X^8 - 1
HOW WOULD LOOK FACTORED?
XY + 4X + 5Y + 20
WHAT ABOUT THIS ONE?
ALGEBRA! brain fart?
2008-06-28 6:35 pm
回答 (8)
2008-06-28 6:41 pm
✔ 最佳答案
1)x^8 - 1
= (x^4 + 1)(x^4 - 1)
= (x^4 + 1)(x^2 + 1)(x^2 - 1)
= (x^4 + 1)(x^2 + 1)(x + 1)(x - 1)
2)
xy + 4x + 5y + 20
= x(y + 4) + 5(y + 4)
= (x + 5)(y + 4)
2008-06-28 6:41 pm
(x^4)^2-1^2=(x^4-1)(x^4+1)=(x^2-1)(x^2+1)(x^4+1)=
(x-1)(x+1)(x^2+1)(x^4+1)
x(y+4)+5(y+4)=(y+4)(x+5)
(x-1)(x+1)(x^2+1)(x^4+1)
x(y+4)+5(y+4)=(y+4)(x+5)
2008-06-28 6:40 pm
The first is already as simple as it can be, unless i read the equation wrong. Did you mean x^(8-1)? In which case, just do the math to give x ^7.
The second equation factors as a quadratic:
(x + 5) (y + 4)
The second equation factors as a quadratic:
(x + 5) (y + 4)
2008-06-28 6:51 pm
x⁸-1 = (x⁴)² - 1²
...this is the difference of squares, so you SHOULD know right away how to factor it:
= (x⁴+1)(x⁴-1)
which is the simplest factoring.
...it can be factored further because the second term, (x⁴-1), is also the difference of squares:
= (x⁴+1)(x²+1)(x²-1)
which is another answer.
...it can be factored even further using the same process. All these answers are correct because the question didn't require x⁸-1 to be *completely* factored.
...this is the difference of squares, so you SHOULD know right away how to factor it:
= (x⁴+1)(x⁴-1)
which is the simplest factoring.
...it can be factored further because the second term, (x⁴-1), is also the difference of squares:
= (x⁴+1)(x²+1)(x²-1)
which is another answer.
...it can be factored even further using the same process. All these answers are correct because the question didn't require x⁸-1 to be *completely* factored.
2008-06-28 6:46 pm
X^8 - 1=
(x^4-1)(x^4+1)=
(x^2-1)(x^2+1)(x^4+1)
(x-1)(x+1)(x^2+1)(x^4+1) Over rational coeff.
Over real
(x-1)(x+1)(x^2+1)[(x^2+1)^2-2x^2]=
(x-1)(x+1)(x^2+1)(x^2+1-x*sqrt(2))(x^2+1+x*sqrt(2))
You may continue on complex coefficients.
(x^4-1)(x^4+1)=
(x^2-1)(x^2+1)(x^4+1)
(x-1)(x+1)(x^2+1)(x^4+1) Over rational coeff.
Over real
(x-1)(x+1)(x^2+1)[(x^2+1)^2-2x^2]=
(x-1)(x+1)(x^2+1)(x^2+1-x*sqrt(2))(x^2+1+x*sqrt(2))
You may continue on complex coefficients.
2008-06-28 6:43 pm
x^8 - 1
=(x^4)^2 -1^2
=(x^4+1)(x^4-1)
=(x^4+1)((x^2^)2-1)
=(x^4+1)(x^2+1)(x^2-1)
=(x^4+1)(x^2+1)(x+1)(x-1)
xy+4x+5y+20
=x(y+4) + 5(y+4)
=(x+5)(y+4)
This is it buddy.... gud luck
=(x^4)^2 -1^2
=(x^4+1)(x^4-1)
=(x^4+1)((x^2^)2-1)
=(x^4+1)(x^2+1)(x^2-1)
=(x^4+1)(x^2+1)(x+1)(x-1)
xy+4x+5y+20
=x(y+4) + 5(y+4)
=(x+5)(y+4)
This is it buddy.... gud luck
2008-06-28 6:42 pm
1st: Use difference of squares ad nauseum:
x^8 - 1
x^4*x^4 - 1*1
(x^4 + 1)*(x^4 - 1)
(x^4 + 1)*(x^2*x^2 - 1*1)
(x^4 + 1)*(x^2 + 1)*(x^2 - 1)
(x^4 + 1)*(x^2 + 1)*(x + 1)*(x - 1)
#2
XY + 4X + 5Y + 20
X(Y + 4) + 5(Y + 4)
(X + 5)*(Y+4)
x^8 - 1
x^4*x^4 - 1*1
(x^4 + 1)*(x^4 - 1)
(x^4 + 1)*(x^2*x^2 - 1*1)
(x^4 + 1)*(x^2 + 1)*(x^2 - 1)
(x^4 + 1)*(x^2 + 1)*(x + 1)*(x - 1)
#2
XY + 4X + 5Y + 20
X(Y + 4) + 5(Y + 4)
(X + 5)*(Y+4)
2008-06-28 6:42 pm
x^8 - 1
(x^4 + 1)(x^2 + 1)(x + 1)(x - 1)
xy + 4x + 5y + 20
(y + 4)(x + 5)
(x^4 + 1)(x^2 + 1)(x + 1)(x - 1)
xy + 4x + 5y + 20
(y + 4)(x + 5)
收錄日期: 2021-05-01 10:46:35
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