Hey.
There is a question that puzzels me on the difference of two squares. In many of the questions on this i the numbers have been able to undergo a square root method but however the numbers 12 and 48 cannot. Why is that? and how is this question done and how can i replicate it to do other questions of the same nature? What are the steps of the question as well? Could someone explain?
It states: Factorise 12y^2 x^6 - 48k^4 m^8
All comments are much appreciated. Thank You!
The Difference of Two Squares?
2008-05-25 3:10 pm
回答 (8)
2008-05-25 3:30 pm
✔ 最佳答案
Jay,Before factoring anything, you should start by seeing if each 'piece' has anything in common. 12 and 48... they have 12 incommon, right? so factor that out.
12(x^6y^2 - 4k^4m^8)
Now... remember your rules of exponents to factor.
x * x = x^2 because we add the exponents there
(x)^2 = x^2 because we multiply the exponents... remember?
Don't forget the difference of two squares. If you look at something to factor, like that one above, and....
1) it's a binomial (2 parts of the equation) and both are perfect squares...
x^6y^2 = x^3y * x^3y and
4k^4m^8 = 2k^2m^4 * 2k^2m^4
2) it's separated by a negative like above...
then, you can write it as (x^3y + 2k^2m^4)(x^3y - 2k^2m^4)
Don't forget to put your 12 back in the front...
12(x^3y + 2k^2m^4)(x^3y - 2k^2m^4)
2016-11-13 6:58 pm
you could, it is basically that do it you will choose complicated numbers. Extending the version of two squares: x² - a² = (x - a)(x + a) = (x - ?a²)(x + ?a²) x² + a² = x² - (-a²) = (x - ?(-a²))(x + ?(-a²)) = (x - ai)(x + ai) notice -a² ? (-a)²
2008-05-25 3:35 pm
it is because 12 nd 48 are not perfect squares like 4,16,64 etc.
square root of 4 is 2, sq root of 16 is 4. but the square root of 12 is "2sq.root3" nd of 48 . it is 4sq. root 3 . so dese r not prfect squares.
in order to solve dis problem, we take out common factor of two terms to make dem perfect squares.
12(y^2 x^6-4k^4 m^8)
12[(yx^3)^2-(2 k^2 m^4)^2]
12[(y x^3-2 k^2 m^4)(y x^3+2 k^2 m^4)]
square root of 4 is 2, sq root of 16 is 4. but the square root of 12 is "2sq.root3" nd of 48 . it is 4sq. root 3 . so dese r not prfect squares.
in order to solve dis problem, we take out common factor of two terms to make dem perfect squares.
12(y^2 x^6-4k^4 m^8)
12[(yx^3)^2-(2 k^2 m^4)^2]
12[(y x^3-2 k^2 m^4)(y x^3+2 k^2 m^4)]
2008-05-25 3:31 pm
12x^2y^6 - 48k^4m^8
= 12(x^2y^6 - 4k^4m^8)
= 12(xy^3 + 2k^2m^4)(xy^3 - 2k^2m^4)
= 12(x^2y^6 - 4k^4m^8)
= 12(xy^3 + 2k^2m^4)(xy^3 - 2k^2m^4)
2008-05-25 3:29 pm
12(y^2 x^6 - k^4 m^8)
=12(y x^3 + k^2 m^4)(y x^3 - k^2 m^4).
Normally each 'nomial' have the same power, whereas
yx^3 has power 4 & k^2m^4 has power 6.
Now you figure out where your squares pop up.
=12(y x^3 + k^2 m^4)(y x^3 - k^2 m^4).
Normally each 'nomial' have the same power, whereas
yx^3 has power 4 & k^2m^4 has power 6.
Now you figure out where your squares pop up.
2008-05-25 3:20 pm
confused by your grammar.
Are you asking why 12 and 48 are not perfect squares?
Factor that equation that contains 4 unknowns? not me.
12(y²x⁶ - 4k⁴m⁶)
.
Are you asking why 12 and 48 are not perfect squares?
Factor that equation that contains 4 unknowns? not me.
12(y²x⁶ - 4k⁴m⁶)
.
2008-05-25 3:18 pm
Man that's a common factor. Pull it out. So its
12( y^2x^6 - 4k^4m^8) Now isn't the bracket of the form
a^2 - b^2? Factorise now.
12( y^2x^6 - 4k^4m^8) Now isn't the bracket of the form
a^2 - b^2? Factorise now.
2008-05-25 3:13 pm
12(y² x^6 - 4 k^4 m^8)
12((yx³ - 2k² m^4)(yx³ + 2k² m^4))
P.S. Factorise is not a word in mathematics.
12((yx³ - 2k² m^4)(yx³ + 2k² m^4))
P.S. Factorise is not a word in mathematics.
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