Question

-6 (a+4)(a-1) (5y2степени+1)(3y2степен-1)

Answer 1

-6(а^2+4а-а)=-6а^2-24а+6а

(5у^2+1)(3у^2-1)=15у^4 - 5у^2 +3у^2 - 1=15у^4 - 2у^2 - 1

Answer 2

To simplify this expression, we can use the distributive property and combine like terms:

-6(a+4)(a-1)(5y^2+1)(3y^2-1)

= -6(a^2 - a + 4a - 4)(5y^2+1)(3y^2-1)

= -6(a^2 + 3a - 4)(5y^2+1)(3y^2-1)

= -6(5y^2+1)(3y^2-1)(a^2 + 3a - 4)

We can factor the quadratic expression a^2 + 3a - 4 by finding two numbers that multiply to -4 and add to 3. These numbers are 4 and -1, so we can write:

a^2 + 3a - 4 = (a + 4)(a - 1)

Substituting this back into our expression, we get:

-6(5y^2+1)(3y^2-1)(a+4)(a-1)

So the final simplified expression is:

-6(a+4)(a-1)(5y^2+1)(3y^2-1) = -6(5y^2+1)(3y^2-1)(a+4)(a-1)