Factorise the following using completing square method?

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

2 Answers
Feb 8, 2018

(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)

Feb 8, 2018

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)