# Math Help - Factorization Question 1

1. ## Factorization Question 1

I am stuck trying to factorize the following question:

$\text{Factorize the following: } a^2c^2+acd+acd+d^2$

The answer in the book is

$= (ac+d)^2$

Here is my attempt to reach this answer

$= aacc+acd+acd+dd$

$= aacc+2acd+dd$

2. ## Re: Factorization Question 1

Notice that $a^2c^2+acd+acd+d^2$ can be written as $a^2c^2+2acd+d^2=(ac)^2+2acd+d^2$
Do you recognize (in general) $x^2+2xy+y^2=(x+y)^2$ in it? ...

EDIT: I changed the variables, because otherwise it can be confusing.

3. ## Re: Factorization Question 1

Originally Posted by Siron
...Do you recognize (in general) $x^2+2xy+y^2=(x+y)^2$ in it?
I recognized it now thanks to the way you laid it out. However I did not see that before because I was trying to factorize the expression by grouping the terms in pairs so that each pair of terms has a common factor.

My question is, if I did not recognize what you pointed out, how do I factorize this expression by grouping?

For example (assuming my working is correct)

$\text{Factorize the following }ab(x^2+y^2)-cd(x^2+y^2)$

$=ab(xx+yy)-cd(xx+yy)$

$=(ab-cd)(xx+yy)$

4. ## Re: Factorization Question 1

Originally Posted by sparky
I recognized it now thanks to the way you laid it out. However I did not see that before because I was trying to factorize the expression by grouping the terms in pairs so that each pair of terms has a common factor.

My question is, if I did not recognize what you pointed out, how do I factorize this expression by grouping?

For example (assuming my working is correct)

$\text{Factorize the following }ab(x^2+y^2)-cd(x^2+y^2)$

$=ab(xx+yy)-cd(xx+yy)$

$=(ab-cd)(xx+yy)$
Yes but $(xx+yy)$ is a poor form to write it in, instead use $(x^2+y^2)$

5. ## Re: Factorization Question 1

yes if you did not recognize , you can factorize
$a^2c^2+acd+acd+d^2=ac(ac+d)+d(ac+d)=(ac+d)(ac+d)=( ac+d)^2$

6. ## Re: Factorization Question 1

Originally Posted by waqarhaider
yes if you did not recognize , you can factorize
$a^2c^2+acd+acd+d^2=ac(ac+d)+d(ac+d)=(ac+d)(ac+d)=( ac+d)^2$
That was tricky, thanks I understand now.