expand the binomial (x+3)^3 ? need help?
回答 (11)
2009-05-03 9:37 am
✔ 最佳答案
(a+b) ^3 = (a + b) (a² + 2ab + b²)So,
(x+3) ^3=(x + 3) (x² + 6x + 9)
On multiplying the terms using FOIL : x^3+9x^2+27x+27
參考: ...
2009-05-07 1:15 am
to expand this, you just write it out the number it is ^.
so it would be
(x+3)(x+3)(x+3)
so it would be
(x+3)(x+3)(x+3)
2009-05-03 12:16 pm
*Multiply first (x+3) by (x+3). Use the FOIL method for this.
(x+3)(x+3)
x²+6x+9
*Multiply (x+3) again by the answer above which is x²+6x+9.Use the distributive property for this and combine similar terms.
(x+3)(x²+6x+9)
x³ + 6x² + 9x
3x² + 18x + 27
-----------------------------
x³ + 9x² + 27x + 27 → Final answer :)
(x+3)(x+3)
x²+6x+9
*Multiply (x+3) again by the answer above which is x²+6x+9.Use the distributive property for this and combine similar terms.
(x+3)(x²+6x+9)
x³ + 6x² + 9x
3x² + 18x + 27
-----------------------------
x³ + 9x² + 27x + 27 → Final answer :)
2009-05-03 11:47 am
(x + 3)^3
= ³C₀(x^3) + ³C₁(3x^2) + ³C₂(3^2 * x) + ³C₃(3^3)
= 1(x^3) + 3/1(3x^2) + [(3 * 2)/(1 * 2)](3^2 * x) + [(3 * 2 * 1)/(1 * 2 * 3)](3^3)
= x^3 + 9x^2 + 27x^2 + 27
= ³C₀(x^3) + ³C₁(3x^2) + ³C₂(3^2 * x) + ³C₃(3^3)
= 1(x^3) + 3/1(3x^2) + [(3 * 2)/(1 * 2)](3^2 * x) + [(3 * 2 * 1)/(1 * 2 * 3)](3^3)
= x^3 + 9x^2 + 27x^2 + 27
參考: Binomial theorem - Topics in precalculus
http://www.themathpage.com/aPreCalc/binomial-theorem.htm
2009-05-03 10:08 am
x^3 + 9x^2 +27x + 27
2009-05-03 9:45 am
=x^3 +9x^2 +27x+27
2009-05-03 9:42 am
x^3+27+9x^2+27x
2009-05-03 9:41 am
(a+b) ^3 = (a + b) (a² + 2ab + b²) then x^3+9x^2+27x+27
2009-05-03 9:40 am
There are formulas that you could use right away to get the answer, but the good old method of multiplying everything out work too:
(x+3)^2=x^2+6x+9
(x+3)^3=(x+3) X (x+3)^2=x^3+9x^2+27x+27
(x+3)^2=x^2+6x+9
(x+3)^3=(x+3) X (x+3)^2=x^3+9x^2+27x+27
2009-05-03 9:39 am
answer from my over price calculator ti89
x^3+9x^2+27x+27
x^3+9x^2+27x+27
收錄日期: 2021-05-01 12:21:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090503013342AA2BwOX