# With the binomial expansion of $(3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$

I know how to expand the binomial expansion but I have no idea how to do this second part.

Hence Express $$(3 - \sqrt 2)^4$$ in the form of $$P+Q\sqrt 2$$

Where P and Q are integers, and then how to state the values of P and Q.

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Try and prove by induction that for each $n$, if $m,p$ are integers, $(m+p\sqrt 2)^n=M+P\sqrt 2$ where $M$ and $P$ are also integers. –  Pedro Tamaroff Apr 19 '13 at 17:10
See Wikipedia entry Binomial theorem. –  Américo Tavares Apr 19 '13 at 17:17

The binomial expansion of $\left(x + 3\right)^4$ is:$$x^4 ~ ~ + ~ ~ \binom{4}{1}x^3\cdot 3 ~ ~+ ~ ~ \binom{4}{2}x^2 \cdot 9 ~ ~ + ~ ~ \binom{4}{3}x\cdot 27 ~ ~ + ~ ~ 81$$You simplify that and get $x^4+12 x^3+54 x^2+108 x+81$. Since the question asks us to use this expansion, we can plug in $-\sqrt 2$ for $x$.

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Hmm so i take it when i plug it in i need to leave it in the for with a -√2, when i do this i get 226.94..+108√2, which does not seam as P is not a whole number. –  jackdh Apr 19 '13 at 17:46

I think easier way is to calculate $$(3-\sqrt{2})^4 = ((3-\sqrt{2})^2)^2 = (11 - 6\sqrt{2})^2=193-132\sqrt{2}$$

Binomial expansion looks like: $$(3-\sqrt{2})^4 = \sum_{k=0}^4\binom{4}{k}3^k(-\sqrt{2})^{4-k}$$

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Would you mind explaining slightly on slightly? –  jackdh Apr 19 '13 at 17:03
forgetting $\binom{4}{k}$? –  Parth Kohli Apr 19 '13 at 17:08
Yes. Thank you. –  UrošSlovenija Apr 19 '13 at 17:10

Using the binomial theorem \begin{align*} (3-\sqrt{2})^4 &= \sum_{k = 0}^4 \binom{4}{k} (3)^k (-\sqrt{2})^{4-k}\\ &= 3^0\binom{4}{0}(-\sqrt{2})^4 + (3^1) \binom{4}{1} (-\sqrt{2})^3 + (3^2)\binom{4}{2}(-\sqrt{2})^2 \\ &\quad + (3^3)\binom{4}{3}(-\sqrt{2}) + (3^4)\binom{4}{4}(-\sqrt{2})^0\\ &= (1)(1)(4) + (3)(4)(-2\sqrt{2}) + (9)(6)(2) + (27)(4)(-\sqrt{2}) + (81)(1)(1)\\ &= 193 - 132\sqrt{2} \end{align*}

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Wait a moment, somethings not right here. –  Suugaku Apr 19 '13 at 17:08
Where? I don't see it. –  Suugaku Apr 19 '13 at 17:23
Oh. The $(3-\sqrt{2})^4$ is missing the $(3$? It is displaying fine for me. Maybe it is an alignment issue and how the tex is being displayed since the second line is so long. –  Suugaku Apr 19 '13 at 17:27
Is that better? –  Suugaku Apr 19 '13 at 17:28
Yes.now it's now ok. –  ABC Apr 19 '13 at 17:29

Here is the binomial formula with two complex numbers $a$ and $b$ : $$(a+b)^n = \sum_{k=0}^n {n \choose k}\ a^{n-k} b^k$$

For $n=4$ you get :

$$\begin{array}{rcl} (a+b)^4 & = & \sum_{k=0}^4 {4 \choose k}\ a^{4-k} b^k \\ & =& {4 \choose 0}\ a^4b^0+{4 \choose 1}\ a^3 b^1+{4\choose 2}\ a^2b^2+{4 \choose 3}\ a^1b^3+{4\choose 4}\ a^0 b^4 \\ &=& a^4+4a^3b+6a^2b^2+4ab^3+b^4\end{array}$$

Now choose $a=3$ and $b=-\sqrt 2$.

You have $a^2=9$; $a^3=27$; and $a^4=81$.

You have $b^2=\left(-\sqrt 2\right)^2=2$; $b^3=b^2b=-2\sqrt 2$ and $b^4=\left(b^2\right)^2=2^2=4$.

Using the above formula, you get : $$\begin{array}{rcl}\left(3-\sqrt 2\right)^2 & =& 81+4\times 27\times\left(-\sqrt 2\right)+6\times 9\times 2+4\times 3\times \left(-2\sqrt 2\right)+4 \\ & = & 81-108\sqrt 2+108-24\sqrt 2+4 \\ & = & 193-132\sqrt 2\end{array}$$

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We know $x^{a\times b}=(x^a)^b$

Also $(a+b)^2=a^2+b^2+2ab$

$$(3-\sqrt2)^4=[(3-\sqrt2)^2]^2$$ Now, $$[11-6\sqrt2]^2=193-132\sqrt2$$

Though also this can be done by binomial expansion. Substitute $x=-\sqrt2$ in the expression you got in 1st part . but still this one would be shorter and less probable to commit a mistake.

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Hmm im not really seeing what you did between [(3−2√)2]2, and [11−62√]2 = 193−1322√ –  jackdh Apr 19 '13 at 17:27
@jack $[(3 - \sqrt{2})^2]^2 = [11 - 6\sqrt2]^2$ because $11 - 6\sqrt2 = (3 - \sqrt{2})^2$ –  Parth Kohli Apr 19 '13 at 17:31