- Solving quadratics by factoring
- Solving quadratics by factoring
- Quadratics by factoring (intro)
- Solving quadratics by factoring: leading coefficient ≠ 1
- Quadratics by factoring
- Solving quadratics using structure
- Solve equations using structure
- Quadratic equations word problem: triangle dimensions
- Quadratic equations word problem: box dimensions
- Solving quadratics by factoring review
Sal solves the equation s^2-2s-35=0 by factoring the expression on the left as (s+5)(s-7) and finding the s-values that make each factor equal to zero. Created by Sal Khan and Monterey Institute for Technology and Education.
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- How would you work out the problem 9k^2 + 45k = 0(215 votes)
- factor out 9k from the expression u get: 9k(k+5)= 0, which means 9k = 0, k + 5 = 0, solving for k you get: k = 0, k = -5(340 votes)
- What would happen if the starting equation didn't equal up to zero, how would you factor those types of problems?(67 votes)
- Let's look at this:
You would have to subtract C from both sides and get:
Then solve the problem normally(3 votes)
- How would you factor this?
It would be (2x-1)(x+3)
3 is a prime number.. so the only two numbers that can be multiplied to get 3 are 3 and 1. Those two numbers do not add up to 5. How could you solve the part for 5x?(22 votes)
- Good question!
Actually, when we are solving these types of problems, we want to take the product of the coefficient of x^2 and the constant term. In this case that would be nothing but -6.
-6 can be written as the product of 6 and -1.
So, the equation would become
2x^2 + 5x -3 = 2x^2 + 6x - x -3
= 2x(x + 3) - (x + 3)
Now, factoring out the (x + 3), the equation becomes:
=> (x + 3)(2x - 1)
Hope this helped!(15 votes)
- (x - 2) (x + 4) = 0(12 votes)
- To solve this, you would use the zero product property. If you make one of the parentheses equal to zero then the whole left side is equal to zero (because zero multiplied by anything is zero). So you'd set the first set of parentheses like so: (x-2)=0. Then to isolate "x", you would add 2 to both sides to get x=2. Then, You would set the other set of parentheses to zero, like so: (x+4)=0. To isolate "x", you would subtract 4 from both sides to get x=-4. So, your final answer would be x=2 or x=-4.
Hope this helped you out.(12 votes)
- At0:49why does he do A + B = -2?(14 votes)
- It's the formula for finding the solutions to the quadratic.
What he is saying is you need 2 numbers that when added together equal -2, but when multiplied equals -35.
Note: since the multiplied is negative, one of the two numbers will be negative and the other will be positive.(16 votes)
- It's really never a good idea to use s and 5 together. I don't even have dyslexia but when I'm learning something you should try to use numbers and variables that don't look similar cause its hard to view.(14 votes)
- I'm pretty confused about this; I wasn't following my teacher when she went over it. So how do I solve (I need to factor it) a problem like this:
5x^2 + 17x - 12(8 votes)
- I've learned this in a mathsmart book that you can buy from costco. What you do is you multiply a and c together (5x^2*-12) and you get -60x^2. Find two numbers that add up to 17x and multiply to -60x^2, those numbers would turn out to be 20x and -3x. Now you write in the equation 5x^2+20x-3x-12 = 5x(x+4)+-3(x+4) = (5x-3)(x+4)(16 votes)
- At2:28, I don't clearly understand how Sal factors out the
( s + 5 ) in the expression s( s + 5 ) - 7 ( s + 5 ) into ( s + 5 )( s - 7 ). How does he do it?(10 votes)
- Sal used the distribution property, if you had A*B - C*B you can change this to B(A-C).
A * B - C * B = B * (A-C)
So s * ( s + 5 ) - 7 * ( s + 5 ) = ( s + 5 ) * ( s - 7 ).(9 votes)
- How would you figure out what value of c makes this expression a perfect square: x^2+x+c? Would you factor?(7 votes)
- You probably know that (x + a)^2 = x^2 + 2ax + a^2. In your case, 2a=1, so a=1/2.
Hence, (x + 1/2)^2 should do the trick. You figure out the c.(6 votes)
We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be positive, one has to be negative. And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common factor of negative 7, so let's factor that out. So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers is equal to zero. If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5. That is one solution to the equation, or you can add 7 to both sides of that equation, and you get s is equal to 7. So if s is equal to negative 5, or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is minus 35. That does equal zero. If you have 7, 49 minus 14 minus 35 does equal zero. So we've solved for s. Now, I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what is that equal to? x times x is x squared, x times b is bx. a times x is plus ax. a times b is ab. So you get x squared plus-- these two can be added-- plus a plus bx plus ab. And that's the pattern that we have right here. We have 1 as a leading coefficient here, we have 1 as a leading coefficient here. So once we have our two numbers that add up to negative 2, that's our a plus b, and we have our product that gets to negative 35, then we can straight just factor it into the product of those two things. So it will be-- or the product of the binomials, where those will be the a's and the b's. So we figured it out. It's 5 and negative 7. 5 plus negative 7 is negative 2. 5 times negative 7 is negative 35. So we could have just straight factored at this point. 2, well, actually this was the case of s. So we could have factored it straight to the case of s plus 5 times s minus 7. We could have done that straight away and would've gotten to that right there. And, of course, that whole thing was equal to zero. So that would've been a little bit of a shortcut, but factoring by grouping is a completely appropriate way to do it as well.