Factorising Polynomials?
2009-03-11 10:39 am
Express p(x)/x-2 in simplest form. When p(x)=x^3 - 3x^2 - 10x + 24. Alegbraic fractions.
回答 (5)
2009-03-11 10:47 am
✔ 最佳答案
p(x) = x³ - 3x² - 10x + 24p(x)/(x - 2) = (x³ - 3x² - 10x + 24)/(x - 2)
p(x)/(x - 2) = x² - x - 12
2009-03-11 6:35 pm
.....................x^2 - x - 12
......__________________
x - 2)x^3 - 3x^2 - 10x + 24
........x^3 - 2x^2
--------------------------------------
................-x^2 - 10x
................-x^2 + 2x
--------------------------------------
........................-12x + 24
........................-12x + 24
--------------------------------------
p(x)/(x - 2)
= (x^3 - 3x^2 - 10x + 24)/(x - 2)
= x^2 - x - 12
= x^2 + 3x - 4x - 12
= (x^3 + 3x) - (4x + 12)
= x(x + 3) - 4(x + 3)
= (x + 3)(x - 4)
......__________________
x - 2)x^3 - 3x^2 - 10x + 24
........x^3 - 2x^2
--------------------------------------
................-x^2 - 10x
................-x^2 + 2x
--------------------------------------
........................-12x + 24
........................-12x + 24
--------------------------------------
p(x)/(x - 2)
= (x^3 - 3x^2 - 10x + 24)/(x - 2)
= x^2 - x - 12
= x^2 + 3x - 4x - 12
= (x^3 + 3x) - (4x + 12)
= x(x + 3) - 4(x + 3)
= (x + 3)(x - 4)
2009-03-11 5:55 pm
x-2 into a=2
2|...1....-3.....-10.....24
............2...... -2.....-24
-------------------------------------
.......1...-1.....-12......0
=x^2-x-12
=(x+3)(x-4)
=(x-2)(x+3)(x-4) answer//
2|...1....-3.....-10.....24
............2...... -2.....-24
-------------------------------------
.......1...-1.....-12......0
=x^2-x-12
=(x+3)(x-4)
=(x-2)(x+3)(x-4) answer//
2009-03-11 5:51 pm
p(x)=x^3 - 3x^2 - 10x + 24
solving p(x) = (x-2)(x-4)(x+3) are the factors
p(x)/x-2=(x-4)(x+3) is the answer..
solving p(x) = (x-2)(x-4)(x+3) are the factors
p(x)/x-2=(x-4)(x+3) is the answer..
2009-03-11 5:46 pm
(x^2 - x + 8) + 8/(x-2)
just divde p(x) by x-2
just divde p(x) by x-2
收錄日期: 2021-05-01 12:08:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090311023910AAXmCP7