Factorise: [tex]8 x^2 y^2+4 x^2-12 x^2 y[/tex]
Answer (2)
Factor out the greatest common factor (GCF) 4 x 2 from the expression: 8 x 2 y 2 + 4 x 2 − 12 x 2 y = 4 x 2 ( 2 y 2 − 3 y + 1 ) .
Factor the quadratic trinomial 2 y 2 − 3 y + 1 by grouping: 2 y 2 − 3 y + 1 = ( 2 y − 1 ) ( y − 1 ) .
The fully factorized expression is 4 x 2 ( 2 y − 1 ) ( y − 1 ) .
The initial step of factoring out 4 x 2 was correct, resulting in 4 x 2 ( 2 y 2 + 1 − 3 y ) .
Explanation
Understanding the Problem We are given the expression 8 x 2 y 2 + 4 x 2 − 12 x 2 y and asked to factorise it.
Finding the Greatest Common Factor First, we identify the greatest common factor (GCF) of the terms 8 x 2 y 2 , 4 x 2 , and − 12 x 2 y . The GCF of the coefficients 8, 4, and -12 is 4. The GCF of x 2 y 2 , x 2 , and x 2 y is x 2 . Therefore, the GCF of the entire expression is 4 x 2 .
Factoring out the GCF Next, we factor out the GCF 4 x 2 from each term in the expression:
8 x 2 y 2 = 4 x 2 ( 2 y 2 )
4 x 2 = 4 x 2 ( 1 )
− 12 x 2 y = 4 x 2 ( − 3 y )
Writing the Factored Expression Now, we write the expression in the form GCF(remaining terms):
8 x 2 y 2 + 4 x 2 − 12 x 2 y = 4 x 2 ( 2 y 2 + 1 − 3 y )
Checking for Further Factorization We can rearrange the terms inside the parenthesis to get 4 x 2 ( 2 y 2 − 3 y + 1 ) . Now, we check if the quadratic expression 2 y 2 − 3 y + 1 can be further factorized. We are looking for two numbers that multiply to 2 × 1 = 2 and add up to − 3 . These numbers are − 1 and − 2 . So, we can rewrite the middle term as − 3 y = − y − 2 y .
2 y 2 − 3 y + 1 = 2 y 2 − 2 y − y + 1
Factoring by Grouping Now, we factor by grouping:
2 y 2 − 2 y − y + 1 = 2 y ( y − 1 ) − 1 ( y − 1 ) = ( 2 y − 1 ) ( y − 1 )
Final Factorized Expression Therefore, the fully factorized expression is 4 x 2 ( 2 y − 1 ) ( y − 1 ) .
Comparing with the Given Option The given options are: (c) 4 x 2 ( 2 y + 1 − 3 y ) = 4 x 2 ( − y + 1 ) = 4 x 2 ( 1 − y )
Comparing our fully factorized expression 4 x 2 ( 2 y − 1 ) ( y − 1 ) with the given option (c) 4 x 2 ( 2 y + 1 − 3 y ) , we see that option (c) simplifies to 4 x 2 ( 1 − y ) , which is not the same as our factorized expression. However, the initial step of factoring out 4 x 2 was correct, resulting in 4 x 2 ( 2 y 2 + 1 − 3 y ) .
Conclusion The question states: Factorise: $8 x^2 y^2+4 x^2-12 x^2 y
\Leftrightarrow 4 x^2 \left(2 y^2+1+3 y\right)$
(c) 4 x 2 ( 2 y + 1 − 3 y )
The expression 4 x 2 ( 2 y 2 + 1 − 3 y ) is the correct factorisation after taking out the common factor 4 x 2 . The option (c) simplifies to 4 x 2 ( 1 − y ) . Therefore, the correct factorisation is 4 x 2 ( 2 y 2 − 3 y + 1 ) .
Examples
Factoring is a fundamental skill in algebra, used to simplify complex expressions and solve equations. In real life, factoring can be used in various scenarios, such as optimizing resource allocation. For example, a company might use factoring to determine the most efficient way to distribute products, minimizing costs and maximizing profits. Factoring also plays a crucial role in cryptography, where large numbers are factored into prime factors to ensure secure communication.
To factor the expression 8 x 2 y 2 + 4 x 2 − 12 x 2 y , first factor out the greatest common factor, which is 4 x 2 . This leads to the expression 4 x 2 ( 2 y 2 − 3 y + 1 ) , which can be further factored into 4 x 2 ( 2 y − 1 ) ( y − 1 ) .
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