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### Course: Get ready for Algebra 2 > Unit 1

Lesson 9: Factoring quadratics with perfect squares- Perfect square factorization intro
- Factoring quadratics: Perfect squares
- Perfect squares intro
- Factoring perfect squares
- Identifying perfect square form
- Factoring perfect squares: negative common factor
- Factoring perfect squares: missing values
- Factoring perfect squares: shared factors
- Perfect squares

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# Identifying perfect square form

Sal shows how we can identify that a trinomial has the "perfect square" form.

## Want to join the conversation?

- Can the answer also be negative?(-5x-2)?(17 votes)
- If you mean (-5x-2)^2, then yes, but -5x-2 is not the same.(4 votes)

- Is possible to factor this quadratic:

25x^2 + 16x + 9 as (a + b)^2 or in anyway?(4 votes)- No it wouldn't work.

Yes... your quadratic has perfect squares at both ends, but the middle term is incorrect.

The terms on the ends of your factors would need to be: (5x + 3)^2

If you multiply this out, you get: 25x^2 + 30x + 9

Hope this helps.(8 votes)

- How is Sal so good at writing with a mouse?(4 votes)
- He uses a stylus (like a pen).(6 votes)

- When in life am I ever going to use this stuff?(4 votes)
- This seems sort of complicated... Why don't we just factor the binomial like we usually would? Like for example, when Sal had 25x^2 + 20x + 4, I did it normally and got the same answer that he did, (5x+2)^2.(4 votes)
- Why is it (Ax)^2 and not just Ax^2? Sorry and thank you to whoever answers! :)(1 vote)
- (Ax)² = (Ax) * (Ax)

Ax² = A * x² = A * x * x(4 votes)

- At1:07the numbers in the brackets are multiplied by each other how does this work?(2 votes)
- By this time you should have learned how to multiply binomials. See this video: https://www.khanacademy.org/math/algebra/introduction-to-polynomial-expressions/multiplying-binomials-2/v/multiplying-binomials(2 votes)

- I'm a bit confused about the second term in the perfect square form.

In this video, Sal said that 20x (the center term of 25x^2 + 20x + 4) is 2.A.B.x, implying that A is the coefficient of 25^2 and x was just added on because that's how the formula works.

However, in the "Practice: Perfect Squares" exercise, a solution says that 14x (the center term of x^2 + 14x + 49) is twice the root of 1x^2 and 49, implying that A is the coefficiant AND the variable of 1x^2.

Can someone please explain which one is correct, if both are, or if I messed something up? I'm not sure if this matters much, but it confused me when I got to Completing the Square in the next unit.(2 votes)- Consider what happens when you multiply: (ax+b)(ax+b)

You get: (ax)^2+abx+abx+b^2 which simplifies to a^2x^2+2ab+b^2

The "a" refers to the square root of the coefficient of the x^2 term. The "b" refers to the square root of the constant term.

For 25x^2 + 20x + 4: a=5 and b=2. 2ab = 2(5)(2) = 20

For x^2 + 14x + 49: a=1 and b=7. 2ab = 2(1)(7) = 7

Hope this helps.(2 votes)

- Hello, I'm studying for an exam and our workbook has this question:

The product of two numbers is 120, and the sum of their squares is 289. The sum of the number is __."

Does anyone know a course/video that talks about these types of problems? Or does anyone know how to solve this? I know what the answer is but I don't really understand how this works.(1 vote)- Use X = one number and Y = 2nd number

Translate the first part of the 1st sentence and you get:

xy=120

Translate the second part of the 1st sentence and you get:

x^2+y^2=289

You now have a system of equations.

-- Solve the 1st equation for one of the 2 variables by dividing by sides: y=120/x

-- Substitute this into the 2nd equation in place of Y to get: x^2+(120/x)^2 = 289

-- Do the exponent: x^2+14400/x^2 = 289

-- Multiply both sides by x^2: x^4 + 14400 = 289x^2

-- Subtract 289x^2 from both sides: x^4-289x^2+14400 = 0

-- Factor: (x^2-225)(x^2-64) = 0

-- Factor more: (x-15)(x+15)(x-8)(x+8)=0

-- Split the factors apart and solve each.

x-15=0 creates x=15

x+15=0 creates x=-15

x-8=0 creates x=8

x+8=0 creates x=-8

To find y: use y=120/x

x=15 creates y=120/15 = 8

x=-15 creates y=120/(-15) = -8

x=8 creates y=120/8 = 15

x=-8 creates y=120/(-8) = -15

Basically, the two numbers are either 8 and 15, or -8 and -15.

Hope this helps.(3 votes)

- To all fellow students asking why we have to learn this and where we will use this:

"Academic" has a few definitions. One of them is "relating to education". Another is "lacking practical relevance; of theoretical interest only".

Relevance is a puzzle. It is completely dependent on a goal---as Booker T. Washington said, all successes in life started out with a goal in mind.

So, education lacks practical relevance based on our goals.

If you want a scientific career, you will need to know all this. It will have "practical relevance".

If you want to work for a grocery store, this can all be of "theoretical interest only"!(2 votes)

## Video transcript

- [Voiceover] We wanna figure
out what AX plus B squared is, and I encourage you to pause the video and figure out what that is in terms of capital-A and capital-B. So let's work through it. This is the same thing as multiplying AX plus B times AX plus B. So let me fill that in. This is AX there, another AX there. I just wrote it in that order to make the color switching
a little bit easier. So AX plus B times AX plus B. Well, what's that going to be equal to? Well, if you take this AX
and you multiply it times that AX, you're going to get AX squared. AX, the entire thing squared. And then if you take, if you take this AX and then multiply it times this B, you're going to get ABX. Then if you take this
B and you multiply it times this AX, you're
going to get another ABX. ABX. And then last but not
least, if you take this B and multiply it times the other B, it's going to be plus B squared. And so what are you left with? Well, you're going to be left with A, I'll write it like this, AX squared, we actually if we want, well, I'll write it in a
different way in a second, and then you have plus, plus two, that's a slightly different color. I'm gonna do that other color. Plus two ABX, and then finally plus B squared. Plus B squared. Now, I said I could write it
in a slightly different way, what I could do is just
rewrite-out AX squared as being the same thing. This is the same thing
as A-squared X-squared, and then I can write out
everything else the same way. Plus two ABX, and then plus B squared. Now, why did I, what's
interesting about doing this? Well, now we can see the pattern for the square of any binomial or binomial like this, so for example, if someone
were to walk up to you and say "alright, I have
a trinomial of the form," let's say they have a
trinomial of the form 25X squared plus 20X plus 4, and they were to tell you to factor this, well, actually, let's just do that. Why don't you pause the video and see if you could factor this as
the product of two binomials. Well, when you look at this, you'd say "well look, there's 25X squared, "that looks like a perfect square. "25X squared, that's the
same thing as five-squared "x-squared," or you could write it as five X squared. This four here, that's a perfect square. That's the same thing as two squared. And let's see, 20, right over here, if we want it to fit this pattern, we would see that A is five and B is two, and so let's see, what
would be two times AB? Well, five times two AB would be 10, and then two times that would be 20. So this right over here is, that is plus two times five. Two times five times two times two X. Times two X, I'll do it in this color. Times two X. So you see that this completely
matches this pattern here where A is equal to five
and B is equal to two. Once again, this is AX,
the whole thing squared, then you have two times A times BX, you see that there, and then finally you have the B squared. So if you wanted to
factor this, you could say "well, this is just
going to the same thing "since we know what A and B are, "this is going to be five X plus two." Five X plus two. Five X plus two, whole thing squared. So the whole point of doing
this is to start recognizing when we actually have perfect squares, especially perfect squares
where the leading coefficient isn't one.