If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Factoring difference of squares: two variables (example 2)

Sal factors 49x^2-49y^2 as (7x+7y)(7x-7y) or as 49(x+y)(x-y). Created by Sal Khan and Monterey Institute for Technology and Education.

Want to join the conversation?

  • mr pink green style avatar for user FlameTail
    How does i solve something like 32 x^3 - 2 w^4 x^3
    (8 votes)
    Default Khan Academy avatar avatar for user
    • starky tree style avatar for user Sandy Knight
      First, look for a common factor between the two terms. Both are divisible by 2 and by x^3. Pull that out of both terms, and write what is left.
      2x^3(16 - w^4)
      Now, factor the part in parentheses as a difference of squares:
      2x^3(16-w^4)
      = 2x^3(4^2 - (w^2)^2)
      = 2x^3(4 - w^2)(4 + w^2)
      Now you can see that the (4-w^2) is also a difference of squares, so factor that part again:
      2x^3(4 - w^2)(4 + w^2)
      2x^3(2 - w)(2 + w)(4 + w^2)
      (11 votes)
  • leaf grey style avatar for user 𝕎𝕙i̶τε 𝕎øℓƒ
    how do i factor this expression?
    4x​^2 - 25?
    please help
    (8 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user saraokeke8
    how to solve x^2+6x+9-y^2
    (8 votes)
    Default Khan Academy avatar avatar for user
    • marcimus pink style avatar for user Alex Tran
      We can't solve this because we have no equation. However, we can factor this expression. Notice that x^2 + 6x + 9 is equal to (x+3)^2. This can be shown by multiplication of the two binomials x+3. Now notice that we can write the new expression as (x+3)^2 - y^2. This is a difference of squares. Let a equal (x+3), and b equal y. Then a^2 - b^2 = (a+b) (a-b). Now replace a and b in the new expression of two multiplied terms. this leaves us with (x+3+y)(x+3-y). Now we are done factoring this expression.
      (8 votes)
  • leaf green style avatar for user Ariel Williams
    This example shows 6x^2-x-12...Can you explain to me step by step how i can solve this problem
    (7 votes)
    Default Khan Academy avatar avatar for user
    • aqualine tree style avatar for user Ted Fischer
      That would not be a "difference of squares" problem...

      Try guessing factor pairs of the form (ax + b)*(cx + d) where a*c = 6 and b*d = -12. (You can see why that is necessary if you FOIL out.) Then check the "cross terms", ad+bc to see which gives you -1 to match the coefficient of the middle term.

      (6x+1)(x-12)
      (2x+ 3)(3x- 4)

      That's actually it! Any other possibility would result in a quadratic with a common factor. For example the first factor in (6x+2)(x-6) has a common factor of 2. Because 2 is not a common factor of the original polynomial, I don't need to consider this one.

      So FOIL them out:
      (6x+1)(x-12) = 6x^2 -71x -12
      (2x+ 3)(3x- 4) = 6x^2 + x - 12

      The first doesn't work. The second ALMOST works, but you need the middle term to be negative. This is easily fixed -- switch the + and - signs. So your factorization is (2x - 3)(3x + 4).
      (3 votes)
  • mr pink red style avatar for user Abbey
    Is there a video that shows how to factor a trinomial in the form of ax^2 + bx + c

    ex. 3p^2 - 2p - 5
    (4 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      There are a variety of ways to solve that. There is a complex method for finding the factors, but we usually reserve that for difficult cases. Here is the method for simple cases:
      Given: ax² + bx + c, factor into two binomials.
      Step 1: Find two numbers that multiply to give a×c, and add to give b. Let us call those two numbers d and g.
      Step 2: Rewrite the expression, replacing "bx" with the sum of those two numbers times x. This would give you:
      ax² + bx + c = ax² + dx + gx + c
      Step 3: Factor the first two and last two terms individually:
      Find GCF of a and d and factor that out of ax² + dx. Let us call the GCF m.
      mx(ax/m + dx/m)
      Now find the GCF of d and g and factor that of gx + c. Let us call this GCF j.
      j(gx/j + c/j)
      Putting your quadratic expressing back together we get:
      mx(ax/m + dx/m) + j(gx/j + c/j)
      If you have done the math correctly, you will always find that
      (ax/m + dx/m) = (gx/j + c/j)
      So let us replace (gx/j + c/j) with (ax/m + dx/m)
      mx(ax/m + dx/m) + j(ax/m + dx/m)
      We can factor out the (ax/m + dx/m) to get:
      (ax/m + dx/m)(mx + j)
      I know this looks hard, but if you do it with actual numbers you'll find it isn't too difficult.
      (5 votes)
  • male robot hal style avatar for user hariharanramesh12
    Sorry for being off topic, but how do I do 3(x + 2)(x – 4) ?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • primosaur seedling style avatar for user projectomega1994
    How would I determine if a Trinomial like 4x^2+12x+9 was a perfect square or not?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • mr pink red style avatar for user andrewp18
      First check if the first and last terms are perfect squares. They are in our case:
      (2x)² = 4x²
      3² = 9
      Now check if the square roots of the first and last terms when multiplied by each other and then doubled, yields the middle term. In other words:
      2x • 3 • 2 = 12x
      Since this matches the second term, we know that are equation fits the form:
      a² + 2ab + b² = (a + b)²
      Where:
      a = 2x
      b = 3
      Comment if you have questions.
      (3 votes)
  • aqualine ultimate style avatar for user hikterion
    well... i came to this 49x^2-49y^2 =49x^2+0+(-49y^2)
    49x^2+7xy-7xy+(49y^2)
    7xy(7x/y)+7xy(-7y/x)
    7xy(7x/y - 7y/x) .....
    it seems easier for me this way, what do you think?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • male robot johnny style avatar for user Thomas B
      While this is an equivalent expression, all you have done here is factor out a 7xy from the original difference of squares. This is neither simpler than the original statement nor particularly helpful for solving an equation in most cases.
      (1 vote)
  • blobby green style avatar for user Leo Wang
    How would i factor 3x-3y-a(x-y)
    (3 votes)
    Default Khan Academy avatar avatar for user
  • piceratops tree style avatar for user ari
    What does difference of squares mean? It is all kind of confusing to me...
    (3 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user Chochlick
      It's difference in value between squares of two numbers. You can visualise it as a difference in surface of two squares of given surfaces.
      The formula used for factoring difference of squares is:
      x^2 - y^2 = (x - y) * (x + y)
      For example:
      64 - 25 = (8-5) * (8+5)
      Or:
      x^2 - 25 = (x - 5) * (x + 5)

      Hope it helped.
      (1 vote)

Video transcript

We need to factor 49x squared minus 49y squared. Now here there's a pattern that you might already be familiar with. But just to make sure you are, let's think about what happens if we multiply a plus b-- where these are just two terms in a binomial-- times a minus b. If you multiply this out, you have a times a, which is a squared, plus a times negative b, which is negative ab-- that's a times negative b-- plus b times a, which is the same thing as ab. And then you have b times negative b, which is negative b squared. So when you do that, you have a negative ab and a positive ab, they cancel out. And you're just going to be left with an a squared minus a b squared. Now, this thing that we have here is exactly that pattern. 49x squared is a perfect square. 49y squared is a perfect square. We can rewrite it like that. We could rewrite this over here as 7x squared minus-- and I'll do it in blue-- minus 7y squared. And so you see it's a pattern. It's a squared minus b squared. So if you wanted to factor this-- if you would just use this pattern that we just derived-- you would say that this is the same thing as a, 7x plus b plus 7y times 7x minus b, minus 7y. And you'd be done. Now there's one alternate way that you could factor this and it'd be completely legitimate. You could start from the beginning and say, you know what? 49 is a common factor here, so let me just factor that out. So you could say it's equivalent to 49 times x squared minus y squared. And you say, oh, this fits the pattern of-- this is a squared minus b squared. So this will be x plus y times x minus y. So the whole thing would be 49 times x plus y times x minus y. And to see that this, right here, is the exact same thing as this right over here, you could just factor 7 out of both of these. You'd factor out a 7 out of that term, a factor 7 out of that term. And when you multiply them, you'd get the 49. So these are-- this or this-- these are both ways to factor this expression.