how do you factorise x^3 -125?
回答 (6)
2010-03-16 11:58 am
✔ 最佳答案
x^3 -125=(x-5)(x^2+5x+25) answer//
2010-03-16 11:52 am
Use the difference of cubes formula, which says that
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
So in your case, the "a" = "x" and the "b" = "5" (since 5^3=125)
Therefore,
x^3 - 125
= x^3 - 5^3
= (x-5)(x^2 + 5x + 25)
[As a check, If you expand that, you'll get back to x^3-125.]
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
So in your case, the "a" = "x" and the "b" = "5" (since 5^3=125)
Therefore,
x^3 - 125
= x^3 - 5^3
= (x-5)(x^2 + 5x + 25)
[As a check, If you expand that, you'll get back to x^3-125.]
2010-03-16 11:52 am
f(x) = x^3 -125
f(5) = 0, so x-5 is a factor.
Use long division to get the quadratic factor. As far as real factors are concerned, you are done.
If not, use the quadratic formula to get the other 2 complex roots.
f(5) = 0, so x-5 is a factor.
Use long division to get the quadratic factor. As far as real factors are concerned, you are done.
If not, use the quadratic formula to get the other 2 complex roots.
2010-03-16 11:52 am
x^3 -125
=x^3 -5^3
= (x-5)(x^2 +5x +25)
=x^3 -5^3
= (x-5)(x^2 +5x +25)
2010-03-16 11:51 am
a^3 - b^3 â¡ (a - b)(a^2 + ab + b^2)
x^3 - 125
= x^3 - 5^3
= (x - 5)(x^2 + 5x + 25)
x^3 - 125
= x^3 - 5^3
= (x - 5)(x^2 + 5x + 25)
2010-03-16 11:50 am
x^(3)-125
The binomial can be factored using the difference of cubes formula, because both terms are perfect cubes. The difference of cubes formula is a^(3)-b^(3)=(a-b)(a^(2)+ab+b^(2)).
(x-5)(x^(2)+5x+25)
The binomial can be factored using the difference of cubes formula, because both terms are perfect cubes. The difference of cubes formula is a^(3)-b^(3)=(a-b)(a^(2)+ab+b^(2)).
(x-5)(x^(2)+5x+25)
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