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It is essentially the same as any other factorization problem, just with the usual order reversed.
Date Posted:
4 years ago
Either square root could take the negative, not just the square root of 9x² i.e it can be equivalently written as (3x - 2y)² as well.
So it's not really correct to say that 'since 4y² is positive, the factor producing 9x² is negative'. Rather, whichever square root you take as positive makes the other negative.
So it's not really correct to say that 'since 4y² is positive, the factor producing 9x² is negative'. Rather, whichever square root you take as positive makes the other negative.
Thanks for the comment, J!
Factorizing polynomials may lead to more than one correct solution. In this case, I used the word know with the same connotation of "We can deduce, for at least one possible solution"
If that offended you, I do apologize!
Factorizing polynomials may lead to more than one correct solution. In this case, I used the word know with the same connotation of "We can deduce, for at least one possible solution"
If that offended you, I do apologize!
Oh I see. No offence caused, just clarifying
Step by step factorisation :
9x² - 12xy + 4y²
= 9x² - 6xy + 4y² - 6xy (split up the -12xy equally into 2)
= 3x(3x - 2y) + 2y(2y - 3x)
= 3x(3x - 2y) - 2y(-(2y - 3x))
= 3x(3x - 2y) - 2y(3x - 2y)
= (3x - 2y)(3x - 2y)
= (3x - 2y)²
Alternatively, you could do this :
9x² - 12xy + 4y²
= (3x)² - 2(3x)(2y) + (2y)²
Notice that this is in the form a² - 2ab + b², where a = 3x and b = 2y
Since (a - b)² = a² - 2ab + b²,
9x² - 12xy + 4y²
= (3x)² - 2(3x)(2y) + (2y)²
= (3x - 2y)²
9x² - 12xy + 4y²
= 9x² - 6xy + 4y² - 6xy (split up the -12xy equally into 2)
= 3x(3x - 2y) + 2y(2y - 3x)
= 3x(3x - 2y) - 2y(-(2y - 3x))
= 3x(3x - 2y) - 2y(3x - 2y)
= (3x - 2y)(3x - 2y)
= (3x - 2y)²
Alternatively, you could do this :
9x² - 12xy + 4y²
= (3x)² - 2(3x)(2y) + (2y)²
Notice that this is in the form a² - 2ab + b², where a = 3x and b = 2y
Since (a - b)² = a² - 2ab + b²,
9x² - 12xy + 4y²
= (3x)² - 2(3x)(2y) + (2y)²
= (3x - 2y)²