Completely factor 3x^3-x^2-12x+4?
回答 (5)
2016-04-06 1:59 am
Let f(x) = 3x³ - x² - 12x + 4
f(2) = 3(2)³ - (2)² - 12(2) + 4 = 0
Then (x - 2) is a factor of f(x).
f(x) = 3x³ - x² - 12x + 4
f(x) = 3x³ - 6x² + 5x² - 10x - 2x + 4
f(x) = (3x³ - 6x²) + (5x² - 10x) - (2x - 4)
f(x) = 3x²(x - 2) + 5x(x - 2) - 2(x - 2)
f(x) = (x - 2)(3x² + 5x - 2)
f(x) = (x - 2)(x + 2)(3x - 1)
f(2) = 3(2)³ - (2)² - 12(2) + 4 = 0
Then (x - 2) is a factor of f(x).
f(x) = 3x³ - x² - 12x + 4
f(x) = 3x³ - 6x² + 5x² - 10x - 2x + 4
f(x) = (3x³ - 6x²) + (5x² - 10x) - (2x - 4)
f(x) = 3x²(x - 2) + 5x(x - 2) - 2(x - 2)
f(x) = (x - 2)(3x² + 5x - 2)
f(x) = (x - 2)(x + 2)(3x - 1)
2016-04-06 2:11 am
3x^3 - x^2 - 12x + 4
= x^2(3x - 1) - 4(3x - 1)
= (3x - 1)(x - 2)(x + 2)
= x^2(3x - 1) - 4(3x - 1)
= (3x - 1)(x - 2)(x + 2)
2016-04-06 2:03 am
=x^2(3x-1) -4(3x-1)
=(3x-1)(x2-4)
=(3x-1)(x-2)(x+2)
=(3x-1)(x2-4)
=(3x-1)(x-2)(x+2)
2016-04-06 1:55 am
=x^2(3x-1) -4(3x-1)
=(3x-1)(x2-4)
=(3x-1)(x-2)(x+2)
=(3x-1)(x2-4)
=(3x-1)(x-2)(x+2)
2016-04-06 1:51 am
factor by grouping, then difference of squares
收錄日期: 2021-05-01 13:06:43
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https://hk.answers.yahoo.com/question/index?qid=20160405175100AAcdiuE