Question

алгебра
3х(х³-4х+6)
(х-3)(2х+1)
(4а-7в) (5а+6в)
(у+2) (у²+у-8)​

Answer 1

$3x(x^3-4x+6) = 3x^4-12x^2+18x\\\\(x-3)(2x+1) = 2x^2+x-6x-3 = 2x^2-5x-3\\\\(4a-7b)(5a+6b) = 20a^2+24ab-35ab-42b^2 = 20a^2-11ab-42b^2\\\\(y+2)(y^2+y-8) = y(y^2+y-8)+2(y^2+y-8) = y^3+y^2-8y+2y^2+2y-16 = \\\\ = y^3+3y^2-6y-16$

Answer 2

Объяснение:

$3x(x {}^{3} - 4x + 6) = 3x {}^{4} - 12x {}^{2} + 18x$

$(x - 3)(2x + 1) = 2x {}^{2} + x - 6x - 3 = 2x {}^{2} - 5x - 3$

$(4a - 7b)(5a + 6b) = 20a { }^{2} + 24ab - 35ab - 42b {}^{2} = 20a {}^{2} - 11ab - 42b {}^{2}$

$(y + 2)(y {}^{2} + y - 8) = y {}^{3} + 3y {}^{2} - 6y - 16$