# If $x = \frac{\sqrt{111}-1}{2}$, calculate $(2x^{5} + 2x^{4} - 53x^{3} - 57x + 54)^{2004}$.

I already have two solutions for this problem, it is for high school students with an advanced level. I would like to know if there are better or more creative approaches on the problem. Here are my solutions:

1. (1st solution): Notice that $$x^{2} = (\frac{\sqrt{111}-1}{2})^{2} = 28 - \frac{\sqrt{111}}{2}$$ $$x^{3} = x \cdot x^{2} = \left( \frac{\sqrt{111}}{2} - \frac{1}{2} \right) \left( 28 - \frac{\sqrt{111}}{2} \right) = 14 \sqrt{111} - 111/4 - 14 + \frac{\sqrt{111}}{4} = \frac{57 \sqrt{111}}{4} - \frac{167}{4}$$ $$x^{4} = (x^{2})^{2} = \frac{ 111 + 4 \cdot 784}{4} - 28 \sqrt{111}$$ $$x^{5} = x^{4} \cdot x = \left( \frac{ 111 + 4 \cdot 784}{4} - 28 \sqrt{111} \right)\left(\frac{\sqrt{111}-1}{2} \right) = \left( \frac{ 3247 }{4} - 28 \sqrt{111} \right) \left(\frac{\sqrt{111}-1}{2} \right)$$ $$= \sqrt{111}\frac{3359}{8} - \frac{15679}{8}$$

So we have $$2x^{5} + 2x^{4} - 53x^{3} - 57x + 54 =$$ $$(\sqrt{111}\frac{3359}{4} - \frac{15679}{4}) + (\frac{ 111 + 4 \cdot 784}{2} - 56 \sqrt{111}) - 53 (\frac{57 \sqrt{111}}{4} - \frac{167}{4}) - 57 (\frac{\sqrt{111}-1}{2}) + 54$$ $$= \sqrt{111} (\frac{3359}{4} - \frac{3359}{4}) - \frac{15679}{4} + \frac{ 111 + 4 \cdot 784}{2} + \frac{53 \cdot 167}{4} + \frac{57}{2} + 54$$ $$= - \frac{15679}{4} + \frac{ 222 + 8 \cdot 784}{4} + \frac{53 \cdot 167}{4} + \frac{114}{4} + \frac{216}{4}$$ $$= - \frac{15679}{4} + \frac{ 5600 + 894 }{4} + \frac{5300 + 3180 + 371 }{4} + \frac{114}{4} + \frac{216}{4}$$ $$= - \frac{15679}{4} + \frac{ 6494 }{4} + \frac{8851}{4} + \frac{114}{4} + \frac{216}{4}$$ $$= -4/4 = -1,$$ and the answer is $$(-1)^{2004} = 1.$$ The above solution requires quite tedious calculation. Below is an alternative solution.

1. (2nd solution): Notice that $$x = \frac{\sqrt{111}-1}{2}$$ is equivalent with $$(2x + 1)^{2} = 111$$ $$4x^{2} + 4x + 1 = 111$$ $$4x^{2} + 4x - 110 = 0$$ $$(2x^{2} + 2x - 55) = 0 \:\: ........ \:\: (1)$$ Multiply $$(1)$$ with $$x^{3}$$ we get $$(2x^{5} + 2x^{4} - 55x^{3}) = 0$$ Multiply $$(1)$$ with $$x$$ we get $$(2x^{3} + 2x^{2} - 55x) = 0$$ Sum both of them and we get: $$2x^{5} + 2x^{4} - 53 x^{3} + 2x^{2} - 55x = 0 \:\: ........ \:\: (2)$$ and now we have the 1st 3 terms of the form that we want to calculate. Substract $$(2)$$ with $$(1)$$ to get: $$2x^{5} + 2x^{4} - 53 x^{3} - 57x + 55 = 0$$ $$2x^{5} + 2x^{4} - 53 x^{3} - 57x + 54 = -1$$ So the answer is $$(-1)^{2004} = 1.$$
• Your second solution is probably the best one there is. – TheSimpliFire May 18 '19 at 12:56
• Forget the brute-force first solution... – Jean Marie May 18 '19 at 13:01
• @JeanMarie I upvote although the 1st solution was not really that tedious. The 2nd solution does not easily come to mind, I peeked into the solution sheet. – Arief May 18 '19 at 14:25
• No real number other than $0,1,-1$ has a neat form when raised to the power of $2004$. A ‘sneaky’ but smart student would use estimations on $x$ to guess between $0$ and $1$. – Szeto May 21 '19 at 23:32

Need long division

Divide $$2x^5+2x^4-53x^3-57x+54$$ by $$2x^2+2x-55$$ to express

$$2x^5+2x^4-53x^3-57x+54=q(x)\cdot(2x^2+2x-5)+r(x)$$ where $$q(x)$$ is the quotient and $$r(x)$$ is the remainder.

$$\implies2x^5+2x^4-53x^3-57x+54=r(x)$$ as $$2x^2+2x-5=0$$

Here $$r(x)=-1$$

• you mean remainder – user10354138 May 18 '19 at 13:20
• – lab bhattacharjee May 18 '19 at 13:23
• How do you know that $2x^{2} + 2x - 55$ should be the divisor? – Arief May 18 '19 at 14:15
• @AriefAnbiya, My idea was to find $$2x^5+2x^4-53x^3-57x+54$$ as $$(2x^2+2x-55)q(x)+r(x)=r(x)$$ as $2x^2+2x-55=0$ – lab bhattacharjee May 18 '19 at 14:17
• Well, it is shorter but it is not different than the 2nd solution.., $q(x) = x^{3} + x - 1$. – Arief May 18 '19 at 14:27

What I did was equivalent to division, or to your second solution, but went as follows.

First I got to $$2x^2+2x=55$$ and multiplied this by $$x^3$$ so that $$2x^5+2x^4=55x^3$$

Now I substituted this into: $$p(x)=2x^5+2x^4-53x^3-57x+55=55x^3-53x^3-57x+54=2x^3-57x+54$$

Next I used $$55x=2x^3+2x^2$$ to give $$p(x)=2x^3-55x-2x+54=2x^3-2x^3-2x^2-2x+54=-2x^2-2x+54=-55+54=-1$$

I find it is sometimes handy to remember that if I am working with an $$x$$ such that $$p(x)=q(x)$$, I can substitute $$p(x)$$ and $$q(x)$$ in any expression involving $$x$$. Formally this works the same as polynomial division by $$p(x)-q(x)$$. However it is sometimes more flexible and easier to apply.

It can turn out to require more lengthy calculations than a more regular method, but it has the advantage of keeping the expressions simple at each stage. If I can't see how to make more progress, I can always resort to division, but I can start with any simpler expression I've derived.

• I upvote. Although still only a bit different with the 2nd solution.. – Arief May 18 '19 at 14:46