Factoring quadratics as (x+a)(x+b)
Sal factors x²-3x-10 as (x+2)(x-5) using the sum-product form: (x+a)(x+b)=x²+(a+b)x+a*b.
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- Will it ever happen that a and b are going to be decimals?(39 votes)
- Yes. It is possible.
Also, if you have a fraction, it can be transformed to decimal. They're just different representations of the same quantity. :)(47 votes)
- I know that when you have something like 7(x^2 + 5) + 6y(x^2 + 5), it can become (6y + 7)(x^2 + 5), but why?(15 votes)
- I have an easy way for you to visualize this, although I know you asked this 3 years ago and are probably smarter than me now but anyways:
Since x^2 + 5 = x^2 + 5
Let's assign the quantity (x^2 + 5) some variable, say, z.
Therefore 7(z) + 6y(z)
Reverse distribute the z and you get z(7+6y)
And then substitute (x^2 + 5) back in for z
(x^2 + 5)(7 + 6y)
Hope that helps :)(20 votes)
- Why is the word quadratic used? Just curious!(13 votes)
- The word quadratic comes from the latin word for square, and this is because in a quadratic, the highest degree is x^2(17 votes)
- How would this work if there ISN'T two numbers that add, subtract, multiply and divide into the desired numbers?(10 votes)
- Then if you decided that the expression is already in simplest form, then it shall be considered unfactorable.(2 votes)
- When the resulting binomials have a positive and a negative number, why does it matter which one's negative and which one's positive? I saw somewhere on this site that suggests that the larger number is always the negative, but when I employ that in practice here, your program says I'm wrong (until I change the negative to the smaller number, instead.) Thanks in advance.(6 votes)
- The placement of the signs does matter!
The rule of thumb is that the larger number gets the sign of the original middle term.
Consider: x^2 -3x -10
We need factors of
-10that add to
-10would be: 2(-5) or -2(5)
Only one set will create the
-3when you add the 2 numbers. Since this is a minus 3, the larger number needs the minus sign. So, you want: 2(-5).
Of course, you can also check this by just adding: 2 + (-5) = -3 (this works). While -2 + 5 = +3 (this doesn't work).
Once you have selected the 2 numbers, use those numbers and their respective signs in your factors: (x+2)(x-5)
Hope this helps.(16 votes)
- Before now we were told that equations in the form of ax^2+bx+c were polynomials 'in standard form'. Now we are hearing the word quadratic - introduced for the first time I think? What are the minimum requirements for an equation to be 'quadratic'?(5 votes)
- A "quadratic" is a polynomial where the highest power is "2". They show up so often, that it's useful to have a separate name for that kind of polynomial.(8 votes)
- I understand this explanation but I just wanted to know if there's a quicker way to do this or is this the quickest way?(2 votes)
- This is the quick way, he spent a long time for a process that will be much quicker when you practice. The basic question is what two numbers multiply to be c and add to be b?
There are other hints that might make more sense of it such as noticing the signs of b and c that can help decide the best places to start.
If C is positive, the sign of B will give us the two signs of the factors, so we add to get B (that way I do not care if they are both negative or both positive)
If C is negative, they are opposite signs and B tells the sign of the biggest, so we subtract and follow put the biggest number with the sign of B.(6 votes)
- How did you know right off the bat that the trinomial 'x^2 -3x - 10' has 2 factors so such that they're in the form of (x+a)(x+b)? I haven't learned about quadratics either, they were never introduced, I'm assuming it has something to do with that?(3 votes)
- A quadratic is formed by multiplying 2 binomial together. So, when factoring, we recreate / find the 2 binomials. You may want to learn the lessons about polynomials and operations on polynomials (like multiplication), then try factoring.(2 votes)
- UPD: I made a mistake. The method presented below doesn't work.
(min3:20) It seems like it is also possible to find a & b in the two equations Sal presented by solving them as a system of equations:
Equation 1: a+b = -3
Equation 2: ab = -10
*Step 1 — Solve for a in the first equation:*
a+b = -3
a = -3-b
*Step 2 — Substitute a in the second equation and solve it:*
ab = -10
(-3-b)*b = -10
-3b-2b = -10
-5b = -10
b = 2
*Step 3 — Now that the b is known, solve for a in any equation:*
a+2 = -3
a = -5
Thus a = -5 and b = 2
*Step 4 — Plug it back into your equation:*
- Sorry, but you have an error.
(-3-b)*b = -3b-b^2, not -3b-2b
All your technique will accomplish it to recreate the original quadratic:
-3b-b^2 = -10
Move terms on left to the right and you have 0=b^2+3b-10
This is the original equation except the "x" is not "b".
Good try, but hopefully you see that it doesn't help you find the 2 factors. You got lucky in that your error happened to find one of the two numbers.(2 votes)
- Will a and b ever be decimals(2 votes)
- Yes. They can be decimals, irrationals and even complex.(2 votes)
- [Voiceover] So we have a quadratic expression here. X squared minus three x minus 10. And what I'd like to do in this video is I'd like to factor it as the product of two binomials. Or to put it another way, I want to write it as the product x plus a, that's one binomial, times x plus b, where we need to figure out what a and b are going to be. So I encourage you to pause the video and see if you can figure out what a and b need to be. Can we rewrite this expression as the product of two binomials where we know what a and b are? So let's work through this together now. I'll highlight a and b in different colors. I'll put a in yellow and I'll put b in magenta. So one way to think about it is let's just multiply these two binomials using a and b, and we've done this in previous videos. You might want to review multiplying binomials if any of this looks strange to you. But if you were to multiply what we have on the right-hand side out it would be equal to, you're going to have the x times the x which is going to be x squared. Then you're going to have the a times the x, which is ax. And then you're going to have the b times the x, which is bx. Actually just let me, I'm not gonna skip any steps here just to see it this time. This is all review, or should be review. So then we have, so we did x times x to get x squared. Then we have a times x to get ax, to get a x. And then we're gonna have x times b, so we're multiplying each term times every other term. So then we have x times b to get bx. So plus bx. b x. And then finally we have plus the a times the b, which is of course going to be ab. And now we can simplify this, and you might have been able to go straight to this if you are familiar with multiplying binomials. This would be x squared plus, we can add these two coefficients because they're both on the first degree terms, they're both multiplied by x. If I have ax's and I add bx's to that I'm going to have a plus b x's. So let me write that down. a plus b x's, and then finally I have the plus, I'll do that blue color, finally I have it. I have plus ab. Plus ab. And now we can use this to think about what a and b need to be. If we do a little bit of pattern matching, we see we have an x squared there, we have an x squared there. We have something times x, in this case it's a negative three times x. And here we have something times x. So one way to think about it is that a plus b needs to be equal to negative three. They need to add up to be this coefficient. So let me write that down. So we have a plus b needs to be equal to negative three. And we're not done yet. We finally look at this last term, we have a times b. Well a times b needs to be equal to negative 10. So let's write that down. So we have a times b needs to be equal to negative 10. And in general, whenever you're factoring something, a quadratic expression that has a one on second degree term, so it has a one coefficient on the x squared, you don't even see it but it's implicitly there. You could write this as one x squared. A way to factor it is to come up with two numbers that add up to the coefficient on the first degree term, so two numbers that add up to negative three. And if I multiply those same two numbers, I'm going to get negative 10. So two numbers that add up to negative three, to add up to the coefficient here. And now when I multiply it, I get the constant term. I get this right over here. Two numbers when I multiply I get negative 10. Well what could those numbers be? Well since when you multiply them we get a negative number, we know that they're going to have different signs. And so let's see how we could think about it. And since when we add them we get a negative number, we know that the negative number must be the larger one. So if I were to just factor 10, 10 you could do that as one times ten, or two times five. And two and five are interesting because if one of them are negative, their difference is three. So if one is negative... So let's see if we're talking about negative 10, you could say negative two times five. And when you multiply them you do get negative 10. But if you add these two, you're going to get positive three. But what if you went positive two times negative five. Now this is interesting because still when you multiply them you get negative 10. And when you add them, two plus negative five is going to be negative three. So we have just figured out our two numbers. We could say that a is two or we could say that b is two, but I'll just say that a is equal to two and b is equal to negative five. And so our original expression, we can rewrite as, so we can rewrite x squared minus three x minus 10. We can say that that is going to be equal to x plus two times x, instead of saying plus negative five which we could say, we could just say, actually let me write that down. I could write just plus negative five right over there because that's our b. I could just write x minus five, and we're done. We've just factored it as a product of two binomials. Now, I did it fairly involved mainly so you see where all this came from. But in the future whenever you see a quadratic expression, and you have a one coefficient on the second degree term right over here, you could say alright well I need to figure out two numbers that add up to the coefficient on the first degree term, on the x term, and those same two numbers when I multiply them need to be equal to this constant term, need to be equal to negative 10. You say okay well let's see, they're gonna be different signs because when I multiply them I get a negative number. The negative one is gonna be the larger one, since when I add them I got a negative number. So let's see, let's say five and two seem interesting. Well negative five and positive two, when you add them you're gonna get negative three, when you multiply them you get negative 10.