Question

• a) (x + 6) (x + 5); 6) (a - 4)(a + 1); в) (2 - y)(y -8) : г) (а - 4)(2a + 1); - д) (2у - 1)(3у + 2) ; e) (5x - 3)(4 - 3x).​

Answer 1

1. (x+6)(x+5)= x²+5x+6x+30= x²+11x+30

2. (a-4)(a+1)= a²+a-4a-4= a²-3a-4

3. (2-y)(y-8)= 2y-16-y²+8y=-y²+10y-16

4. (a-4)(2a+1) = 2a²+a-8a-4= 2a²-7a-4

5. (2y-1)(3y+2)= 6y²+4y-3y-2= 6y²+y-2

6. (5x-3)(4-3x)= 20x-15x²-12+9x=-15x²+29x-12

Answer 2

Let's expand each of the given expressions:

a) (x + 6)(x + 5): Using the distributive property, we can expand this expression as follows: (x + 6)(x + 5) = x(x) + x(5) + 6(x) + 6(5) = x^2 + 5x + 6x + 30 = x^2 + 11x + 30

b) (a - 4)(a + 1): Using the distributive property, we can expand this expression as follows: (a - 4)(a + 1) = a(a) + a(1) - 4(a) - 4(1) = a^2 + a - 4a - 4 = a^2 - 3a - 4

c) (2 - y)(y - 8): Using the distributive property, we can expand this expression as follows: (2 - y)(y - 8) = 2(y) - 2(8) - y(y) + y(8) = 2y - 16 - y^2 + 8y = -y^2 + 10y - 16

d) (a - 4)(2a + 1): Using the distributive property, we can expand this expression as follows: (a - 4)(2a + 1) = a(2a) + a(1) - 4(2a) - 4(1) = 2a^2 + a - 8a - 4 = 2a^2 - 7a - 4

e) (2y - 1)(3y + 2): Using the distributive property, we can expand this expression as follows: (2y - 1)(3y + 2) = 2y(3y) + 2y(2) - 1(3y) - 1(2) = 6y^2 + 4y - 3y - 2 = 6y^2 + y - 2

f) (5x - 3)(4 - 3x): Using the distributive property, we can expand this expression as follows: (5x - 3)(4 - 3x) = 5x(4) - 5x(3x) - 3(4) + 3(3x) = 20x - 15x^2 - 12 + 9x = -15x^2 + 29x - 12

These are the expanded forms of the given expressions.