Question

(x+3)(x+1)(x+5)(x+7)=-16

Answer 1

Объяснение:

$(x+3)*(x+1)*(x+5)*(x+7)=-16\\(x+1)*(x+7)*(x+3)*(x+5)=-16\\(x^2+8x+7)*(x^2+8x+15)=-16.$

Пусть x²+8x+7=t        ⇒

$t*(t+8)=-16\\t^2+8t+16=0\\t^2+2*t*4+4^2=0\\(t+4)^2=0\\t+4=0\\t=x^2+8t+7=-4\\x^2+8t+11=0\\D=20\ \ \ \ \sqrt{D}=2\sqrt{5} \\x_1=-4-\sqrt{5}\ \ \ \ x_2=-4+\sqrt{5} .$

Ответ: x₁=-4-√5      x₂=-4+√5.

Answer 2

To solve the equation $(x+3)(x+1)(x+5)(x+7) = -16$, you can follow these steps:

1. Expand the left side of the equation:

$(x+3)(x+1)(x+5)(x+7) = (x+3)(x^2+6x+7)(x+7)$

2. Multiply the polynomials:

$x(x^2+6x+7) + 3(x^2+6x+7)(x) + 7(x^2+6x+7) = -16$

3. Distribute and simplify each term:

$x^3+6x^2+7x + 3x^3+18x^2+21x + 7x^2+42x+49 = -16$

4. Combine like terms:

$4x^3 + 31x^2 + 70x + 49 = -16$

5. Move all the terms to one side of the equation to set it equal to zero:

$4x^3 + 31x^2 + 70x + 49 + 16 = 0$

6. Combine constants on the left side:

$4x^3 + 31x^2 + 70x + 65 = 0$

Now you have a cubic equation. Unfortunately, cubic equations can be challenging to solve algebraically, but you can use numerical methods or software to find approximate solutions.