Questions: Write the function in the form f(x)=(x-k) q(x)+r(x) for the given value of k. f(x)=-3 x^3+8 x^2+13 x-4, k=2+sqrt3 f(x)= Demonstrate that f(k)=r. f(2+sqrt3)=

Solution Steps

We need to express the polynomial function $f(x) = -3x^3 + 8x^2 + 13x - 4$ in the form $f(x) = (x - k) q(x) + r(x)$ for $k = 2 + \sqrt{3}$. Then, we need to demonstrate that $f(k) = r$.

To express $f(x)$ in the desired form, we perform polynomial division of $f(x)$ by $x - (2 + \sqrt{3})$.

1. Divide the leading term: $$\frac{-3x^3}{x} = -3x^2$$
2. Multiply and subtract: $$(-3x^2)(x - (2 + \sqrt{3})) = -3x^3 + 3(2 + \sqrt{3})x^2$$ $$-3x^3 + 8x^2 + 13x - 4 - (-3x^3 + 6x^2 + 3\sqrt{3}x^2) = 2x^2 - 3\sqrt{3}x^2 + 13x - 4$$
3. Combine like terms: $$2x^2 - 3\sqrt{3}x^2 = (2 - 3\sqrt{3})x^2$$
4. Repeat the process: $$\frac{(2 - 3\sqrt{3})x^2}{x} = (2 - 3\sqrt{3})x$$ $$((2 - 3\sqrt{3})x)(x - (2 + \sqrt{3})) = (2 - 3\sqrt{3})x^2 - (2 - 3\sqrt{3})(2 + \sqrt{3})x$$ $$(2 - 3\sqrt{3})x^2 - (4 + 2\sqrt{3} - 6\sqrt{3} - 9)x = (2 - 3\sqrt{3})x^2 - (4 - 4\sqrt{3} - 9)x$$ $$2x^2 - 3\sqrt{3}x^2 + 13x - 4 - (2x^2 - 3\sqrt{3}x^2 - 4x + 4\sqrt{3}x + 9x) = 13x - 4 - (-4x + 4\sqrt{3}x + 9x)$$ $$13x - 4 - (-4x + 4\sqrt{3}x + 9x) = 13x - 4 + 4x - 4\sqrt{3}x - 9x = (13 + 4 - 9)x - 4 - 4\sqrt{3}x = 8x - 4 - 4\sqrt{3}x$$
5. Combine like terms: $$8x - 4 - 4\sqrt{3}x = (8 - 4\sqrt{3})x - 4$$
6. Repeat the process: $$\frac{(8 - 4\sqrt{3})x}{x} = 8 - 4\sqrt{3}$$ $$(8 - 4\sqrt{3})(x - (2 + \sqrt{3})) = (8 - 4\sqrt{3})x - (8 - 4\sqrt{3})(2 + \sqrt{3})$$ $$(8 - 4\sqrt{3})x - (16 + 8\sqrt{3} - 8\sqrt{3} - 12) = (8 - 4\sqrt{3})x - 4$$ $$8x - 4 - 4\sqrt{3}x - 4 = 0$$

Substitute $k = 2 + \sqrt{3}$ into $f(x)$: $$f(2 + \sqrt{3}) = -3(2 + \sqrt{3})^3 + 8(2 + \sqrt{3})^2 + 13(2 + \sqrt{3}) - 4$$ Calculate each term: $$(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$$ $$(2 + \sqrt{3})^3 = (2 + \sqrt{3})(7 + 4\sqrt{3}) = 14 + 8\sqrt{3} + 7\sqrt{3} + 12 = 26 + 15\sqrt{3}$$ $$-3(2 + \sqrt{3})^3 = -3(26 + 15\sqrt{3}) = -78 - 45\sqrt{3}$$ $$8(2 + \sqrt{3})^2 = 8(7 + 4\sqrt{3}) = 56 + 32\sqrt{3}$$ $$13(2 + \sqrt{3}) = 26 + 13\sqrt{3}$$ Combine all terms: $$f(2 + \sqrt{3}) = -78 - 45\sqrt{3} + 56 + 32\sqrt{3} + 26 + 13\sqrt{3} - 4$$ $$= -78 + 56 + 26 - 4 + (-45\sqrt{3} + 32\sqrt{3} + 13\sqrt{3})$$ $$= 0$$

Final Answer