More examples of factoring quadratics with a leading coefficient of 1

Can't get enough of Sal factoring simple quadratics? Here's a handful of examples just for you!
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More examples of factoring quadratics with a leading coefficient of 1

Discussion and questions for this video
Can both a and b equal the same for example x^2+20x+100 can both a and b equal 10?
Yes, a and b can be equal. In the example you wrote, that expression is a special case. It is the square of a binomial. (x+10)*(x+10) = (x+10)^2. x+10 is a binomial that is being squared. I hope I helped.
it is the number that is multiplied by a variable
how come the constant term is the product of a and b?
We are looking for a and b for this situation:
(x+a)(x+b) = 0
When we multiply the left side out we get
x^2 + bx + ax + ab = 0
So x^2 + (a+b)x + ab = 0
Therefore ab is the constant term.
(And a + b is the coefficient of the x term.)
Can the 1 and the 9 switch so it's (x+9)(x+1)? Or does it have to be (x+1)(x+9)?
yes they can
remember the commutative property of multiplication
Why would you need to factor a quadratic in the first place?
So that we can........
1.) Find the value of a missing variable, such as x.
2.) Cancel out something that is on the numerator and the denominator to simplify rational expressions.
For instance if we have something like....
x²-2x-80/x²+16x+64
(x-10)(x+8)/(x+8)(x+8)→As we can see we can now cancel out an (x+8) here to simplify it to.....
x-10/x+8
There are a lot of videos on factoring by grouping. We have to learn to do it by trial and error. Are there videos for that too? How would you find 8x^2 -14x+3?
Even simpler (so as to avoid the fractions), if the quadratic is in the form:
`ax^2 + bx + c`
where a is not 1, you can do the following:

Multiply `a` and `c` to get a new number (say `d`). Now, you must find two numbers that add up to `b`, and multiply to `d.`
`x + y = b
x*y = d = a*c`

So in your example above, you would see that `d = a*c = 8*3 = 24.` Two numbers that add up to `b = -14` and multiply to `d = 24` are `x = -12`, and `y = -2.`

Then you can simply group and factor to get the correct answer.
What if the numbers are prime and have no factors?
All quadratics can be factored, but not all of them can be factored with rational numbers or even real numbers. If a quadratic cannot be factored into rational factors, it is said to be irreducible. However, it is always possible to factor a quadratic, if you allow irrational or complex factors.

Here's how to factor ANY quadratic expression in the form: ax² + bx+c.
Let d = b² - 4ac
(If d is not a positive perfect square, then the quadratic is "irreducible".)
The factors are:
a [x + ¹⁄₂ₐ (b + √d)] [x + ¹⁄₂ₐ (b - √d)]
If it makes for convenient numbers, you may use the distributive property to multiply a into one (but not both) of the factors.
You cannot work out what x is in the expression x^2-5x+24. This is because expressions cannot be solved as they do are not equal to anything. E.g they are like statements.

If x^2-5x+24 was equal to a number or another expression then we would be able to work out the value of x.

For example:
x+1 (x could be equal to any number)
BUT
x+1 = 3 (x can only be equal to 2)

I hope that helps
How do you factor a polynomial when the highest power is greater than ^2?
A constant is a number with no variable. It is called a constant because it's value does not change.
i want to learn how to factor polynomials with 4&5 degree Quartic and Quintic expressions... can someone plz help me find the video? thanks
Unlike quadratic or even cubic equations, there is no easy way to factor quartic and quintic expressions. That is an exceedingly advanced topic. With the exception of a form of quartic expression called a "biquadratic", factoring quartic and quintic expressions requires the most prodigious of mathematical skills.
This is a really good question. It confused me for a long time, because "quad" usually means "four".

A quadratic equation is one where one of the terms is raised to the second power, and there are no terms raised to higher powers. There are probably a few more restrictions if we wanted to be precise. So a quadratic equation might look like this:
```
ax² + bx + c = d
```
A more concrete example might be:
``` 3x² + 2x + 1 = 0```
or ```-2x² = 8```

I always wondered why it was called quadratic when it goes up to the second power, and not the fourth. It turns out the same word is used for square as for four. I think this is because if you want the area of a square, you multiply the length by itself. The same goes for cubic. A cubic has a term raised to the third power, and that is what you have to do to get the volume of a cube.

It stops there as far as I know. I'm not sure what something raised to the fourth power would be called. Maybe a tesseratic?
Should I work on polynomials before quadratics? I felt I needed quadratics for some parts of functions, but trooped through. I've felt sometimes I should have gone through polynomials before most everything and polynomials is misplaced in the playlist, but I'm not sure. Should I work through polynomials before quadratics first? I don't want to just know how to do the problems, I want to be able to visualize it well and have a conceptual knowledge.
Since a quadratic expression is a special form of polynomial, you are actually working on both. For most polynomial problems you will be given three terms and asked to factor it as the product of two binomials or one binomial with another polynomial. If you understand that at any time a coefficient can be replaced with another variable then it will be easier for you to understand how to factor all polynomials. You will often see the equation Ax^2 + Bx + C = 0 and that just shows the general form of all quadratic equations. When you understand how B is related to A and C then it is easier to factor these types of expressions.

There are multiple ways to visualize polynomials because they can be used to calculate real world problems. An easy visualization is that of a rectangle with a variable length and width and the area is equal to the length times the width. The two binomials define the width and length while the final polynomial represents the area.

Here is a tool to examine factoring quadratic expressions. It doesn't address polynomials with multiple variables, but is a good way to see the relationship of the a and b in the (x + a)(x + b).
http://www.khanacademy.org/cs/quadratic-factorization/1386235304
Do the numbers have to be whole, or could you end up with an answer like (x+2.5)(x+5.7)?
Yes, that could certainly happen. The quadratic expression x^2 + 4x + 3.75 factorises into (x + 1.5)(x + 2.5), for example. But (at least at first) you will probably find that the examples you are given factorise into integers.
Neha,
To factor 7x^2-31x-20
First multiply the first and last numbers. 7*20 = 140.
Factor 140 to its primes 7*5*2*2
Now find two numbers that multiply to 140 and subtract to 31
You need a number that is more than 31 so start with
7*5=35 and 2*2 = 4 And we got it first guess 35-4 = 31
We know the x values are
(7x ± ? )(1x ± ?)
You know the 7 need multiplies by the 5 so the second ? is 5
(7x ± ? )(1x ± 5) The other ? must be 4
(7x ± 4 )(1x ± 5)
The 20 is a -20 so one sign must be + and the other must be -
The middle factor is -31 not +31 so the 7*5 must be negative so use -5
(7x ± 4 )(1x - 5) And as we already said the other sign must be +
(7x + 4 )(x - 5)

7x^2-31x-20 factors to (7x + 4)( x - 5)

I hope that helps make it click for you.
If the leading coefficient is -1, is it a MUST to factor out the -1?
Not necessarily a "must", but it is much easier if you do so.
at 14:46 we do not have to multiply negative 1 to both (x-3)(x+8) ?
Remember that multiplication is associative: a(bc) = (ab)c.

So (-1) ( (x-3)(x+8) ) = ( (-1)(x-3) ) (x+8) = (-x + 3)(x+8)
Is it OK to switch the order of the equation? For instance, instead of x*2+10x+9, could you solve it as 9+10x+x*2? Or would that just be harder? Would it also work that way for other problems? Can you still factor it in this form?
Theoretically, that isn't incorrect, but, as mathematicians like to be neat freaks, it's better to write x² + 10x + 9. In some cases it's just easier to solve.
There are other techniques you will learn soon that deal with that situation.
The process of factoring 3rd degree or 4th degree polynomial is quite long for one answer.
Factoring 3rd degree Polynomials:
http://m.wikihow.com/Factor-a-Cubic-Polynomial
Factoring 4th degree Polynomials:
http://mathcentral.uregina.ca/QQ/database/QQ.09.05/kyle2.html

I hope this helps!
At 14:49, can you do (-x-3)(x+8) instead of -1(x-3)(x+8)?
No. The distributive property is not being properly used.

You are trying to multiply -1 by (x-3). Using the distributive property, you'll get -x+3 as a result. So it should be (-x+3)(x+8).
Isn't there a box method you could use to find the same answer?
The box method is the same thing as what was done in this video except that it has more visualization so it takes less time to find the answer.
At 14:47 and 16:27, did Sal mean -1 ( ( x + a ) + ( x + b ) )
Given:

``` -x^2 - 5•x + 24```

When attempting to factorize, it's preferable to have your x^2 positive.

So factor out a `-1`:

``` -1•(x^2 + 5•x - 24)```

For factoring `(x^2 + 5•x - 24)`, we need to find number pairs that add to the middle coefficient, and whose product equals the last term:

```n1 n2 sum product
-1 24 23 -24
-2 12 10 -24
-3 8 5 -24 << found terms```

Group with new terms:

``` (x - 3)•(x + 8)```

From above, we still need to account for the originally factored `-1`:

``` [-1•(x - 3)•(x + 8)]```
I have a question on solving but I can't find a video that deals with the topic (probably because I don't know what this type of problem is called) Could someone point me in the right direction or solve this step by step for me?

Solve for x
x^4-13x^2+36=0
answers given on the sheet are {3, -3, 2, -2}
my first thought was to reduce to x^2-13x+6=0 and solve from there using the quadratic formula but my answers don't match the answers on the handout and I only got two when we are supposed to find four, how do you get four?

Thanks!
simplify your life and from the very beginning, substitute a different variable in to stand for x^2 the whole time until you've got it solved and are ready for the final answer.
Say, g=x^2. So, just deal with this equation: g^2-13g+36=0.
Factor: Two numbers that multiply to be 36 and add up to be -13 are -9 and -4.
(g-9)(g-4) = 0. Now, set each part equal to zero.
g-9=0 ; g-4=0
so, g=9 ; g=4. NOW, bring back the x^2 instead of g again.
x^2 = 9 ; x^2 = 4
so, x can be -3, 3, -2, or 2.
Hi! I just wanted to ask about: What is the meaning of "quadratic"?
A polynomial or equation in general is considered a "quadratic" when it has a degree of 2 (meaning the highest exponent a variable in the equation is being raised to is 2).
Once you have a quadratic formula in (x-a)(x-b) form, you don't need to use FOIL. If you do, you just get the original equation. The answers are a and b.
The last term is called a constant because when x changes, the last term doesn't change since it doesn't have x in it. It's value stays *constant*.
This is honestly VERY confusing. I'm in high school, failing my Algebra I class, and i'm using Khan Academy to help...But how do variables like a and b and x and y add together? They're different variables! Please help!
Different variables (ex. x and y) can't be combined because they aren't *like terms*
_Example_
7x + 2x - 3y
= 9x - 3y *Can't be simplified further*

Let me know if you need this explained any further!
Quadatus is latin for square. Quadratic pertains to a square. If you have a term in an expression or equation that is to the second degree or squared, it is called quadratic.
Suppose you have x^2 - 11x + 24.
In this case a=-8 and b=-3.
So the factoring results in (x - 8)(x - 3).
Now, what values of x make the expression equal to zero?
When x = 8, we have (8 - 8)(8 - 3) = (0)(5) = 0
When x = 3, we have (3 - 8)(3 - 3) = (-5)(0) = 0
Therefore the solution is: x = 8 and x = 3.
Do *un-factorable* polynomials exist?

For example:

```x^2+8x+36```
That is a great question. The anwer is actually yes. There are un-factorable (or irreducible) polynomials and the reason why can go very deep into mathematics. It is something that people study is college and beyond. Another simple polynomial that can not be factored is x^2+1. To see why suppose it factored as (x+a)(x+b). Then we would need a+b=0 and ab=1. The first equations tells us that either both 'a' and 'b' are 0 or one and not the other is negative. From there we see that both equations can not be true at the same time (did that make sense?) The same kind of thing is true about your polynomial.

It turns out that you can always factor polynomials with degree greater than 2 into smaller polynomials but there are many degree 2 polynomials that can not be factored. If you want to know more about this you will probably have to learn about roots of equations and then maybe the imaginary numbers like the square root of -1.
So, what would a real life problem for using this be? How would you use it with other parts of math? Please ask me a question if you are confused about what I mean here! Thanks!
That, I believe, is the standard form of a quadratic equation.

The quadratic formula is used to solve equations like these and is x = [-b plus or minus sqrt(b^2 - 4 * a * c)] / (2 * a)
Most likely, but factoring out the -1 is easy, and the tools we have for factoring are known to work on positive x-squared's.

In the real world all behavior that is not specifically forbidden is allowed... in mathematics all that is not specifically allowed is forbidden.

So I would first have to verify that the factoring procedure I was using was certain to work with negative leading coefficients.
Puneet,
4x³ + 8x² is not a quadratic expression.
A quadratic expression has a term with exponent of 2 and no term with any exponent above 2.
Quadratic expressions are called quadratic because quadratus is Latin for "square"; The highest term is squared in a quadratic expression.
At 2:58 how is Sal so positive that a=1 and b=9? Couldn't a=9 and b=1 instead? Or does switching the numbers of a and b matter?
That's right. The order here is arbitrary. Sal made a choice, but your choice is equally valid.
How would I solve something like 2x^2+5x+3=0?
Doing this way gives me 3 and 1, does it not matter if the x^2 has something in front of it?