Question

How do you factor ${x^2} + 6x + 9 - {y^2}$?

Answer 1

Hint: We will first make use of the identity which states that ${(a + b)^2} = {a^2} + {b^2} + 2ab$. After that we will use another identity which states that \[{a^2} - {b^2} = (a - b)(a + b)\].

Complete step by step solution:
We are given that we are required to factor ${x^2} + 6x + 9 - {y^2}$.
Now, since we know that we have an identity given by the following expression:-
$ \Rightarrow {(a + b)^2} = {a^2} + {b^2} + 2ab$
Replacing a by x and b by 3, we will then obtain the following expression with us:-
$ \Rightarrow {(x + 3)^2} = {x^2} + {3^2} + 2 \times 3 \times x$
Simplifying the calculations on the right hand side of the above mentioned expression, we will then obtain the following equation with us:-
$ \Rightarrow {(x + 3)^2} = {x^2} + 6x + 9$
Putting this in the expression originally given to us, we will then obtain the following equation with us:-
$ \Rightarrow {x^2} + 6x + 9 - {y^2} = {(x + 3)^2} - {y^2}$
Now, since we know that we have a formula given by the following expression:-
\[ \Rightarrow {a^2} - {b^2} = (a - b)(a + b)\]
Replacing a by (x + 3) and b by y, we will then obtain the following expression with us:-
$ \Rightarrow {(x + 3)^2} - {y^2} = (x + 3 - y)(x + 3 + y)$
Now, we can write this as the following expression:-
$ \Rightarrow {x^2} + 6x + 9 - {y^2} = (x + 3 - y)(x + 3 + y)$

Note: The students must commit to memory the following formulas that we used in the solution given above:-
1. \[{a^2} - {b^2} = (a - b)(a + b)\]
2. ${(a + b)^2} = {a^2} + {b^2} + 2ab$
We may also prove both of the formulas as follows:-
The first one:-
Let us consider the right hand side of this formula for once.
$ \Rightarrow $R. H. S. = (a – b) (a + b)
Now, we know that we have the fact given by the following expression with us:-
$ \Rightarrow $(a + b) (c + d) = a (c + d) + b (c + d)
Replacing b by – b, c by a and d by b in the above mentioned equation, we will then obtain the following expression with us:-
$ \Rightarrow $(a + b) (a - b) = a (a + b) - b (a + b)
Simplifying it further by using the distributive property which states that a (b + c) = ab + ac, we will then obtain the following expression with us:-
$ \Rightarrow \left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} - ab + ab$
Clubbing the like terms in the right hand side of the above equation, we will then obtain the following expression with us:-
$ \Rightarrow \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
The second one:-
We will use binomial theorem to prove this which states that:-
$ \Rightarrow {\left( {a + b} \right)^n}{ = ^n}{C_0}{a^0}{b^n}{ + ^n}{C_1}{a^1}{b^{n - 1}}{ + ^n}{C_2}{a^2}{b^{n - 2}} + .........{ + ^n}{C_n}{a^n}{b^0}$
Replacing n by 2 in the above formula, we have:-
$ \Rightarrow {\left( {a + b} \right)^2}{ = ^2}{C_0}{a^0}{b^2}{ + ^2}{C_1}{a^1}{b^1}{ + ^2}{C_2}{a^2}{b^0}$
Simplifying it, we will have:-
$ \Rightarrow {(a + b)^2} = {a^2} + {b^2} + 2ab$