Question

B)(x²-3)(x+2)+(x²+3)(x-2)=4 срочноооо

Answer 1

Слылсьыждмладйдчьбыбсьс

Answer 2

Sure, let's solve the equation for x:

B)(x²-3)(x+2) + (x²+3)(x-2) = 4

First, let's expand the terms:

(x²-3)(x+2) = x²(x) + x²(2) - 3(x) - 3(2) = x³ + 2x² - 3x - 6 (x²+3)(x-2) = x²(x) - x²(2) + 3(x) - 3(2) = x³ - 2x² + 3x - 6

Now, the equation becomes:

B)(x³ + 2x² - 3x - 6) + (x³ - 2x² + 3x - 6) = 4

Now, combine like terms:

Bx³ + 2Bx² - 3Bx - 6B + x³ - 2x² + 3x - 6 = 4

Combine the x³ and x² terms:

(B + 1)x³ + (2B - 2)x² + (3 - 3B)x - (6B + 6) = 4

Now, since this is an equation, we want to set it equal to zero:

(B + 1)x³ + (2B - 2)x² + (3 - 3B)x - (6B + 6) - 4 = 0

Simplify further:

(B + 1)x³ + (2B - 2)x² + (3 - 3B)x - 10B - 10 = 0

Now, you might have some specific values for B in mind, but without knowing its value, it is not possible to find a specific solution for x.

If you have a particular value for B, you can substitute it into the equation and solve for x. Otherwise, the equation is in its factored form and can't be solved further without more information.