Factorise expression?
factorise
2x^2 - 3xy - 2y^2 - 2x - 11y - 12
回答 (4)
2x² - 3xy - 2y² - 2x - 11y - 12 = (2x + y)(x - 2y) - 2x - 11y -12
2x² - 3xy - 2y² - 2x - 11y - 12 = (2x + y + a)(x -2y + b)
2x² - 3xy - 2y² - 2x - 11y - 12 = 2x² - 3xy - 2y² + (a + 2b)x + (-2a + b)y + ab
x term on each side: a + 2b = -2 …… [1]
y term on each side: -2a + b = -11 …… [2]
[1]*2 + [2]: 5b = -15 ⇒ b = -3
[1] - [2]*2: 5a = 20 ⇒ a = 4
Check: constant term on each side = ab = -12
Hence, 2x² - 3xy - 2y² - 2x - 11y - 12 = (2x + y + 4)(x -2y - 3)
2x^2 - 3xy - 2y^2 - 2x - 11y - 12
= (x - 2 y - 3) (2 x + y + 4)
The expression starts with 2x^2, so the two factors must contain 2x and x.
Then there's -2y^2, so the factors contain 2y and y, one added and the other subtracted.
There are two ways we could combine them:
(2x ± 2y )(x ± y )
(2x ± y )(x ± 2y )
Of these choices the only way to get -3xy is
(2x + y )(x - 2y )
Finally, the constants have a product of -12. When multiplied with y and -2y, they give a sum of -11y. The only way to get an odd sum is to multiply the y by an odd factor, and the only way to get a negative sum is to multiply the -2y by the positive factor.
The only odd factors of 12 are 1 and 3, so check each of those and you'll find (y)(-3) + (-2y)(4) = -11y.
Place those into the factors:
(2x + y + 4)(x - 2y - 3)
and then verify that the multiplication of the x's with the constants produces -2x.
收錄日期: 2021-04-24 08:03:08
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