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Question 1209704: Find a monic quartic polynomial f(x) with rational coefficients whose roots include x = 1 + \sqrt{5} - 7 + \sqrt{3}. Give your answer in expanded form.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to find the monic quartic polynomial:
1. **Identify the roots and their conjugates:**
We are given the root $x = 1 + \sqrt{5} - 7 + \sqrt{3} = -6 + \sqrt{5} + \sqrt{3}$. Since the polynomial has rational coefficients, the conjugate roots must also be present. These are:
* $-6 + \sqrt{5} - \sqrt{3}$
* $-6 - \sqrt{5} + \sqrt{3}$
* $-6 - \sqrt{5} - \sqrt{3}$
2. **Form quadratic factors:**
We can group these roots into pairs to create quadratic factors. Let's pair the roots as follows:
* Pair 1: $-6 + \sqrt{5} + \sqrt{3}$ and $-6 + \sqrt{5} - \sqrt{3}$
* Pair 2: $-6 - \sqrt{5} + \sqrt{3}$ and $-6 - \sqrt{5} - \sqrt{3}$
For Pair 1, the quadratic factor is:
\begin{align*} \label{eq:1}(x - (-6 + \sqrt{5} + \sqrt{3}))(x - (-6 + \sqrt{5} - \sqrt{3})) &= (x + 6 - \sqrt{5} - \sqrt{3})(x + 6 - \sqrt{5} + \sqrt{3}) \\ &= ((x+6-\sqrt{5}) - \sqrt{3})((x+6-\sqrt{5}) + \sqrt{3}) \\ &= (x+6-\sqrt{5})^2 - (\sqrt{3})^2 \\ &= (x^2 + 36 + 5 + 12x - 2x\sqrt{5} - 12\sqrt{5}) - 3 \\ &= x^2 + 12x + 38 - 2\sqrt{5}(x+6)\end{align*}
For Pair 2, the quadratic factor is:
\begin{align*} (x - (-6 - \sqrt{5} + \sqrt{3}))(x - (-6 - \sqrt{5} - \sqrt{3})) &= (x + 6 + \sqrt{5} - \sqrt{3})(x + 6 + \sqrt{5} + \sqrt{3}) \\ &= ((x+6+\sqrt{5}) - \sqrt{3})((x+6+\sqrt{5}) + \sqrt{3}) \\ &= (x+6+\sqrt{5})^2 - (\sqrt{3})^2 \\ &= (x^2 + 36 + 5 + 12x + 2x\sqrt{5} + 12\sqrt{5}) - 3 \\ &= x^2 + 12x + 38 + 2\sqrt{5}(x+6)\end{align*}
3. **Multiply the quadratic factors:**
Now, multiply the two quadratic factors:
\begin{align*} &(x^2 + 12x + 38 - 2\sqrt{5}(x+6))(x^2 + 12x + 38 + 2\sqrt{5}(x+6)) \\ &= (x^2 + 12x + 38)^2 - (2\sqrt{5}(x+6))^2 \\ &= (x^2 + 12x + 38)^2 - 20(x+6)^2 \\ &= (x^4 + 144x^2 + 1444 + 24x^3 + 76x^2 + 912x) - 20(x^2 + 12x + 36) \\ &= x^4 + 24x^3 + 220x^2 + 912x + 1444 - 20x^2 - 240x - 720 \\ &= x^4 + 24x^3 + 200x^2 + 672x + 724\end{align*}
Therefore, the monic quartic polynomial is $f(x) = x^4 + 24x^3 + 200x^2 + 672x + 724$.
Final Answer: The final answer is $\boxed{x^4+24x^3+200x^2+672x+724}$
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