Question

Factorize the expression \[{x^2} - {y^2} + 6y - 9\]

Answer 1

Hint: In order to solve this question first, we have to make a perfect square of the y variable and the constant term given in the term. Then using the identities of the multiplication of sum of two numbers and subtraction of two numbers after further solving we are able to split that whole term in two-parts that two-part are the factors of the given term.

Complete step-by-step solution:
Given,
\[{x^2} - {y^2} + 6y - 9\]
On taking negative sign from the equation after variable \[x\]
After taking a common negative sign the expression looks like.
\[{x^2} - {y^2} + 6y - 9 = {x^2} - \left( {{y^2} - 6y + 9} \right)\] ……(i)
Making the perfect square by the y terms and the constant term.
The y term can be written as
\[{y^2} - 6y + 9 = {\left( {y - 3} \right)^2}\]
On putting these values in equation (i)
\[{x^2} - {y^2} + 6y - 9 = {x^2} - {\left( {y - 3} \right)^2}\]
Using the identity of \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
After using this property expression looks like
\[{x^2} - {y^2} + 6y - 9 = \left( {x - \left( {y - 3} \right)} \right)\left( {x + \left( {y - 3} \right)} \right)\]
On further calculations
\[{x^2} - {y^2} + 6y - 9 = \left( {x - y + 3} \right)\left( {x + y - 3} \right)\]
Final answer:
The factor of the given expression is
\[{x^2} - {y^2} + 6y - 9 = \left( {x - y + 3} \right)\left( {x + y - 3} \right)\]

Note: Students are confused because of looking into two variables. But there is a hint also that is a perfect square in one of the variables and able to use the identity of exponential. Many of the students are not able to make a perfect square so they are unable to factorize. If the equation of the expression is only in one variable then we factorize that polynomial by different – different methods. The methods to solve polynomials are directly splitting the middle term. By the formula which is used to find the roots of the quadratic equation. For cubic and biquadratic equations we find some roots by hit and trial methods. Then divide those factors to make them quadratic and solve that further.