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## Algebra 2

### Course: Algebra 2>Unit 4

Lesson 3: Dividing polynomials by linear factors

# Factoring using polynomial division: missing term

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+6) is a factor of (x³+9x²-108) in order to completely factor the polynomial. We just need to be careful because the polynomial has no x-term.

## Want to join the conversation?

• Would this polynomial have been completely factored if Sal had written the expression as (x+6)^2 * (x-3)?
• It's the same, but one of them is shorter
• at , couldn't you write (x+6)(x+6)(x-3) as (x+6)^2(x-3) to simplify it, or would that turn the polynomial into a quadratic?
• what wpould happen in this case if there is a remainder?
• If there was a nonzero remainder, it would mean the polynomial is not divisible by x+6. So you would have to find a different factor.
• Can you write is as (x+6)^2(x-3)?
• The expressions are equal, and I don't think anybody would say that that isn't a product of linear factors, so you can.
(1 vote)
• could it also be = to (X-3)*(X+6)^2
• Yes that is the same thing as what Sal wrote
• What does P(X) or F(X) mean?
• Hi. Both represent a function, in your case, "P" and "F" respectively. "P(x)" can be read as "P in function of x" that is, x is the input and P(x) the output. Hope this helps.
(1 vote)
• what If the question is like, "x to the six power, 15x to the third power, 45x and 200"? ps this is a equation that I made up and may not have a solution.
(1 vote)
• You should say if they are being added or subtracted. x^6+15x^3+45x+200 would be quite different a graph than x^6+15x^3-45x+200.

Also, all polynomials have solutions, they just may be non real. Although, if the degree of the polynomial is odd that guarantees at least one real answer.

For yours they are indeed all nonreal
(1 vote)
• How would you multiply this out to check your answer?
(1 vote)
• We use the distributive law a bunch of times, and collect like terms when we can.

(𝑥 + 6)(𝑥 + 6)(𝑥 − 3)

= (𝑥(𝑥 + 6) + 6(𝑥 + 6))(𝑥 − 3)

= (𝑥² + 6𝑥 + 6𝑥 + 36)(𝑥 − 3)

= (𝑥² + 12𝑥 + 36)(𝑥 − 3)

= 𝑥²(𝑥 − 3) + 12𝑥(𝑥 − 3) + 36(𝑥 − 3)

= 𝑥³ − 3𝑥² + 12𝑥² − 36𝑥 + 36𝑥 − 108

= 𝑥³ + 9𝑥² − 108
(1 vote)
• Could he have done (x+6)^2 instead of (x+6)(x+6)?
(1 vote)
• Yes, both equations give the same result of X^2+12x+36. So it doesn't really matter what equation he uses.
(1 vote)
• At the end of the video(), if we wrote (x+6)^2 * (x-3) instead of (x+6)(x+6)(x-3), would it still be linear factors?