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a2 - b2 = (a + b)(a - b).

Q: What is the formula to a factor a difference of perfect squares?

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There is a formula for the "difference of squares." In this case, the answer is (6v + w)(6v - w)

It is the difference of two squares which is: (x+2y)(x-2y)

That looks simple enough for me; but if you want to factor it, you can use the formula for the difference of two squares.

There is a formula for the "difference of squares." In this case, the answer is (x2 - 5)(x2 + 5)

perfect * * * * * Not strictly. they could both be multiples of sqaures. For example, factorise 3x3 - 48x. Neither term is a square but they do become squares when the common factor, 3x, is separated out. Also, when rationalising surds, one would use the difference of two squares but (at least) one of the terms is not a square.

Related questions

There is a formula for the difference of two squares. The sum of two squares doesn't factor.

The formula to factor the difference of two squares, a2 - b2, is (a + b)(a - b).

The "difference of squares" has a formula. (11x + 1)(11x - 1)

The "difference of squares" has a formula. (3cd2 + 4ef)(3cd2 - 4ef)

There is a formula for the "difference of squares." In this case, the answer is (3A + B)(3A - B)

There is a formula for the "difference of squares." In this case, the answer is (y + 9)(y - 9)

There is a formula for the "difference of squares." In this case, the answer is (x + 10)(x - 10)

There is a formula for the "difference of squares." In this case, the answer is (2x - 3)(2x + 3)

There is a formula for the "difference of squares." In this case, the answer is (4x + 3)(4x - 3)

There is a formula for the "difference of squares." In this case, the answer is (5c - 14d)(5c + 14d)

There is a formula for the "difference of squares." In this case, the answer is (2x + 5)(2x - 5)

There's a formula for the "difference of squares." (x + 1)(x - 1)