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## Algebra 2

### Course: Algebra 2>Unit 4

Lesson 3: Dividing polynomials by linear factors

# Factoring using polynomial division

If we know one linear factor of a higher degree polynomial, we can use polynomial division to find other factors of the polynomial. For example, we can use the fact that (x+2) is a factor of (4x³+19x²+19x-6) in order to completely factor the polynomial.

## 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.
• At , 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.
• Well, you have two real choices. You can factor by grouping:

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:

With any luck, you'll be able to factor by grouping.
• what i there is a remainder?
• 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.
• At he says linear factors, but doesn't explain what that exactly means....
• "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.
• How do you get the 12 from the equation?
• 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.
• 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.
• At about , 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?
• Why does Sal multiply 4x^2+11x-3 with x+2 again? I don't understand why...
• 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.