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Solving quadratics by factoring: leading coefficient ≠ 1

Sal solves 6x²-120x+600=0 by first dividing by 6 and then factoring.

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  • starky sapling style avatar for user twizlerpuppy
    What do you do if you have a leading coefficient that is not 1 and you can't simplify all of the numbers like he did in the video?
    For example:
    5x^2+16x+12=0
    (148 votes)
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    • aqualine ultimate style avatar for user Quantum
      You simply multiply the coefficient of x^2 by the constant. Then ab = constant * coefficient of x^2. While a + b = well the coefficient of x.
      So 5x^2+16x+12=0 . 12 * 5 = 60 so ab = 60. While a + b = 16
      5x^2 + 10x + 6x + 12.........It's easy from here.
      (38 votes)
  • aqualine ultimate style avatar for user minaraza1
    what does coefficient mean again?
    (14 votes)
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    • stelly blue style avatar for user Kim Seidel
      The simplest definition of a coefficient is that it is the number in front of the variable(s) in a term.
      For example, consider the polynomial: 5x^2 + 3x - 8
      The 5 is the coefficient for the term 5x^2.
      The 3 is the coefficient for the term 3x.
      Hope this helps.
      (28 votes)
  • blobby green style avatar for user Eliza Randall
    so say i had the problem h^3+4h^2-12h-672 i can find what you factor out of the first two terms but i can't figure out what to take out to the second two terms to match the part in the parenthesizes you take out an h^2 from the first two and you left with (h+4) but i can't figure out the second part
    (5 votes)
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    • purple pi purple style avatar for user doctorfoxphd
      This looks like an example of factoring by grouping. The tip-off is the 4 terms and the leading exponent of 3.
      h³ + 4h² - 12h - 672
      so if you factor out the h², you get (h + 4) as you said.
      After that, there is a problem with this method in this example. (h + 4) is not actually a factor of this polynomial, but it would have to be in order for there to be a way for us to find it again in the second set of terms. So, possibly you wrote the example down wrong? Or if you made it up, you would have to have something like - 48 as the final term. Then it would be h³ + 4h² - 12h - 48
      Then you could follow this method:
      (h³ + 4h²) - (12h + 48)
      h²(h + 4) - 12 (h + 4)
      So the factors would be (h + 4) and (h²- 12) and the roots would be -4, +2sqrt(3) and -2sqrt(3)
      Now, with your h³ + 4h² - 12h - 672, if you graph this polynomial, there seem to be one positive root and two imaginary roots -- the positive root is 8
      That means there is a factor of (h - 8), leaving h² + 12h + 84 as the other (quadratic) factor
      This looks factorable, but is not. It has the two imaginary roots
      h = - 6 + 4sqrt3 i
      h = - 6 - 4sqrt3 i
      So don't feel bad that you couldn't factor by grouping--this isn't a good victim for that method.
      (14 votes)
  • male robot hal style avatar for user Anthony Rivera
    At to the end, how come you didn't add the 6 with your solution to make it look like 6(x+-10)(x+-10)?
    (8 votes)
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    • mr pink green style avatar for user David Severin
      Since you are finding solutions, not the equation, the 6 does not have any meaning because as Sal did in the beginning, 0/6 = 0. If you were trying to factor it as an equation, then you are correct in that f(x) = 6(x-10)(x-10) or f(x) = 6 (x-10)^2. This shows the whole quadratic function, not just the doubled up solution.
      A solution is when f(x) = 0.
      (7 votes)
  • marcimus pink style avatar for user Ashna
    how would we factor out the expression y=3x^2+12-15 ?
    (4 votes)
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    • stelly blue style avatar for user Kim Seidel
      First - Did you mean: y=3x^2+12x-15, with an x on the 12?
      If yes, start factoring by factoring out a GCF=3
      y=3(x^2+4x-5)
      Then, factor the trinomial by finding 2 factors of -5 that add to 4. See if you can finish the factoring.
      Comment back if you get stuck or you want to check your result.

      Remember, you can always check your factors using multiplication to see if they create the original polynomial.
      (6 votes)
  • male robot hal style avatar for user Anken Aoudia
    i got same result by only regrouping (factoring 6 from the left side of the equation) is it correct ?
    (4 votes)
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  • blobby green style avatar for user ineedmathshelp
    what if you have x² + 6x + 8 = 0
    (2 votes)
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  • duskpin seedling style avatar for user Brenn.io
    What about something like 9m^2-60m+100, where you cannot factor out a GCF? I know there's a way to solve it, but it's complicated and my brain isn't working and I was wondering if there was a video around here that deals with those specific types of problems
    (3 votes)
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  • duskpin sapling style avatar for user bri.joh.128
    i am still confused on what a quadratic is, what is it?
    (2 votes)
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  • duskpin ultimate style avatar for user Maya1123
    Is there a way to determine if a quadratic equation can't be factored without using the quadratic formula?
    (3 votes)
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Video transcript

- [Voiceover] We have six x squared, minus 120 x, plus 600, equals zero. Like always, pause this video, and see if you can solve for x, if you could find the x values that satisfy this equation. Alright, let's work through this together. The numbers here don't seem like outlandish numbers. They seem like something that I might be able to deal with, and I might be able to factor, so let's try to do that. The first thing I like to do is see if I can get a coefficient of one, on the second degree term, on the x-squared term. It looks like actually all of these terms are divisible by six. So if we divide both sides of this equation by six, I'm still going to have nice integer coefficients. Let's do that. Let's divide both sides by six. If we divide the left side by six, divide by six, divide by six, divide by six. And I divide the right side by six. If I do that, and clearly if I do the same thing to both sides of the equation, then the equality still holds. On the left-hand side, I am going to be left with x squared, and then negative 120, divided by six. That is, let's see. 120 divided by six is 20. So that's minus 20 x. Then 600 divided by six, is 100. So plus 100, is equal to zero divided by six. Is equal to zero. So let's see if we can factor, if we can express this quadratic as a product of two expressions. The way we think about this, and we've done it multiple times, if we have something, if we have x plus a, times x plus b, and this is hopefully review for you. If you multiply that out, that is going to be equal to, that equals to x squared, plus a, plus b, x, plus a b. What we want to do is see if we can factor this into an x plus a, and an x plus b. A plus b, needs to be equal to negative 20. That needs to be a plus b. And then a times b, right over here, that needs to be equal to the constant term. That needs to be a times b, right over there. Can we think of two numbers, that if we take their product, we get positive 100, and if we take their sum, we get negative 20? Well since their product is positive, we know that they have the same sign. They're both going to have the same sign. They're either both going to be positive, or their both going to be negative, since we know that we have a positive product. Since their sum is negative, well they both must both be negative. You can't add up two positive numbers, and get a negative. So they both must be negative. Let's think about it a little bit. What negative numbers, when I add them together I get negative 20, when I multiply it, I get 100? Well you could try to factor 100. You could say, well negative two times negative 50, or negative four times negative 25. But the one that might jump out at you is this is negative 10, times, I'll write it this way, negative 10, times negative 10, and this is negative 10, plus negative 10. So in that case, both our a and our b, would be negative 10. And so we can rewrite the left side of this equation as, we can rewrite it as, x, and I'll write it this way at first, x plus negative 10, times, x plus negative 10 again. X plus negative 10, and that is going to be equal to zero. So all I've done is I've factored this quadratic. Another way, these are both the same thing as x minus 10. I could rewrite this as x minus 10, squared, is equal to zero. The only way that the left-hand side is going to be equal to zero, is if x minus 10 is equal to zero. You could think of this as taking the square root of both sides. It doesn't matter if you're taking the positive or negative square root, or both of them. The square root of zero, is zero. So, we would say, that x minus 10 needs to be equal to zero. So x , adding 10 to both sides of this, you have x is equal to 10, is the solution to this quadratic equation, up here.