Question

Factorise the following using completing square method?

`9x^2 -9(a+b)x +(2a^2+5ab+2b^2)`

Answer 1

`(3x-2a-b)(3x-a-2b)`

Explanation:

.

`9x^2-9(a+b)x+(2a^2+5ab+2b^2)`

This seems to be in the form of:

`(alpha-beta)^2=alpha^2-2alphabeta+beta^2`

Let's factor out `9`:

`9[x^2-(a+b)x+(2a^2+5ab+2b^2)/9]`

Now, it looks like it has the following burried in the expression:

`(alpha-beta)^2=(alpha^2-2alphabeta+beta^2)`

`alpha^2=x^2`, `-2alphabeta=-(a+b)x`, `beta^2=??`

Since `alpha=x`, then:

`-2alphabeta=-2xbeta`

`-2zbeta=-(a+b)x`

Let's divide both sides by `-2x`:

`beta=(a+b)/2`

Now, if we write the expression as:

`9[x^2-2x(a+b)/2+((a+b)/2)^2-((a+b)/2)^2+(2a^2+5ab+2b^2)/9]`

the expression is unchanged and we can write it as:

`9[(x-(a+b)/2)^2-(a^2+2ab+b^2)/4+(2a^2+5ab+2b^2)/9]=`

`9[(x-(a+b)/2)^2-(9a^2+18ab+9b^2-8a^2-20ab-8b^2)/36]=`

`9[(x-(a+b)/2)^2-(a^2-2ab+b^2)/36]=`

`9[(x-(a+b)/2)^2-((a-b)^2)/36]=`

`9[(x-(a+b)/2)^2-((a-b)/6)^2]=`

`9(x-(a+b)/2-(a-b)/6)(x-(a+b)/2+(a-b)/6)=`

`9(x-(3a+3b+a-b)/6)(x-(3a+3b-a+b)/6)=`

`9(x-(4a+2b)/6)(x-(2a+4b)/6)=`

`9(x-(2a+b)/3)(x-(a+2b)/3)=`

`(3x-2a-b)(3x-a-2b)`

Answer 2

`9x^2-9(a+b)x+(2a^2+5ab+2b^2) = (3x-2a-b)(3x-a-2b)`

Explanation:

Given:

`9x^2-9(a+b)x+(2a^2+5ab+2b^2)`

In the following we premultiply by `4=2^2` to avoid working with fractions, complete the square, then use the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

with `A=(6x-(3a+3b))` and `B=(a-b)`.

We find:

`4(9x^2-9(a+b)x+(2a^2+5ab+2b^2))`

`=36x^2-36(a+b)x+(8a^2+20ab+8b^2)`

`=(6x)^2-2(6x)(3a+3b)+(9a^2+18ab+9b^2)-(a^2-2ab+b^2)`

`=(6x-(3a+3b))^2-(a-b)^2`

`=(6x-(3a+3b)-(a-b))(6x-(3a+3b)+(a-b))`

`=(6x-4a-2b)(6x-2a-4b)`

`=4(3x-2a-b)(3x-a-2b)`

So:

`9x^2-9(a+b)x+(2a^2+5ab+2b^2) = (3x-2a-b)(3x-a-2b)`