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$$4\left(5+3\sqrt2\over 2\right)^4-16\left(5+3\sqrt2\over 2\right)^3-17\left(5+3\sqrt2\over 2\right)^2+27\left(5+3\sqrt2\over 2\right)-3$$

Please solve the following equation without using calculator.

Substituting $\left(5+3\sqrt2\over 2\right)$ to x must be the first step, but then I don't know how to factor it.

3 Answers 3

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We know that $$4x^4-16x^3-17x^2+27x-7 =(4x^2-20x+7)(x^2+x-1) $$ $$=(x- \frac{5+3\sqrt {2}}{2})(x-\frac {5-3\sqrt {2}}{2})(x-\frac {-1 +\sqrt {5}}{2})(x+\frac {-1-\sqrt {5}}{2}) $$

Observe that $\frac {5+3\sqrt {2}}{2} $ is a root of this polynomial. Also observe that the above polynomial is $4$ less than the required polynomial in question. Thus, the answer is $4$. Hope it helps.

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    How did you go from $(4x^2-20x+7)(x^2+x-1)$ to $(x- \frac{5+3\sqrt {2}}{2})(x-\frac {5-3\sqrt {2}}{2})(x-\frac {-1 +\sqrt {5}}{2})(x+\frac {-1-\sqrt {5}}{2})$ ? – 2017-02-16
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    @KelvinAndersen Just use the [quadratic formula](https://en.m.wikipedia.org/wiki/Quadratic_formula). – 2017-02-16
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    Use $x={b \pm {\sqrt {b^2 - 4ac}}\over 2a}$ ? How do you input it? – 2017-02-16
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Let $f(x)=4x^4-16x^3-17x^2+27x-3$. Then $g(x)=f(x)-4=(4x^2 - 20x + 7)(x^2 + x - 1)$. Since $a=(5+3\sqrt{2})/2$ satisfies $4a^2-20a+7=0$ we know that $g(a)=0$ and hence $f(a)=4$. So the answer is $4$.

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    How did you factor $4x^4-16x^3-17x^2+27x-7$ to $(4x^2−20x+7)(x^2+x−1)$ ? – 2017-02-16
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    One writes $g(x)=(ax^2+bx+c)(x^2+dx+f)$ and compares the coefficients. This gives easy equations in $a,b,c,d,f$, which I wanted to solve over the integers (which makes things even easier, e.g., $a=4, bc=-7$ over the integers has not many possibilities). – 2017-02-16
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    How did you get $b=-20$ $c=-7$ $d=1$ ? – 2017-02-16
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    We have $cf=-7$, hence $c=7,f=-1$, or $c=-7,f=1$ or $c,f$ permuted; then just continue with the other equations like that, and you will find solutions (need not be unique in general). – 2017-02-16
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Let $\alpha=\displaystyle\frac{5+\sqrt{3}}2, \beta=\displaystyle\frac{5-\sqrt{3}}2$.

$$\alpha+\beta=5; \alpha\beta=\frac 74$$ $\alpha,\beta$ are roots of the quadratic $$x^2-(\alpha+\beta)+\alpha\beta=0\\ x^2-5x+\frac 74=0\\ 4x^2-20x+7=0$$ i.e. $$\overbrace{4\alpha^2-20\alpha+7}^{f(\alpha)}=0$$ Now consider the following:

$$\begin{array} &&&&&\\ \alpha^2\cdot f(\alpha):&4\alpha^4&-2 0\alpha^3 &+7\alpha^2 & & &=0\\ \alpha\cdot f(\alpha): & &\;\;\;4\alpha^3&-20\alpha^2 &+7\alpha & &=0\\ - f(\alpha): & & &-4\alpha^2 &+20\alpha&-7&=0\\ +4: & & & & &+4&=4\\ \hline \text{Adding}: &4\alpha^4&-16\alpha^3 &-17\alpha^2 &+27\alpha &-3&=\color{red}4\\ \hline \end{array}$$