What is the rationalising factor of the given number?

2+7+210

2 Answers
Aug 13, 2018

6+5+72410

Explanation:

I will assume that you are looking for a radical conjugate, that is an expression which when multiplied by the given expression gives a rational product.

Given:

2+7+210

First we should check whether the expression 7+210 can be simplified.

Note that in https://socratic.org/s/aTtiPKas I found that given:

p+qr with p,q,r>0

then if p2q2r is a perfect square s2 then:

p+qr=2p+2s2+2p2s2

In our example, putting p=7, q=2, r=10, then:

s=p2q2r=7222(10)=4940=9=3

So:

7+210=2(7)+2(3)2+2(7)2(3)2

7+210=202+82

7+210=5+2

So:

2+7+210=2+5+2

Since this involves two square roots, a suitable radical conjugate is formed by multiplying the variants of the expression 2±5±2 apart from the one we already have.

(2+52)(252)(25+2)

=(2+52)((25)2(2)2)

=(2+52)((445+5)2)

=(2+52)(745)

=7(2+52)45(2+52)

=(14+7572)(85+20410)

=6572+410

Note that this is negative, so, let's negate it to get the slightly more attractive radical conjugate:

6+5+72410

As a check, let's multiply this by 2+5+2 and see what we get:

(2+5+2)(6+5+72410)

=2(6+5+72410)+5(6+5+72410)+2(6+5+72410)

=(12+25+142810)+(65+5+710202)+(62+10+1485)

=31

Aug 13, 2018

Here's another way to simplify...

Explanation:

One way of simplifying 7+210 involves considering the quartic polynomial with integer coefficients of which it is one of the zeros.

The other zeros will be the variants:

7+210, 7210, and 7210

The quartic of which these are zeros can be expressed as:

(x27)240

=x414x2+9

=x46x2+98x2

=(x23)2(22x)2

=(x222x3)(x2+22x3)

=(x222x+25)(x2+22x+25)

=((x2)2(5)2)((x+2)2(5)2)

=(x25)(x2+5)(x+25)(x+2+5)

Hence zeros: x=±2±5

The greatest of these is x=2+5, which must be the greatest of ±7±210, i.e. 7+210

So:

7+210=2+5

From this point we can proceed as my other answer to multiply:

(2+52)(252)(25+2)