Factorization
The process of expressing a polynomial expression as the product of its factor is called factorization. In this case, the part of the expression is called the factor of the polynomial.
Example:
If the polynomial \(x^2 + 5x + 6\) is expressed as \((x+3) (x+2)\) then \((x+3) \text{ and } (x+2)\) are the factors of \(x^2 + 5x + 6\)
Important Formulae
Here are the important formulae that can be used while solving the problems.
\(
\begin{aligned}
& \textbf{Formulae} \\
& 1. (a + b)^2 = a^2 + 2ab + b^2 \\
& 2. (a - b)^2 = a^2 - 2ab + b^2 \\
& 3. (a^2 + b^2) = (a + b)^2 – 2ab \text{ or, } (a - b)^2 + 2ab \\
& 4. (a^2 – b^2) = (a + b)(a - b) \\
& 5. (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \\
& 6. (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \\
& 7. (a^3 + b^3) = (a + b)(a^2 - ab + b^2) \\
& 8. (a^3 - b^3) = (a - b)(a^2 + ab + b^2) \\
\end{aligned}
\)
Example Questions with Solutions
Tips and Tricks
Some Tips and Tricks for Factorizing the given expressions.
- Look for a common factor.
- Analyze the expression to see if we can use formulae.
- Check your steps thoroughly to see if anything went wrong.
Example Questions with Solutions
Here are some example questions.
\(
\begin{aligned}
& \textbf{ Q1. Factorize } \mathbf{(x + 2)^3} \\
& \text{Solution:} \\
& = (x + 2)^3 \\
& = x^3 + 3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2 + 2^3\\
& \therefore x^3 + 6x^2 + 12x + 8
\end{aligned}
\)
\(
\begin{aligned}
& \textbf{ Q2. Factorize } 8x^4 + 27x \\
& \text{Solution:} \\
& = 8x^4 + 27x \\
& = x(8x^3 + 27) \\
& = x\lbrace(2x)^3 + 3^3\rbrace \\
& = x[(2x + 3)\lbrace(2x)^2 + 2x \cdot 3 + 3^2\rbrace] \\
& \therefore x(2x + 3)(4x^2 + 6x + 9) \\
\end{aligned}
\)
\(
\begin{aligned}
& \textbf{ Q3. Factorize } (\frac{x^3}{y^3} - \frac{y^3}{x^3}) \\
& \text{Solution:} \\
& = \frac{x^3}{y^3} - \frac{y^3}{x^3} \\
& = (\frac {x}{y})^3 – (\frac{y}{x})^3 \\
& = (\frac{x}{y} - \frac{y}{x}) \lbrace (\frac{x}{y})^2 + \frac{x}{y} \cdot \frac{y}{x} + (\frac{y}{x})^2\rbrace \\
& \therefore (\frac{x}{y} - \frac{y}{x}) (\frac{x^2}{y^2} + \frac{y^2}{x^2} + 1) \\
\end{aligned}
\)
Factorization of the expression in the form of \((a^4 + a^2b^2 + b^4)\)
Here are the general steps for the factorization of the expression in the form of \((a^4 + a^2b^2 + b^4)\).
\(
\begin{aligned}
& \text{Steps} \\
& = a^4 + a^2b^2 + b^4 \\
& = (a^2)^2 + (b^2)^2 + a^2b^2\\
& = (a^2 + b^2)^2 – 2a^2b^2 + a^2b^2\\
& = (a^2 + b^2)^2 – (ab)^2\\
& \therefore (a^2 + ab + b^2)( a^2 - ab + b^2)
\end{aligned}
\)
Example Questions with Solution
Here are some example questions and solutions.
\(
\begin{aligned}
& \text{Q1. Factorize } (x^4 + x^2y^2 + y^4) \\
& \text{Solution:} \\
& = x^4 + x^2y^2 + y^4 \\
& = (x^2)^2 + (y^2)^2 + x^2y^2\\
& = (x^2 + y^2)^2 – 2x^2y^2 + x^2y^2\\
& = (x^2 + y^2)^2 – (xy)^2\\
& \therefore (x^2 + xy + y^2)( x^2 - xy + y^2)
\end{aligned}
\)
\(
\begin{aligned}
& \text{Q2. Factorize } (x^4 + 5x^2y^2 + 4y^4) \\
& \text{Solution:} \\
& = x^4 - 5x^2y^2 + 4y^4 \\
& = (x^2)^2 + (2y^2)^2 - 5x^2y^2 \\
& = (x^2 + 2y^2)^2 – 2 \cdot x^2 \cdot 2y^2 - 5x^2y^2 \\
& = (x^2 + 2y^2)^2 – 4x^2y^2 - 5x^2y^2 \\
& = (x^2 + 2y^2)^2 – 9x^2y^2 \\
& = (x^2 + 2y^2)^2 – (3xy)^2 \\
& \therefore (x^2 + 3xy + 2y^2)( x^2 - 3xy + 2y^2)
\end{aligned}
\)
\(
\begin{aligned}
& \text{Q3. Factorize } (y^4 + \frac{1}{y^4} + 1) \\
& \text{Solution:} \\
& = y^4 + \frac{1}{y^4} + 1 \\
& = (y^2)^2 + (\frac{1}{y^2})^2 + 1 \\
& = (y^2 + \frac{1}{y^2})^2 – 2 \cdot y^2 \cdot \frac{1}{y^2} + 1 \\
& = (y^2 + \frac{1}{y^2})^2 – 2 + 1 \\
& = (y^2 + \frac{1}{y^2})^2 – 1^2 \\
& \therefore (y^2 + \frac{1}{y^2} + 1)(y^2 + \frac{1}{y^2} - 1)
\end{aligned}
\)
Class 9: Mathematics
