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 quadratics with common factor (old)

An old video where Sal factors 8k²-24k-144 by first taking out a common factor of 8 and then using the sum-product pattern. Created by Sal Khan and Monterey Institute for Technology and Education.

Want to join the conversation?

  • old spice man green style avatar for user Marc Cutillo
    This may seem like a basic question, but how does he rewrite the equation as 8*(k+3)*(k-6)? don't you have to distribute the 8 onto both the k+3 and k-6 since he factored them out in the first place? It's not the same as the distributive property of multiplication? Help?
    (9 votes)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user Patrick Stetz
      If the '8' was distributed on to both the expression would look like this ...
      (8k + 24)*(8k - 48)
      and if you multiplied these two out you would get ...
      (64k^2 - 384k + 192k - 1152) ----> (64k^2 -192k -1152)
      which does not equal the original: 8k^2 - 24k -144
      (7 votes)
  • leaf blue style avatar for user dkalfatennis
    Can't you factor this equation without simplifying the value of x^2?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • starky tree style avatar for user Sandy Knight
      Yes, you can factor the trinomial without first factoring out any common factor. It may be much tougher to find the right combination of factors to make it work, though.
      Also, in the end, for the equation to be factored completely, the factor will have to come out sooner or later. Here is the problem worked both ways:
      8k^2 - 24k - 144= 0
      8(k^2 - 3k - 18) = 0
      8(k-6)(k+3) = 0

      8k^2 - 24k - 144= 0
      (8k + 24)(k - 6) = 0 *** This is tough to find the right factors.
      8(k+3)(k-6)= 0
      (7 votes)
  • orange juice squid orange style avatar for user Jack Hedberg
    If the equation were set equal to zero, couldn't you have just divided both sides by 8 and simply gotten the answer (k+3)(k-6)=0? I'm assuming that because it's an expression and not equation, we left the 8 in and didn't do algebraic operations. Am I correct?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      You can't just "set it equal to zero". Suppose k = 7
      8(k+3)(k-6) = 8(7+3)(7-6) = 80
      But (k+3)(k-6) = (7+3)(7-6) = 10
      So, you cannot just ignore the constant except for those points where the rest of the expression equals zero (since any number times 0 equals 0).

      Even if it were an equation, it might not be zero that it equaled.
      Suppose 8(k+3)(k-6) = 176, dividing both sides by 8 leave (k+3)(k-6) = 22.

      So, you must be VERY cautious about the constant in front of the other factors. You can only divide it out when the expression actually does equal zero (which is not the usual case).
      (6 votes)
  • piceratops tree style avatar for user ari
    What is the difference between a trinomial and a polynomial?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user Jay
    What happens if there are no 2 numbers compatible as the product of the last number, and sum of the second?
    (5 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      Not all trinomial quadratics can be rationally factored. For example, it is not possible to rationally factor 2x²+15x-3
      So, the answer is that you cannot factor such a trinomial. There is a test to determine whether it is possible to factor a trinomial quadratic expression. The test doesn't tell you what the factors are, but only whether there are any rational factors at all.
      For a quadratic in the form of: ax² + bx + c
      and a, b and c are all rational real numbers:
      There can be rational factors if and only if b²-4ac is either 0 or a perfect square.
      For example, the above quadratic, 2x²+15x-3 has
      a = 2, b=15, c=−3
      Thus, b²-4ac = 15² − 4(2)(−3) = 249
      This is not a perfect square or 0, thus it is impossible to rationally factor this polynomial.
      Now let us test 10x²-51x-91 (this is not easy to factor, so let us check whether it is even possible).
      a = 10, b= -51, c = -91 thus
      b²-4ac = (-51)² - 4(10)(-91) = 2601 + 3640 = 6241
      √6241 = 79
      this is a perfect square, so it can be factored`
      (0 votes)
  • blobby green style avatar for user wmichael04
    HELP
    What do you do for an equation that = 0 like
    p^2 + 5p - 84 = 0
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Christina Rutter
    but how would foil something like (x-2)(x3)(x-5)
    (3 votes)
    Default Khan Academy avatar avatar for user
  • mr pants teal style avatar for user donovanmahon
    What happens when they aren't common factors?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • piceratops tree style avatar for user Conner Baird
      When there isn't a GCF...
      If a=1, then find a 2 two numbers that multiply to C, and add to B.
      EX:
      x^2 + 11x + 24 8 & 3 works!
      (x + 8)(x + 3) is your answer!

      If a≠1, then you can group!
      EX.
      5x^2 - 17x + 6, -2x and -15x will replace B
      (5x^2 - 2x)(-15x + 6)
      x(5x-2)-3(5x-2)
      (x-3)(5x-2) is your answer!

      Hope this helped, there is probably a lesson on this.
      (1 vote)
  • old spice man green style avatar for user Justin Lee

    how did you come up with +3 and -6?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • piceratops sapling style avatar for user Josh
    Sorry if this isn't supposed to be under this video,
    But I have been searching Kahn Academy for a tutorial on factoring trinomials WITHOUT a common factor and that are NOT a perfect square. Such as:
    2x^2 + 5x - 3 = 0
    Could anyone lead me to a tutorial for this type of problem?
    Thank you!
    (2 votes)
    Default Khan Academy avatar avatar for user
    • hopper happy style avatar for user Rashel
      You can factor this by grouping.
      1 Multiply 1st and last term (2*3=6)
      2 Look at what the number is for the middle term (5)
      3 What 2 numbers are a product of 6 and add up to 5?
      6*-1=-6
      6-1=5
      4 Replace these 2 numbers for middle term (5) and attach an x to each number
      5 Combine like terms (terms that you can be factored)
      6 Then solve

      2x^2+6x-1x-3
      2x(x+3) -1(x+3)
      (2x-1)(x+3)
      I put the negative in front of the 1 so the numbers in the parentheses would be identical.
      Is this answer correct?
      Expand it out and if you get the original equation, then you have the correct answer.
      (2x-1)(x+3)
      2x^2+6x-1x-3
      Combine like terms
      2x^2+5x-3

      https://www.khanacademy.org/math/algebra/multiplying-factoring-expression
      Look at videos under section:
      Factor by grouping
      (3 votes)

Video transcript

Factor 8k squared minus 24k minus 144. Now the first thing we can do here, just eyeballing each of these terms, if we want to simplify it a good bit is all of these terms are divisible by 8. Clearly, 8k squared is divisible by 8, 24 is divisible by 8, and 144-- it might not be as obvious is divisible by 8-- but it looks like it is. 8 goes into one and 144, 8 goes into 14 one time. 1 times 8 is 8. Subtract, you get a 6. 14 minus 8 is 6. Bring down the 4. 8 goes into 64 eight times. So it goes into 144 18 times. So let's just factor out an 8 of this. And then that will simplify our expression. It will actually give us a leading 1 coefficient. So this will become 8 times k squared minus 24 divided by 8 is 3k minus 18. Now we have to factor this business in here. And remember if anything has the form x squared plus bx plus c, where you have a leading 1 coefficient-- this is implicitly a one-- we have that here in this expression in parentheses. Then we literally just need to-- and we can do this multiple ways-- but we need to find two numbers whose sum is equal to the coefficient on x. So two numbers whose sum is equal to negative 3 and whose product is equal to the constant term. And whose product is equal to negative 18. So let's just think about the factors of negative 18 here. Let's see if we can do something interesting. So it could be 1. And since it's negative, one of the numbers has to be positive, one has to be negative 1 and 18 is if it was positive. And then one of these could be positive and then one of these could be negative. But no matter what if this is negative and this is positive then they add up to 17. If you switch them, then they add up to negative 17. So those won't work. So either we could write it this way, positive or negative 1, and then negative or positive 18 to show that they have to be different signs. So those don't work. Then you have positive or negative 3. And then negative or positive 6, just to know that they are different signs. So if you have positive 3 and negative 6, they add up to negative 3 which is what we need them to add up to. And clearly, positive 3 and negative 6, their product is negative 18. So it works. So we're going to go with positive 3 and negative 6 as our two numbers. Now, for this example-- just for the sake of this example-- We'll do this by grouping. So what we can do is we can separate this middle term right here as the sum of 3k negative 6k. So I could write the negative 3k as plus 3k minus 6k. And then let me write the rest of it. So we have k squared up here, plus 3k minus 6k, which is the same thing as this over here. And then we have minus 18. And then all of that's being multiplied by 8. Now we're ready to group this thing. We can group these first two terms, they're both divisible by k. And then we can group-- let me put a positive sign-- let's group these second two terms. So then we have 8 times-- I'll write brackets here instead of drawing double parentheses. Brackets are really just parentheses that look a little bit more serious. Now let's factor out a k from this term right here. I'm going to do this in a different color. Let's factor out the k here. So this is k times k plus 3. And then we have plus. And then over here it looks like we could factor out a negative 6. So let's factor out-- I'm going to do this in a different color-- let's factor out a negative 6 over here. So plus negative 6 times k plus 3. So now it looks like we can factor out a k plus 3. There's a k plus 3 times k, and then we have a k plus 3 times the negative 6. So let's factor that out. So we have this 8 out front, that's not changing. So let me write that in the brackets. We're factoring out a k plus 3. So then we have the k plus 3 that we factored out. And then inside of that we just have left this k. Instead of writing plus negative 6, I could just write k minus 6. We factor out the k plus 3 and we're done. And then we can rewrite this. The way we wrote it here it's 8 times the product of k plus 3 times k minus 6. But we know from the properties of multiplication, this is the exact same thing as 8 times k plus 3 times k minus 6. And we are done.