2(x - 1)(2x + y)
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∙ 8y ago2xy - 4x plus 8y - 16 equals x plus 4 in parentheses multiplied by 2y minus 4 in parentheses. So, the factors are ( x + 4) and ( 2y - 4) .
2y^2+5y+2=(2y+1)(y+2)
(2y - 5)(y - 2)
6xyz(3x + 2y + z)
4xy - 2y(x + 4) = 4xy - 2xy - 8y = 2xy - 8y = 2y(x - 4)
4x2 - 4x + 2xy - 2y = 4x(x - 1) + 2y(x - 1) (4x + 2y)(x - 1).
2 (2x2 + 2x + xy + y ) it's 2 (2x + y)(x + 1) if you're doing A+
2xy - 4x plus 8y - 16 equals x plus 4 in parentheses multiplied by 2y minus 4 in parentheses. So, the factors are ( x + 4) and ( 2y - 4) .
2y + 1
2y^2+5y+2=(2y+1)(y+2)
(2y - 5)(y - 2)
6xyz(3x + 2y + z)
4xy - 2y(x + 4) = 4xy - 2xy - 8y = 2xy - 8y = 2y(x - 4)
2x^2y^2(3x^2y^3 - 2xy + 1)
4x2 - 4x + 2xy - 2y = 4x(x - 1) + 2y(x - 1) = (x - 1)(4x + 2y) = 2(x - 1)(2x + y)
If that's -18y2, the answer is (3x - 2y)(x + 9y)
You can find this by grouping the factors, then looking for common factors:2my + 7x + 7m + 2xy = (2my + 2xy) + (7x + 7m) = 2y(m+x) + 7(m+x).Since the factor m+x also happens to be a common factor, you can use the distributive law once more: 2y(m+x) + 7(m+x) = (2y+7)(m+x).Sometimes you have to try several possibilities; and of course, not all polynomials with 4 or more factors can be factored this way - or at all.