В)(y^2-3y-2)(y^2+3y-2) г)(a+2b+3c)(2a-b+2c)

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вопрос

В) (y^2 - 3y - 2)(y^2 + 3y - 2)

First, let's multiply the two binomials:
y^2 y^2 = y^4
y^2 3y = 3y^3
y^2 (-2) = -2y^2
-3y y^2 = -3y^3
-3y 3y = -9y^2
-3y (-2) = 6y
-2 y^2 = -2y^2
-2 3y = -6y
-2 * (-2) = 4

Putting it all together, we get:
y^4 + 3y^3 - 2y^2 - 3y^3 - 9y^2 + 6y - 2y^2 + 6y - 4

Simplifying the expression, we get:
y^4 - 11y^2 + 12y - 4

Therefore, (y^2 - 3y - 2)(y^2 + 3y - 2) simplifies to y^4 - 11y^2 + 12y - 4

г) (a + 2b + 3c)(2a - b + 2c)

First, let's multiply the two binomials:
a 2a = 2a^2
a (-b) = -ab
a 2c = 2ac
2b 2a = 4ab
2b (-b) = -2b^2
2b 2c = 4bc
3c 2a = 6ac
3c (-b) = -3bc
3c * 2c = 6c^2

Putting it all together, we get:
2a^2 - ab + 2ac + 4ab - 2b^2 + 4bc + 6ac - 3bc + 6c^2

Simplifying the expression, we get:
2a^2 + 4ab + 6ac - ab - 3bc - 2b^2 + 4c^2

Therefore, (a + 2b + 3c)(2a - b + 2c) simplifies to 2a^2 + 4ab + 6ac - ab - 3bc - 2b^2 + 4c^2.