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## Integrated math 3

### Course: Integrated math 3 > Unit 3

Lesson 3: Dividing polynomials by linear factors- Dividing polynomials by linear expressions
- Dividing polynomials by linear expressions: missing term
- Divide polynomials by linear expressions
- Factoring using polynomial division
- Factoring using polynomial division: missing term
- Factor using polynomial division

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# Factoring using polynomial division

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Identify the main points of the video, and summarize them in 40-60 words at a 6th grade level. Optimize the text for SEO. Don't use the words "in this video" - just get to the point. Use active voice. Avoid the words magic, adventure, dive, lowdown, fun, and world.
The video breaks down the process of dividing polynomials by linear factors. It starts with a given polynomial and a known factor, then uses polynomial division to rewrite the expression as a product of linear factors. The video emphasizes understanding the steps and the reasoning behind each one.

## Want to join the conversation?

- i dont understand i think my brain fried, i have watched this video more than 3 times i am a lost boy in neverland.(15 votes)
- At0:29, he says the polynomial wouldn't be so easy to factor, so my question is how do you factor this polynomial if you don't have the given linear factor.(8 votes)
- Well, you have two real choices. You can factor by grouping:

https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-advanced-polynomial-factorization-methods/v/factoring-5th-degree-polynomial-to-find-real-zeros

Another method is called the "rational root theorem"... which I wasn't able to find on this site. If someone else finds it, please link it. It's a pain to use, and it doesn't work to find irrational or imaginary roots... but it does find all of the rational roots (plus a couple extra)... Here's how it works:

For the equation: 4x^3 + 19x^2 + 19x - 6, take the last coefficient, and divide it by the lead coefficient.

6/4

Try all possible combinations of the factors of the numerator, with all possible factors of the denominator:

6/4, 3/4, 2/4, 1/4, 6/2, 3/2, 2/2, 1/2, 6, 3, 2, 1

OR: 3/2, 3/4, 1/2, 1/4, 3, 1, 6, 2

OR (in order): 6, 3, 2, 3/2, 1, 3/4, 1/2, 1/4

Then divide the polynomial by x +- (each one of those numbers listed above). When you're done dividing, if you don't get a remainder, then congratulations! You found a root! If you get a remainder, then it wasn't really a root.

Pro-tip: If you do use this method, learn synthetic division, as it really speeds up all the dividing you have to do:

https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-synthetic-division-of-polynomials/v/synthetic-division

With any luck, you'll be able to factor by grouping.(11 votes)

- what i there is a remainder?(7 votes)
- Winding up with a remainder is just that. Theres no way to make it a "nice" polynimial. It'd basically be a polynomial plus some rational function.

A quick example, I'm just making these up, (x^4+2x^3-3x^2+5x-7)/(x-2) would get you x^3+4x^2+5x+15 + 23/(x-2). the 23/(x-2) is that rational function I mentioned. Hopefully you notice x-2 is in the denominator and is the divisor int he division problem. This is always the case, and the numerator is at most one degree less than the denominator.(7 votes)

- At0:14he says linear factors, but doesn't explain what that exactly means....(4 votes)
- "Linear factors" is just a phrase for a factor that looks like (ax+b).

At the end of the video we see the factors are (x+2), (x+3), and (4x-1), which all follow that format.(11 votes)

- How do you get the 12 from the equation?(4 votes)
- By multiplying a*c of the quadratic 4x^2+11x-3 and then finding factors of a*c that when added together gets you +11, which happens to be +12 and -1. That’s how you would break up +11, then start grouping, and go from there to factor. Hope that answers your question.(3 votes)

- At about1:30, he says about that you could use the quadratic formula to factor into a linear expression if the quadratic was hard to factor, so my question is how do you factor this polynomial if it's hard to factor using the quadratic formula to put it in linear factors?(4 votes)
- I'm sorry, but I'm still confused about how you break down the equation after finding the quadratic for the polynomial that you get after dividing.(3 votes)
- 2:33Why does Sal multiply 4x^2+11x-3 with x+2 again? I don't understand why...(2 votes)
- So it started with 4x^3 + 19x^2 + 19x - 6. then he divided that third degree polynomial by x+2 which turned it into 4x^2 + 11x - 3. you could look at this as saying 4x^2 + 11x - 3 times x + 2 equals 4x^3 + 19x^2 + 19x - 6. basically 6/3 = 2 so 3 * 2 = 6, that kind of thinking.

When you factor something you are splitting one value into two (or more) smaller ones that when multiplied together make the original. again, you can factor 6 into 2 and 3, so 6 = 2*3. factoring 4x^3 + 19x^2 + 19x - 6 got 4x^2+11x-3 and x+2. If you try and multiply those two together you should get the original.

Let me know if that didn't help.(2 votes)

- What do you do if there is a remainder?(1 vote)
- If you are factoring and the term you have chosen to divide by has a remainder, it is not a factor and you have to choose another.(2 votes)

## Video transcript

- [Instructor] We are
told the polynomial p of x is equal to four x to the
third plus 19 x squared plus 19 x minus six has a
known factor of x plus two. Rewrite p of x as a
product of linear factors. So pause this video and see
if you can have a go at that. All right, now let's
work through it together. So if they didn't give
us this second piece of information that as a known factor of x plus two, this polynomial
would not be so easy to factor. But because we know we have a known factor of x plus two, I could divide that into our expression and
figure out what I have left over, and then see if
I can factor from there. So let's do that. Let's divide x plus two
into our polynomial. So it's four x to the third power plus 19 x squared plus 19 x minus six. And we've done this
multiple times already. We look at the highest-degree terms. X goes into four x to the third. Four x squared times, I
put that in the x squared or the second-degree column. Four x squared times x
is four x to the third. Four x squared times two is
eight x, eight x squared. And then I wanna subtract
these from what I have up here. So I'll subtract. And then I'm going to be
left with 19 x squared minus eight x squared is 11 x squared. And then I will bring
down, bring down the 19 x. So plus 19 x. And so once again, look
at x and 11 x squared. X goes into 11, x squared 11 x times. So that's plus 11 x. 11 x times x is 11 x squared. 11 x times two is 22 x. Need to subtract these from
what we have in that teal color. And we are left with 19
minus 22 of something is negative three of that something. In this case, it's negative three x. And then we bring down that negative six. And then we look at, once again, at the x and the negative three x. X goes into negative three
x negative three times. And so negative three times
x is negative three x. Negative three times two is negative six. And then if we wanna
subtract what we have in red from what we have in magenta. So I could just multiply
them both by negative. And so everything just cancels
out and we have no remainder. And so we can rewrite p of x now. We can rewrite p of x as being equal to x plus two times all of this business, four x squared plus 11 x minus three. Now we're not done yet because we haven't expressed it as a product
of linear factors. This one over here is
linear but four x squared plus 11 x minus three,
that's still quadratic. So we have to factor that further. And let's see, there's a couple of ways we could approach it. We could use, well, we could try with this and the quadratic formula or
we could factor by grouping. And to factor by grouping,
and the whole reason we have to factor by grouping is we have a leading coefficient
here that is not one. And so we need to think of two numbers whose product is equal to
four times negative three. So we have to think of
two numbers, let's just call them a and b. A times b needs to be equal to four times negative three, which is negative 12. And a plus b needs to be equal to 11. And so the best that I
can think of, they have to be opposite signs 'cause their product is a negative. So If I had negative, if
I had, if I had positive 12 and negative one, that works. If a is equal to negative
one and b is equal to positive 12, that works. And so what I wanna do is I wanna take this first-degree term
right over here, 11 x, and I wanna split it into a
12 x and a negative one x. So let's do that. So I can, let's just focus on this part, on this part right now
and then I'll put it all back together at the end. So I can rewrite all of this
business as four x squared. And instead of writing the 11 x there, I'm gonna use this blue
color, I'm gonna break it up as a 12 x. So plus 12 x and then minus one x. Notice these two still add up to 11 x. And then I have my minus three. And then, let's see,
out of these first two what can I factor out? Let's see, I can factor out a four x. So I can rewrite these first two as, and if this is unfamiliar
to you I encourage you to review factoring by
grouping on Khan Academy. So if we factor out a
four x that's going to be, it's going to be, we're
gonna be left with an x here. And then we're gonna be
left with a three over here. And then these second
two terms, if we factor out a negative one, so
I'll write negative one, times, we're going to
have an x plus three. And so then we can factor
out the x plus three. So let's do that. I'm running out of colors. So I factor out the x
plus three and I am left with x plus three times, times four x. And I'm gonna keep these colors the same so you know where I got 'em from. Four x minus one. This is a very colorful
solution that we have over here. And there you have it. I factored the second part
into these two factors. And so now I can put it all together. I can rewrite p of x as a
product of linear factors. P of x is equal to x plus
two times x plus three times four x minus one. Four x minus one. And we are done.