Factoring mathematics?
6x^2 - 8x - 8
I came up with 2(3x^2 - 4x - 4)
Can it be factored any more?
回答 (7)
✔ 最佳答案
6x^2 - 8x - 8
=2(3x^2 - 4x - 4)
=2(3x+2)(x-2) answer//
6x² - 8x - 8 = 0
x² - 4/3x = 4/3
x² - 2/3x = 4/3 + (- 2/3)²
x² - 2/3x = 12/9 + 4/9
(x - 2/3)² = 16/9
x - 2/3 = 4/3
= x - 2/3 - 4/3, = x - 6/3, = x - 2
= x - 2/3 + 4/3, = x + 2/3, = 3x + 2
Answer: 2(x - 2)(3x + 2) are the factors.
Proof (F.O.I.L.):
= 2(x - 2)(3x + 2)
= 2(3x² + 2x - 6x - 4)
= 2(3x² - 4x - 4)
= 6x² - 8x - 8
because the numeric section is 6, there are in uncomplicated words 2 possiblitites: 6 and a million or 3 and 2. when you consider that 6 and a million can't produce a a million (for the coefficient of the 2d time period), then we ought to apply the three and 2. because the coefficient of the middle is a plus one, the mixture must be a plus 3 with a minus 2. for this reason, the answer must be (2x + 3)(x - 2), which yields 2x^2 +3x -4x -6 which in reality does produce 2x^2 - x - 6 = 0, If 2x + 3 = 0, then x must be -a million If x - 2 = 0, then x must be 2
2 (3x ² - 4 x - 4)
2 (3x + 2)(x - 2)
There will always be two factors with x for a quadratic equation, though the may be identical, and you can always find them with the quadratic formula. If you find two roots, say a and b with the quadratic formula, the factors are (x-a) and (x-b)
6x^2 - 8x - 8
= 2(3x^2 - 4x - 4)
= 2(3x^2 + 2x - 6x - 4)
= 2[(3x^2 + 2x) - (6x + 4)]
= 2[x(3x + 2) - 2(3x + 2)]
= 2(3x + 2)(x - 2)
收錄日期: 2021-05-01 12:06:08
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