a)
[tex]A(x) = x^4 - 7x^2 + 6x = x(x^3 - 7x + 6) = f(x) B(x)\\x(x^3 - 7x + 6) = 0 \iff x = 0 \lor x^3 - 7x + 6 = 0\\f(x) = 0 \iff x = 0\\B(2) = 2^3 - 7 \cdot 2 + 6 = 8 - 14 + 6 = 0\\\frac{B(x)}{x - 2} = x^2 + 2x - 3 \implies B(x) = (x - 2)(x^2 + 2x - 3) = g(x) h(x)\\(x - 2)(x^2 + 2x - 3) = 0 \iff x - 2 = 0 \lor x^2 + 2x - 3 = 0\\g(x) = 0 \iff x = 2\\\Delta = 2^2 - 4 \cdot 1 \cdot (-3) = 4 + 12 = 16\\\sqrt{\Delta} = \sqrt{16} = \{-4,\ 4\}\\x_1 = \frac{-2 - 4}{2} = -3\\x_2 = \frac{-2 + 4}{2} = 1\\[/tex]
[tex]h(x) = 0 \iff x = \{-3,\ 1\}\\[/tex]
[tex]x \in \{-3,\ 0,\ 1,\ 2\}[/tex]
b)
[tex]2x^3 + 8x = 5x^2 + 20 \ \ \ \ \ | - 5x^2 + 20\\A(x) = 2x^3 - 5x^2 + 8x - 20 = x^2(2x - 5) + 4(2x - 5) = (x^2 + 4)(2x - 5) = f(x)g(x)\\(x^2 + 4)(2x - 5) = 0 \iff x^2 + 4 = 0 \lor 2x - 5 = 0\\f(x) = x^2 + 4\\\Delta = 0^2 - 4 \cdot 1 \cdot 4 = -16\\\Delta < 0 \implies x _0 \in \varnothing\\g(x) = 2x - 5\\2x - 5 = 0 \ \ \ \ \ | +5\\2x = 5 \ \ \ \ \ | : 2\\x = \frac{5}{2} = 2\frac{1}{2} \\g(x) = 0 \iff x = 2\frac{1}{2}[/tex]
[tex]x = 2\frac{1}{2}[/tex]
c)
[tex]A(x) = 4x^4 + 12x^3 + 13x^2 + 6x + 1\\A(-1) = 4 \cdot (-1)^4 + 12 \cdot (-1)^3 + 13 \cdot (1)^3 + 6 \cdot (-1) + 1 = 4 - 12 + 13 - 6 + 1 = 0\\\frac{A(x)}{x + 1} = 4x^3 + 8x^2 + 5x + 1 \implies A(x) = (x + 1)(4x^3 + 8x^2 + 5x + 1) = f(x)B(x)\\ (x + 1)(4x^3 + 8x^2 + 5x + 1) = 0 \iff x + 1 = 0 \lor 4x^3 + 8x^2 + 5x + 1 = 0\\x + 1 = 0 \ \ \ \ \ | -1\\x = -1\\f(x) = 0 \iff x = -1\\B(-1) = 4 \cdot (-1)^3 + 8 \cdot (-1)^2 + 5 \cdot (-1) + 1 = -4 + 8 - 5 + 1 = 0\\[/tex]
[tex]\frac{B(x)}{x + 1} = 4x^2 + 4x + 1 \implies B(x) = (x+1)(4x^2 + 4x + 1) = g(x)h(x)\\(x+1)(4x^2 + 4x + 1) = 0 \iff x + 1 = 0 \lor 4x^2 + 4x + 1 = 0\\x + 1 = 0 \ \ \ \ \ | -1\\x = -1\\g(x) = 0 \iff x = -1\\\Delta = 4^2 - 4 \cdot 4 \cdot 1 = 16 - 16 = 0\\\sqrt{\Delta} = \sqrt{0} = 0\\x_0 = \frac{-4}{2 \cdot 4} = \frac{-4}{8} = -\frac{1}{2} \\h(x) = 0 \iff x = -\frac{1}{2}[/tex]
[tex]x = \{-1,\ -\frac{1}{2} \}[/tex]