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## Algebra 2

### Course: Algebra 2>Unit 4

Lesson 4: Polynomial Remainder Theorem

# Proof of the Polynomial Remainder Theorem

The PRT (Polynomial Remainder Theorem) may seem crazy to prove, but Sal shows how you can do it in less than six minutes!

## Want to join the conversation?

• How does the polynomial remainder theorem behaves when you have to deal with polynomials of more than one variable? •   This theorem has not been extended to divisions involving more than one variable. A more general theorem is:
If f(x) is divided by ax + b (where a & b are constants and a is non-zero), the remainder is f(-b/a).
Proof:
Let Q(x)be the quotient and R the remainder.
f(x) = Q(x)*(ax+b) + R
Substituting the solution of 0 = ax + b we have
f(-b/a) = Q(x)*(a*(-b/a)+b) + R
f(-b/a) = Q(x)*(-b + b) + R
f(-b/a) = Q(x)*0 + R
f(-b/a) = R
• At , why isn't the remainder placed over the divisor as it is in other videos? Is there a difference between the two answers or the way they're are written? Such as having added (6/(3x-1)) or having added 6. •  It's because if you move (x - 1) from the right side of the formula to the left by dividing, you also have to divide 6 by that monomial. The equation then changes to:
(3x^2 - 4x + 7)/(x - 1) = (3x - 1) + 6/(x - 1)
• Can this formula be used when the divisor is a quadratic equation? • we use x = a in the theorem .............but that suugests we have to divide the polynomial by 0....
as x - a = 0 then • Does the theorem only work when the remainder is a constant? What if the remainder is a polynomial having several x terms? • Would the same theorem apply to f(x)/(x+a) , with the remainder would be equal to f(-a)? • I just thought of something else.

In Wikipedia it says that the polynomial remainder theorem states that the remainder of the division of a polynomial f(x) by a linear polynomial x - r...

That's my question. Does the polynomial remainder theorem only work when the divisor is a linear polynomial? Is it possible for it to be a quadratic or something else? Or the dividend has to change (for example be of higher degree) for the theorem to work for quadratic polynomials (or other higher-degree polynomials).

Thanks again in advance for any answers and for the work of finding/thinking of an answer ^^ • I remember when I first started learning remainder theorem, these were the initial questions that used to wander in my mind. So here's the answer:
The reminder theorem is only true when the divisor is a linear polynomial. That means it cannot be utilized when the divisor is something else and if the degree of the divisor polynomial is more than 1 , the sole way to find the remainder is polynomial long division. However if you are able to reduce the divisor polynomial to linear polynomial. Then you can use the remainder theorem.
• Hello everyone!
I have already asked many questions here but I really want to know the reason for things. I appreciate every answer I get from you guys, thanks ^^

My question, which I can already guess must be pretty obvious, is why in the x - r the r has to be a negative r? I don't understand why. Probably it would be understandable with something that was taught in some lesson I haven't seen.

If you can answer me this question, I'd really appreciate it. Or if you cant tell me on what lesson - if there is one - he explains the reason for this.

Thanks again, and everyone have a great weekend ^^ • It does not HAVE to be negative, it could be either one. There is a good reason that we often choose the negative. If we have x-a=0, we end up with x=a. If it were positive, we would end up with x=-a which is not as neat. We see this all the time such as in vertex form of quadratic y=a(x-h)^2+k or in general with functions such as given a function f(x), g(x)=f(x-2) shifts the function two units to the right (positive direction) and g(x)=f)(x+2) shifts it to the left.
• For the long division, when you got 6 as a remainder, isn't it supposed to be 6/x-1 for the remainder? • I did not understand what the proof was. Will someone please explain?
(1 vote) • f(x)=q(x)*(x-a)+r.
f is a polynomial,q is the quotient when f is divided by (x-a), (x-a) is what it is being divided by, and r is the remainder.
We plug in a as x:
f(a)=q(a)*(a-a)+r
(a-a) is 0, so that leads us to:
f(a)=q(a)*0+r
Anything multiplied by 0 is 0, so that leads us to:
f(a)=0+r
Anything added to 0 is that thing, and the home stretch!
f(a)=r