -1x^2 +81 =0 >> Now we factorize
x^2(x^2-81)-1(x^2-81) =0
(x^2 - 1)(x^2 - 81)=0
Hope it helps.
D
Deleted member 4993
Guest
X4-82X2+81 = 0
Alright here we go:
This is called AC, B Method/Factoring middle term.
AC = 81
B = -82
Now we have to find two numbers that give us the product of 81 and sum of -82 and the 2 numbers are (-81; -1).
Now we will substitute the middle term with those two factors there so we will have:
x^4 -81x^2 -1x^2 +81 =0 >> Now we factorize
x^2(x^2-81)-1(x^2-81) =0
(x^2 - 1)(x^2 - 81)=0
Hope it helps.
The answer above is not complete.
x2 - 1 can be factored further down to (x + 1)(x - 1), and
x2 - 81 can be factored further down to (x + 9)(x - 9)
Then we have,
x4 - 82x2 + 81 = (x2 - 1)(x2 - 81) = (x + 1)(x - 1)(x + 9)(x - 9)
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
I hope you see now that you cannot "take out x3" because there is no x3 in "82x2" or "81".I haven't done this in a while...
I'm attempting to simplify X4-82X2+81 = 0
I took out X3 and reduced it to X3(X-81)(X-1) = 0
If I messed up, I'd appreciate any help. Thanks!
It might help to, temporarily, let \(\displaystyle u= x^2\) so that \(\displaystyle x^4= (x^2)(x^2)= u^2\) and your four degree equation becomes the quadratic u2- 82u+ 81= 0 which can be solved for u by factoring, using the quadratic formula, or completing the square.
TeacherSatya
New member
- Joined
- Oct 19, 2014
- Messages
- 10
The answer above is not complete.
x2 - 1 can be factored further down to (x + 1)(x - 1), and
x2 - 81 can be factored further down to (x + 9)(x - 9)
Then we have,
x4 - 82x2 + 81 = (x2 - 1)(x2 - 81) = (x + 1)(x - 1)(x + 9)(x - 9)
You r right, how silly of me hahaha. (y)