Example 1: Factoring quadratic expressions

Factoring trinomials with a leading 1 coefficient
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Example 1: Factoring quadratic expressions

Discussion and questions for this video
He factors (t+3) out. You can treat t*(t + 3) + 5*(t+3) as though it has parenthesis around it (t*(t + 3) + 5*(t+3)). If you do that you can see that you have two things (t and 5) multiplied by (t+3) so you can factor out a (t+3). This leaves you with a (t+3) multiplied by what is left inside the parenthesis which is (t+5). So the expression ends up being (t+3)(t+5)
where is the other t+3 gone can anyone explain please
You didn't specify at which time but I guess that you meant somewhere at 3:30 where we have:
(t²+3t) + (5t+15)
t(t+3) + 5(t+3) → We factored out t from (t²+3t) and 5 from (5t+15)
(t+3) (t+5) → now this is part that you were asking for, you're wondering where did the other (t+3) go, actually if you look at t(t+3) + 5(t+3) carefully you might realize that they both have (t+3) so we can just factor that out and add their coefficients in another parenthesis as another variable.
And that is why (t+3) (t+5) is our final answer!
I have a question?

HOW DO YOU SOLVE THIS!

Example:
3/2x+4=-9

Thanks, Please help!
3/2x + 4 = -9. To solve for x, we can do a few things. I would suggest one method though, let me show you...
3/2x + 4 = -9
subtract 4 from both sides.
3/2x = -13
Now, multiply both sides by 2/3.
(2/3)*3/2x = (2/3)*(-13/1)
6/6x = -26/3
6/6 = 1, so
x = -26/3
and we are done. Tell me if that helped or not.
When doing the KhanAcademy lessons for Factoring Polynomials 1 (https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_factoring/e/factoring_polynomials_1) I was given this problem to factor.

x2−10x+24 (x to the second power minus by ten x plus twenty four)

I wrote down this answer and it solves the right way,

(x + 2)(x - 12)

But KhanAcademy said that I got it wrong and that the answer was actually,

(x−6)(x−4)

Can anybody tell me where I went wrong? I just don't get it. They both solve the same way.
So the equation they want is x^2-10x+24 Correct?

If you multiplied yours out, it would be x^2-10x-24.
(x+2)(x-12) = x*x-12x+2x-24 = x^2-10x-24 because 2 times -12 is -24.

While the answer provided
(x-6)(x-4) = x*x-6x-4x+24 = x^2-10x+24. Negative 6 times negative 4 is positive 24.Which is what they wanted.

You got really close, just gotta watch that minus sign on the last term.
Why do the (t+3)'s you factored out of the trinomial cancel each other, and if they did wouldn't they leave a 1 instead of a a single (t+3) remaining? sorry if this was a badly worded question btw
what if the problem is x^2-2x-8. the factors of 8 dont add up to 2? like in the video 5 and3 add up to 8
sifuentesj,
-4*2 = -8
and -4+2 = -2
So x²-2x-8 can be factored.

There are other methods such as "completing the square" and the quadratic formula that can be used when you can't easily factor the expression.
Here is a video on completing the square
https://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/solving-quadratic-equations-by-completing-the-square

And on the quadratic formula https://www.khanacademy.org/math/algebra/quadratics/quadratic_formula/v/using-the-quadratic-formula

I hope that helps make it click for you.
This question is a bit off topic but can anyone explain how I would factor an expression such as this one
x^2+(3y+4)x+(2y+3)(y+1) I have been stuck on this question for a while and any help would be useful, Thank you in advance.
The sum of two numbers is 3y + 4 and the product of two numbers is (2y + 3)(y + 1)
So this equation can be factored as (x + 2y + 3)(x + y + 1)
This problem is of the form (x + a)(x + b) where a = 2y + 3 and b = y + 1
To factor the expression ax² + bx + c, first solve the equation ax² + bx + c = 0 to find the roots of the polynomial (this can always be accomplished using the quadratic formula). Suppose ß and µ are roots of said equation. Then both (x - ß) and (x - µ) must be factors of the expression. We may then write ax² + bx + c = a(x - ß)(x - µ).

Note: ß and µ need not be distinct, and they may even be complex conjugates of each other.
I want to see an example of factoring a trinomial in the type of ax(squared) + bx + c where a is greater than 1 and a, b, c do not have a common factor greater than 1.
Julie - Chuck's answer leads you to the quadratic formula, which is a great one-size-fits-all solution to polynomials like those in this video. But it doesn't answer your question: if you're looking for a way to factor that's analogous to the way above, where "a" is not 1, check out this video:

https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-quadratic-expressions/v/factoring-trinomials-with-a-common-factor

The quadratic formula becomes more important when you encounter quadratic polynomials that you can't factor easily - often roots and irrational values result.
at 2:27 he mentions grouping, what is grouping
Taking out separate thing and putting them into groups
What if at 3:40 you had t(t+3) + 5(t-3). How do you deal with one of the values being negative?
At the end, I seem to have lost my step. What did you do? I was lost.
ok, now what if the question is x^2 -x -8 = 0 ?
I'm just having trouble coming up with the two numbers for the factoring...
Sal is an awesome teacher explaining it in short times and no questions needed for his methods
You find factors for the last term and then add them together and the ones that add together to equal the middle coefficient are the ones that you put after x.

Eg. t^2+8t+15
The factors of 15 are 5 and 3, and 1 and 15. 5 and 3 add together to make 8, which is the middle coefficient, so the factored equation is (x+3)(x+5)
Expand the polynomial, to get x^2 + 2x - 3 = 4
Set the equation equal to zero, and then factor.
if a polynomial cannot be factored completely and you use the quadratic formula to solve for the variable should you simplify and factor the beginning polynomial as far as you can before you use the quadratic formula?
Thanks!
so the final answer, the thing you would write down and box is "(t+3)(t+5)" right?
What if the leading coefficient is -1?
It does not make a difference. Just do the same thing that you normally do. Ex: -x^2+5x-6
=> (-x+2)(x-3).
Please Help! How would I factor the quadratic 6x^2-45x-24=0, thanks.
First lets notice that all of the coefficients are divisible by 3. We can factor that out to get 3(2x^2-15x-8).
So now we have ax^2 + bx + c and we're looking for something with the form of (mx + p)(nx + q) where a = mn, b = pn + mq, and c = pq.
a = 2 so m and n have to be 1 and 2. We now have 3(2x+p)(x+q).
b = p + 2q and c = pq. c also equals -8 so p and q must be { -1, 1, -2, 2, -4, 4, -8, 8}. If you test out all of the option you'll see that 8 * -1 = -8 (c) and -1 + 2 * 8 = -15 (b).
Our answer is 3(2x - 1)(x + 8)
I understand everything except at 3:28, when t(t+3) + 5(t+3) equals (t+3)(t+5), I do not know how Sal got this, plz someone explain to me....

By the way Sal is awesome!!! LOL
How does he get from t(t + 3) + 5(t + 3) to (t + 3)(t + 5)

I'm interested why and how the factoring out of t + 3 works, exactly which places are factored out.
Answering my own question with the help of my GF, important is the following math rule:
a * b + a * c = a (b + c)

*Why is this?*
(a * b) + (a * c)
____________ * a = (b + c) * a
a

So to get from t(t + 3) + 5( + 3) to (t + 3)(t + 5) we can do the following:

t(t+3) + 5(t+3)
____________ * (t+3) = (t+5) * (t+3)
(t+3)
Am I performing a correct arithmetic operation if I factor like following:
(x+4)(4+4)=8x+32
I haven't seen anything like it, so I got a bit confused.
No. Use FOIL First Inside Outside Last to do it backwards

3x(2x) -10x +3x - 5 =
6x^2 -7x -5
A trinomial follows the form a^2+2ab+b^2, where a and b do not equal 0. When factored, the trinomial equals (x+a)(x+b). A and b are the variables commonly used to write out the formula for factoring a trinomial, but really any two symbols can be used.
I need some homework help.
The problem is 20t^2 - 12t - 32
I know the answer is 4(5t-8)(t+1) but I don't understand how my teacher got to that answer.
It is kind of a homework question, practice question. I have worked on it for two days!! x(4x-5) =114. I cannot find anything that factors out. 4x-5x-114=0 What adds to 5 and when multiplies gives you 114??
Hi Patty,
Nikola is correct in the factoring but I wanted to point out where I think you were having trouble. It sounds like you were forgetting about the 4 in front of the x^2 when you were looking for factors. Remember that for Ax^2 + Bx + C = 0, when we are looking to break down the expression, we multiply A*C to get a value. In this case, A*C = -456. We then break down the -456 into two factors that, when added together, give us B, in this case, -24 and 19 are the two factors we want. We then rewrite the original equation, using our two factors to replace B:
4x^2 - 24x + 19x - 114 = 0
We can then group and factor:
(4x^2 - 24x) + (19x - 114) = 0
From the first group we can factor out a 4x and from the second group we can factor out a 19:
4x(x - 6) + 19(x - 6) = 0
At this point, since we have like terms (x-6), we can combine the 4x and the 19:
(4x + 19)(x - 6) = 0
From there, if you are solving for x, you would set each factor equal to zero:
x - 6 = 0
x = 6
or
4x + 19 = 0
4x = -19
x = -19/4
So your roots would be x = -19/4 or x = 6
Hope that helps.
Again, How do you factor a quadratic expression that looks like 6x^2-23+15
See video titled: Factor by grouping and factoring completely
The number in front of the variable. An example is 9x where 9 is the coefficient and x is the variable.
how do i get the coefficient, and the numbers,variables, to have a similar factor?
need a more clearer question, what the topic? You can multiply an entire equation by some factor to match up other factors, say in elimination.

x^2+4x+5 can be multiplied by say 2 to get 2[x^2+4x+5] = 2x^2+8x+10 or by whatever you want to change the coeffients to.
The FOIL Method is a process used in algebra to multiply two binomials. The lesson on the Distributive Property, explained how to multiply a monomial or a single term such as 7 by a binomial such as (4 + 9x).
The leading term, first term, `x^2` has no explicit coefficient since, by convention, we do not write the one, but it is there.
At 3:45, how does Mr Khan go from (tsq.+3t)+(5t+15) to (t+3)*(t+5)?
A coefficient is the constant number in front of a variable that is multiplying by it. So if you have "5x", 5 is the coefficient.
I'm only confused about problems where "a" (in ax^2 + bx + c) does not have a value of 1. How do you take this into account when factoring?

Without using the quadratic formula.
If you have ax^2 + bx + c = 0 and you want to factor, the first thing I would do is divide both sides by a. Here is an example: 4x^2 + 12x + 9 = 0. Dividing both sides by 4 yields: x^2 + 3x + (9/4) = 0. You might be able to recognize this as a perfect square: The square root of 9/4 is 3/2 and (3/2)x + (3/2)x = (6/2)x = 3x. So we have x^2+3x+(9/4) = (x + 3/2)^2
ok, what if there is a variable in the middle like X to the second,minus x, minus 20?
You can still factor x2-x-20. the -x would be the same as -1x. I prefer plugging these things into the quadratic equation for then you get exact answers without the guess and check method.
How do you do this is the question is t squared + 8t MINUS 15?
equals what? you have to complete the square or use the quadratic formula if you wanted to find t
Well, to be fair to teachers, they do have questions asked all the time. . . . but as for the 'taught it better' part, that depends on the teacher. A bit of a redundant statement, even by my standards. And the awesome thing is that we can ask our questions in the comments and know that we won't be shunned by the world for asking a redundant question, or a simple question that many people know but a few just can't grasp.
what if you had something like 2x(x+2)-3(x-2). How would you further simplify it?
You would have to expand it out using the distributive property. Then you could simplify it into a trinomial. That particular problem cannot be factored using real numbers, so a trinomial would be its simplest form.
Hi everyone, for factoring quadratic expressions how do you know what numbers to put in the parenthesis. I understand that you find numbers that multiply to whatever x and add up to the other number but does anyone have tips for finding out these numbers??
You can factor the last number (15 in the example) to see which of its factors add up to the middle number (8 in this case). The factors of 15 are {15,1} and {5,3}. Since 5 +3 = 8, you can try them to see if they work.
(t + 5)(t+3).= (t^2 + 8t +15)
Also, since the middle number was positive, we didn't need to consider the negative factors of 15, namely {-15,-1} and {-5,-3}. If the middle number was negative (-8) then {-5,-3} would have been the answer since
(t - 5)(t - 3).= (t^2 - 8t +15)
At 4:00, is that the answer for this problem? I still don't get it how does that when mutiplied do we get t squared? Is it the the one in purple with green or the one you were talking about t +3 times t+5. I still don't understand.
The 1 doesn't have to be put since there is only one "t", but if there were 3 t's, then you would need to put 3t, because 1t=t, you do not need to worry about the leading coefficient.
Sal is just showing us the same problem in "color Coded" form. he is showing us that if you replaced a and b with the real numbers in this problem you will get the same "form" of answer.
How would you factor an expression with numbers that are more than squared? Something along the lines of X to the 3rd + bX squared + cX. How would that work? Btw just as a side note...how do you get a computer to type in exponents?
Ok, I run into an issue when trying to factor my own equation: X^2+4x-3.

I cannot seem to get both the product of -3 and the sum of 4. I am running into many problems like this. does the method in this video not apply here??
You are correct in that this method does not work for all quadratics. Your expression's roots are not rational numbers so you would need to solve for the roots with the quadratic formula.
What do I do if the exponent with the first monomial is a 3?
Ex: 6x^3-13x^2-28x
Thank you! That just never even clicked in my mind! Glad I found out before EOC's though!
what if you have a problem such as 9x^2+72x+142=0 where none of the numbers you mutiply together to get 142 and have a sum of 72
Ok, so I have a question. If i had a problem like this: x^2+x-90, what would i do? because the middle term only has a variable x. Can someone help please?
ok, so here goes nothing! so we know that we have to find a value for a and b. a times b should be equal to 15 and their sum should be equal to 8. so a and b are 5 and 3. this makes the equation: t(t+3)+5(t+3). so since there are two (t+3)'s then we can factor it out to this: (t+3)(t+5) and that would be the answer! does that help? hope ie does!
I'm working on Factoring Quadratics 2 and I got a question that looks like this: -2x^3-18x^2+20x
I'm confused about the -18 and the +20. I tried watching a video to know how to factor that part, but all of the videos either show two negatives or two positives. Can anyone help? Thanks!
```-2•x^3 - 18•x^2 + 20•x

-2•x•(x^2 + 9•x - 10)

[-2•x•(x + 10)•(x - 1)]```
I'm starting to get the concept of the factorizing idea. But what if I had to imply on harder polynomials, for example this one in my book:
81p^4-18p^2q^2+q^4
The answer is (9p^2-q^2)^2
or
(9p^2-q^2)(9p^2-q^2)

Note that you can treat the p^4 as (p^2)^2 and q^4 as (q^2)^2.

It may be simpler to look at it this this way.
Make the definition of
p^2 = a
and the definition of
q^2 = b.

Then the original expression can be written as 81a^2 - 18ab + b^2.

This factors into (9a - b)(9a - b).

Then make the substitution a = p^2 and b = q^2. This will give you
(9p^2 - q^2)(9p^2 - q^2)
or
(9p^2 - q^2)^2
Mathematical convention. a and b are coefficients. The accepted general form of a quadratic equation is `ax^2 + bx + c = 0`
Using x and y would be confusing because the expression already has avariable named x.
I am working on Factoring Trinomials using FOIL. However, I am having a problem finding a video to explain that. Can you point me in the right direction? Thanks!
I am sorry, but Sal doesn't have a video on this topic. I found someone else who does a really good job on this. http://www.youtube.com/watch?v=uFenBVfPxio
Find factors of 15 (2 numbers that multiply to create 15) and those same 2 numbers need to add to -8
-3 and -5 work
This let's you factor the polynomial into: (T - 3) (T - 5)
I would like to know what video would inform me on how to do a problem like y=(x+9)². It is a quadratic equation.
I don't get from 3:35 to the end. how did he factor it out to where from t(t+3) + 5(t+3) to the answer being (t+3)(t+5) ? Please Help!
okay this follows the distributive property, he distributed the t+3 out because
(t+3)(t+5) can be rewritten as t(t+3) + 5(t+3) if you distribute the t+3...did that make sense? its like when you have 15+30 and you can make it (5)(3)+(5)(6) = (5)(3+6)
I understand how to factor these equations but I don't understand when the leading coefficient is greater than 1. For example, a problem I received in class was 4x^2+12x-7. How would this be solved? Someone please help me. Thanks.
Okay, so i understand this, but what if the second number in the trinomial was posotive and the third one was negative? How would you solve the problem x squared +14x-49???
This is a Great video but the entire thing was confusing to me can you help me Khan
Lots of people here would love to help you, but please ask a more specific question. What was the first thing in the video that confused you?
is the a and b step really necessary i feel that there is a easier way to factoring if so please tell me
Once you become better at factoring quadratics, you'll realize that the "a" and "b" step is practically useless, as you asked. Basically, for simple quadratics without a coefficient in front of the squared variable, you just have to find two numbers that multiply to equal the constant, or number without the variable. Also, the two numbers must be added to equal a sum that is the coefficient in front of the second term. Hopefully, that will make sense.
Perhaps that quadratic equation cannot be factored. Try using the Quadratic Formula instead and see what kind of solutions you get for x. If they involve irrational numbers, then the equation is "prime" and cannot be factored.
i can't use what sal is doing in the problem in my homework..

2a^2-5a-4^2
There are a couple of ways to approach this, but factoring always starts by:

1) Look for a GCF (greatest common factor)
2a^2-5a-16 There are no common factors.

2) Look for a common pattern (a^2-b^2)=(a+b)(a-b); a^2+2ab+b^2=(a+b)^2; a^2-2ab+b^2=(a-b)^2
These are three patterns you just have to brute force memorize called difference of squares and perfect square patterns.
None of these patterns work.

3) Use a generic factoring method (ac method, slide and divide, grouping, educated guess and check, etc.)
ac method: 2•-16=-32, so look for factor pairs that multiply to equal -32 and add to equal -5
1+-32=-31
2+-16=-14
4+-8=-4
There are no other factor pairs of -32 so this will not factor evenly (aka it is 'prime')

4) For more advanced factoring of quadratics you can use the quadratic formula to get irrational and imaginary factors. a*x^2+b*x+c will factor as a(x-r1)(x-r2) where r1=(-b+√(b^2-4ac))/(2a) and r2=(-b-√(b^2-4ac))/(2a)

[-(-5)±√((-5)^2-4(2)(-16))]/[2(2)]
[5±√(25+128)]/4
[5±√153]/4
[5±√(3•3•17)]/4
[5±3√17]/4

2(a+[5+3√17]/4)(a-[5-3√17]/4)
how do you solve 4x2 - 6x - 30 ???
I've tried factoring it but can't seenm to find an answer
There are bunch of ways to solve quadratic equations. Depending on if you have a binomial (expression with two terms) or a trinomial (expression with three terms).

Binomial:

Difference of two squares
a^2 - b^2 =(a+b)(a-b)
ex. 16x^2 - 9x^2 =
(4x)^2 - (3x)^2=
(4x+3x)(4x-3x)

Difference of two cubes:
a^3 - b^3 = (a-b)(a^2+ab+b^2)
ex. 27x^3 + 1 = (3x)^3 + 1^3
= (3x + 1)((3x)^2 – (3x)(1) + 1^2)
= (3x + 1)(9x^2 – 3x + 1)

Sum of two cubes:
a^3 + b^3 = (a+b)(a^2-ab+b^2)
Works the same way as Difference of two cubes


Trinomial:

FOIL
You already know this method

Quadratic Formula
(-b±√(b^2-4ac))/2a

Completing the square:
This is a bit hard to explain without a diagram so here is a website for help. http://www.purplemath.com/modules/sqrquad.htm
Do you have videos that show you how to factor this type of polynomial ---> x3-8 or 10x3-25
Does the method shown in the Factoring quadratic expression video only work if the term with an exponent is squared? Could it work if it was an equation with an exponent of 4?
And what about this? Can you factor this out by grouping?
4x^2+4x+13
Cause i always see examples that i can solve very quickly, but i haven't see an example what i can't factor by grouping. Is that possible or i just can't see the factors in my example?
My question for my my friend, is where does your name come from? Ive never heard it before and it is interesting...
What would I do if I had a c that couldn't be divided by my b?
Like in b^2+*b=7?
Use the quadratic formula. Answers to a^2x+bx+c=0 are (-b plus or minus sqrt(b^2-4ac)) / (2a).
This one: b^2+b-7=0. => (-1 plus or minus sqrt(1-4(1)(-7))) / (2) => (-1+sqrt(29))/2 and (-1-sqrt(29))/2 are your answers.
I need help determining when to factor and when to apply the quadratic equation. For instance, in the problem:
Solve for x:
2x^2+6x−140=0
My first inclination was to apply the quadratic formula, which gave me x= (-3+or- 2(70)^1/2)/2
And that's OK, but not pretty.
But when I entered it, it was wrong. So I tried again and got the same answer. So I asked for a hint and they started in factoring. And I ran with the whole factoring thing and got the nice, pretty answer x=-10 or x=7.
Why did the two methods get me different answers?
And how can I tell when looking at a problem which one to use?
say the sum is 4 x^3 y^3 - 12 x^4 y^5 + 16 x^6 y^6
for the multiplication and addition part im unable to get the right numbers
please help me out
My teacher in the past two years dont help and i been watching you for one hour and i alraedy get it? I need u as my teacher
At 1:58, Sal says you can use this method if the coefficient is 1.
What if the coefficient of x term raised to the second power, is greater than one?
Like: 2x^2 + 2x + 1
Can you use this same factoring method?
You can do something similar.
2X^2 +10X + 8.
Multiply A and C. 2*8 = 16.
Find factors of AC (16) that add to equal B (10). These are 8 and 2.
Divide each factor by A. 8/2 and 2/2.
Have each coefficient of X as 1
(X+ )(X+ )
and put your factors in
(X+8/2)(X+2/2)
Multiply this by A
2(X+8/2)(X+2/2). This reduces to
2(x+4)(x+1).
The same thing that happens when A is 1, but multiplying and dividing by 1 don't change your values.
Can x^2+24+143 be factored, having trouble with the operation.
You can save some time by noticing that you only have to check for divisibility up to the square root of the number. For example, when checking factors of 143, we only have to go up to 11 because sqrt(143)~=11.9. All of the factors of 143 that are above 11.9 will be flushed out when you test the lower numbers.

In other words, you can stop checking factors when the result you get is smaller then the number you divided by.
A good shortcut would be finding the factors of the last monomial or variable which you can add or subtract and the answer would be the numeral coefficient in the middle.
That sounded more complicated
the terminology you used is wrong. The last thing in that expression, the +15, isn't a monomial or a variable, it's a "term". A monomial is an expression with only two terms. This expression has three terms, so it's a trinomial.
How do you do the problem a^3-9a+3a^2-27? How do you factor it. I'm really confused
Continuing from Kate Ballard's explanation, the first quantity can be factored more using the difference of two squares. The final answer should be (a-3)(a+3)(a+3) and this can be changed to (a-3)(a+3)^2.