Question

How do you factor $4{x^2} - 20xy + 25{y^2}$?

Answer 1

Hint: We will first split the middle term 20xy to be written in the linear combination of 10xy twice and thus, we will have some factors common from first two and last two terms and thus we have the factors.

Complete step by step solution:
We are given that we are required to factor $4{x^2} - 20xy + 25{y^2}$.
Now, we can split the middle term ‘20xy’ and we will then obtain the following expression with us:-
$ \Rightarrow 4{x^2} - 10xy - 10xy + 25{y^2}$
Now, we can observe that we can take 2x common out of the first two terms and then, we will then obtain the following expression with us:-
$ \Rightarrow 2x(2x - 5y) - 10xy + 25{y^2}$
Now, we can observe that we can take – 5y common out of the last two terms and then, we will then obtain the following expression with us:-
$ \Rightarrow 2x(2x - 5y) - 5y(2x - 5y)$
Now, we can observe that we can take (2x – 5y) common out of the last two terms and then, we will then obtain the following expression with us:-
$ \Rightarrow (2x - 5y)(2x - 5y)$
Now, we have the same factor (2x – 5y) twice, we will then obtain the following expression with us:-
$ \Rightarrow {(2x - 5y)^2}$

Thus, we have the required factors with us which is (2x – 5y) twice.

Note: The students must note that we have an alternate way to do the same method. Thus, we have the following way which is known as the way of completing square.
Alternate Way:
We are given that we are required to factor $4{x^2} - 20xy + 25{y^2}$.
We can write the given equation as follows:-
$ \Rightarrow {\left( {2x} \right)^2} - 2\left( {2x} \right)\left( {5y} \right) + {\left( {5y} \right)^2}$ …………..(1)
Now, we know that we have a formula given by the following expression:-
$ \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ for any two numbers a and b.
Replacing a by 2x and b by 5y, we will then obtain the following equation with us:-
$ \Rightarrow {\left( {2x - 5y} \right)^2} = {\left( {2x} \right)^2} - 2\left( {2x} \right)\left( {5y} \right) + {\left( {5y} \right)^2}$
Using this in equation number 1, we have the required.
Thus, we have the required factors as required that is (2x – 5y) twice.