- Identifying quadratic patterns
- Identify quadratic patterns
- Factorization with substitution
- Factorization with substitution
- Factoring using the perfect square pattern
- Factoring using the difference of squares pattern
- Factor polynomials using structure
Sal factors 25x^4-30x^2+9 as (5x^2-3)^2. Created by Sal Khan and Monterey Institute for Technology and Education.
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- at "0:44" isn't Sal supposed to write (((+-5)^)^) just like he did for 9?(65 votes)
- Good observation!
(At first I instinctively replied "No, he wasn't. What he wrote is correct.", but that is because I was too focused on your un-mathematical notation. ("^" does not equal "²", you need to write it as "^2" if you don't have access to the "²" symbol. The symbol "^" stands for to-the-power-of, not for to-the-power-of-two.)
Also, (±5²)² would equal 625, which is not the 25 we need.
But he could (and should) have wrote ±5x². So you are right.(71 votes)
- At1:27where did the 2 come from?(8 votes)
- Sal is using the pattern created by squaring a binomial.
Here's the pattern: (a+b)^2 = a^2 + 2ab + b^2
Here's where the 2 comes from... use FOIL and multiply (a+b)(a+b) ...
(a+b)(a+b) = a^2 + ab + ab + b^2
Notice... the 2 middle terms match. When you add them you get 2ab.
That's where the 2 comes from.
Hope this helps.(11 votes)
- Couldn't the answer also be (-5x^2+3)^2 ...? :)(7 votes)
- yes, it could.
in fact he can also write (-5x^2)^2 'cause if the power is even base's sign become positive. u right :)(1 vote)
- I don't get how 25x^4 equals (5x^2)^2; how did Sal change the 25x^4 to (5x^2)^2 and how do you do the same with similar problems?(3 votes)
- Lets say you have something like:
z(x + y)
In this case, you would distribute the z to x and y:
z(x + y) = zx + zy
Likewise, you can go backwards and factor out a z when you have:
zx + zy = z(x + y)
If you have:
Then you distribute z to both variables:
(xy)^z = (x^z) * (y^z)
In this case we have:
We can factor out a square (because both 25 and x^4 are perfect squares:
25x^4 = [√(25x^4)]^2 = (5x^2)^2
Notice when you distribute the exponent of 2 back to the 5 and x^2 we get:
(5x^2)^2 = 25x^2^2 = 25x^4 Back where we started. Comment if you have questions.(6 votes)
- How would you solve an equation with a 4th degree but doesn't factor out?(4 votes)
- There is an equation called the quartic equation that can be used to solve 4th degree polynomials. Check out the wikipedia article on it for more (http://en.wikipedia.org/wiki/Quartic_function)(4 votes)
- what if one of the variables had an odd exponent, how would you solve then
ex. 3s^7+24s(5 votes)
- If you factor out 3s instead of s you'd get
3s(s^6 + 8) = 3s((s^2)^3 + 2^3) which is now 3s times the sum of cubes:
3s(s^2 + 2)(s^4 - 2s^2 +4)(1 vote)
- At0:48, he says that it is (5x^2)^2 , but at0:57, he says that (3x^2) can also be (-3x^2). Can't the (5x^2)^2 also be negative?(4 votes)
- Yes, Sal could have written
25𝑥⁴ = (±5𝑥²)², just like 9 = (±3)²
That way we would get two choices for the middle term:
−30𝑥² = 2 ∙ 5 ∙ (−3) ∙ 𝑥²
−30𝑥² = 2 ∙ (−5) ∙ 3 ∙ 𝑥²
Thereby, we also have two choices for the factorization:
25𝑥⁴ − 30𝑥² + 9 = (5𝑥² − 3)²
25𝑥⁴ − 30𝑥² + 9 = (−5𝑥² + 3)²
which makes sense, because squaring 5𝑥² − 3 will give us the same result as squaring its opposite, namely −(5𝑥² − 3) = −5𝑥² + 3(3 votes)
- What if the last term is not a perfect square?
- What you are looking at is in the form of the quadratic u^2 - 4u - 45, which factors to:
(u - 9)(u + 5), so your polynomial factors to:
(x^2 - 9)(x^2 + 5), which factors further to:
(x - 3)(x + 3)(x^2 + 5).(3 votes)
- if you can factor a 4th degree expression we can also factor an expression like
3x^8- 3y^8. would it be 3x^4 y^4 (x^4-y^4)(3 votes)
- You can take it a few steps further:
3x^8 - 3y^8 =
3(x^8 - y^8) =
3(x^4 + y^4)(x^4 - y^4) =
3(x^4 + y^4)(x^2 + y^2)(x^2 - y^2) =
3(x^4 + y^4)(x^2 + y^2)(x + y)(x - y)
If you use complex numbers, you may even be able to factor the remaining 4th and 2nd degree terms. But I guess that's going a bit too far in this context.(1 vote)
We need to factor 25x to the fourth minus 30x squared plus 9. And this looks really daunting because we have something to the fourth power here. And then the middle term is to the second power. But there's something about this that might pop out at you. And the thing that pops out at me at least is that 25 is a perfect square, x to the fourth is a perfect square, so 25x to the fourth is a perfect square. And 9 is also perfect square, so maybe this is the square of some binomial. And to confirm it, this center term has to be two times the product of the terms that you're squaring on either end. Let me explain that a little bit better So, 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now, what is 30x squared? What happens if we take 5 times plus or minus 3? So remember, this needs to be two times the product of what's inside the square, or the square root of this and the square root of that. Given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square. So we can just rewrite this as this is equal to 5x squared-- let me do it in the same color. 5x squared minus 3 times 5x squared minus 3. And we saw in the last video why this works. And if you want to verify it for yourself, multiply this out. You will get 25x to the fourth minus 30x squared plus 9.