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## Algebra 2

### Course: Algebra 2>Unit 4

Lesson 4: Polynomial Remainder Theorem

# Remainder theorem: finding coefficients

Sal determines the value of coefficient c in p(x)=x^3+2x^2+cx+10, in order for (x-5) to be a factor of p. The key is using the factor theorem.

## Video transcript

- So for what value of c, or values of c, is x minus five? Let me write it this way. For what c is x minus five a factor of p of x? I encourage you to pause the video and try to work through it. That p is actually a lower case p. P of x. All right, so I'm assuming you have attempted this. We just have to realize, okay, if x minus five is a factor of p of x, that means you could write p of x, let me do this in that green color, that means you could write p of x as being equal to x minus five times some other business, times some other polynomial. So times some other polynomial. I don't know, let's just call it g of x maybe, g of x, and this would be a 2nd degree polynomial. So what would p of five have to be? Well p of five would have to zero. Five would have to be a root of this polynomial. You see it right over here. If you replace this with the five, then this is going to be a five, and this is going to be a five. It doesn't matter what g of five is. This is going to be zero. So if x minus five is a factor, if and only if p of five is equal to zero. You could say that five is a root of the polynomial p. P of five is equal to zero. So let's just set p of five equal to zero and then try to solve for c. All right, so we're going to get five to the 3rd power. So this right over here, five to the 3rd, so p of five. P, let me just write it down, p of five is equal to five to the 3rd is 125 plus two times five squared. So that's two times 25, plus 50, plus c times five, plus five c plus 10. That needs to be equal to zero. Now let's see, if we were to add 125 to 50 to 10, we are going to get 175, 185. So we get 185 plus five c is equal to zero. Subtract 185 from both sides, you get five c is equal to -185, or c is equal to -185 over five, which is going to be, let's see, it's -185 over five, which is equal to, let's see, five goes into 180, let's see, it goes into, well it's five, let's see, it's going to be 30. Five times 30 is 150. Then we'll have another 35 to go, so it's going to be 37, -37. Is that right? Five times 30 is 150, five times seven is 35, yep, -37. And we're done. If this was x to the 3rd plus 2 x squared minus 37 x plus 10, then x minus five would be a factor of this polynomial.