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### Course: Algebra 2>Unit 3

Lesson 5: Factoring using structure

# Factoring using the perfect square pattern

Sal factors 25x^4-30x^2+9 as (5x^2-3)^2. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• at "" isn't Sal supposed to write (((+-5)^)^) just like he did for 9?
• 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.
• At where did the 2 come from?
• 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.
• Couldn't the answer also be (-5x^2+3)^2 ...? :)
• 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 :)
• How would you solve an equation with a 4th degree but doesn't factor out?
• 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)
• At , he says that it is (5x^2)^2 , but at , he says that (3x^2) can also be (-3x^2). Can't the (5x^2)^2 also be negative?
• 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
or
−30𝑥² = 2 ∙ (−5) ∙ 3 ∙ 𝑥²

Thereby, we also have two choices for the factorization:
25𝑥⁴ − 30𝑥² + 9 = (5𝑥² − 3)²
or
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
• 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?
• 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:
(xy)^z
Then you distribute z to both variables:
(xy)^z = (x^z) * (y^z)
In this case we have:
25x^4
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.
• what if one of the variables had an odd exponent, how would you solve then
ex. 3s^7+24s
• 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)
• What if the last term is not a perfect square?

x^4-4x^2-45
• 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).
• in how did he know to multiply it by 2 to get negative thirty?
• 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)