更新1:
I haven't done maths in ages and I don't have a clue how to work this out.
How do you factorise 4x^2-y^2?
回答 (11)
2015-08-30 6:51 pm
4x² - y²
= (2x)² - y²
= (2x + y)(2x - y)
= (2x)² - y²
= (2x + y)(2x - y)
2015-08-30 6:52 pm
Hello Carys, recall a^2 - b^2 = (a+b)(a-b)
So (2x)^2 - y^2 = (2x+y)(2x-y) as a = 2x and b = y
So (2x)^2 - y^2 = (2x+y)(2x-y) as a = 2x and b = y
2015-08-30 6:51 pm
How do you factorise 4x^2-y^2?
Note that this is a special pattern: a difference of squares (two perfect squares subtracted). A difference of squares is a special product in the form of (a^2 - b^2) and is factored into the form (a - b)(a + b), where here a = 2x and b = y (you can tell by taking the square root of both those terms)...so...
4x^2 - y^2
= (2^2)x^2 - y^2 <-- by exponent property: (ab)^n = a^n b^n..
= (2x)^2 - (y)^2 <-- seems familiar? That's because a = 2x and b = y...
= (2x - y)(2x + y) <-- answer...
Note that this is a special pattern: a difference of squares (two perfect squares subtracted). A difference of squares is a special product in the form of (a^2 - b^2) and is factored into the form (a - b)(a + b), where here a = 2x and b = y (you can tell by taking the square root of both those terms)...so...
4x^2 - y^2
= (2^2)x^2 - y^2 <-- by exponent property: (ab)^n = a^n b^n..
= (2x)^2 - (y)^2 <-- seems familiar? That's because a = 2x and b = y...
= (2x - y)(2x + y) <-- answer...
2015-09-02 12:58 pm
4x^2 -y^2
=(2x)^2-y^2
=(2x-y)(2x+y)
=(2x)^2-y^2
=(2x-y)(2x+y)
2015-08-31 7:04 am
(2x-y)(2x+y)
2015-08-31 2:31 am
You don't "factorise" anything. There is no such thing as factorisation.
You factor. Now 4 x^2 - y^2 = (2x)^2 - y^2. Do you recognise that this is
a difference of squares of the form a^2 - b^2 = (a - b)(a + b), where a = 2x and b = y? Obviously not before posting your question. Otherwise you would have been able to simply write 4x^2 - y^2 = (2x - y)(2x + y)
and call it "End of story".
You factor. Now 4 x^2 - y^2 = (2x)^2 - y^2. Do you recognise that this is
a difference of squares of the form a^2 - b^2 = (a - b)(a + b), where a = 2x and b = y? Obviously not before posting your question. Otherwise you would have been able to simply write 4x^2 - y^2 = (2x - y)(2x + y)
and call it "End of story".
2015-08-30 8:27 pm
4x^2 – y^2 = 0
4x^2 = (2x)^2
(2x)^2 – y^2 = 0
Apply the algebraic property: a^2 – b^2 = (a+b) (a-b)
(2x + y) (2x-y)
4x^2 = (2x)^2
(2x)^2 – y^2 = 0
Apply the algebraic property: a^2 – b^2 = (a+b) (a-b)
(2x + y) (2x-y)
2015-08-30 7:08 pm
Use the perfect square formula: a^2-b^2=(a+b)x(a-b)
So 4x^2-y^2=(2x+y)x(2x-y)
The x in between the two parentheses indicating multiplication, not another variable. Hope this helped.
So 4x^2-y^2=(2x+y)x(2x-y)
The x in between the two parentheses indicating multiplication, not another variable. Hope this helped.
2015-08-30 7:00 pm
4x^2-2xy+2xy-y^2
2x(2x-y)+y(2x-y)
(2x+y)(2x-y)
2x(2x-y)+y(2x-y)
(2x+y)(2x-y)
2015-08-30 6:57 pm
first of all, its FACTOR.
you should recognize that because it doesn't have three terms, only two in which both are squared, it will be a "difference of squares" problem.
when you have a difference of squares problem, its easy memorization.
take the square root of each term, the square root of the first term (4x^2) is 2x
the square root of the second term(y^2) is y.
Then you just put it in this form (A+B)(A-B) where A represents the square root of the first term and B represents the square root of the second term.
answer is: (2x+y)(2x-y) also you can always check your answer by FOIL-ing it out (multiplying it out) and seeing if you get the original problem.
you should recognize that because it doesn't have three terms, only two in which both are squared, it will be a "difference of squares" problem.
when you have a difference of squares problem, its easy memorization.
take the square root of each term, the square root of the first term (4x^2) is 2x
the square root of the second term(y^2) is y.
Then you just put it in this form (A+B)(A-B) where A represents the square root of the first term and B represents the square root of the second term.
answer is: (2x+y)(2x-y) also you can always check your answer by FOIL-ing it out (multiplying it out) and seeing if you get the original problem.
2015-08-30 6:53 pm
[ 2x- y ] [ 2x + y ]
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