Thread: Remainder theorem

1. Remainder theorem

Hi,

I hope someone can help. I'm trying to solve for 8b.

Is it wrong to assume that I can just sub in the zero of either factors, (x+3) or (x+5), into x^2+9x-7, in order to get the value of r? I'm using my knowledge of the remainder theorem. I thought that by plugging in either -3 or -5 for x on the right-hand side, that I would be able to find the value of the remainder, but this method is wrong apparently. Can someone please explain to me why? My method worked for question 8a and 8c, so I find it rather odd why it doesn't work for 8a. I would really appreciate it if you explained in the context of the remainder theorem.

The remainder according to the textbook is x-22.

Please help!!

2. Re: Remainder theorem

If the remainder were always a unit, then you could do that. However, the remainder is not always a unit. The method you are using will ONLY work when the degree of $r$ (as a function of $x$) is zero.

For b), multiply out and solve for $r$. See what you get.

3. Re: Remainder theorem

I do get the correct answer when I multiply out.

What were you saying that the method only works when the degree of r is zero? How do you know whether the degree is zero?

4. Re: Remainder theorem

Originally Posted by otownsend
I do get the correct answer when I multiply out.

What were you saying that the method only works when the degree of r is zero? How do you know whether the degree is zero?
You calculate the remainder by multiplying out and see if it has a degree of zero.

5. Re: Remainder theorem

Accidental double post.

6. Re: Remainder theorem

The problem that SlipEternal is pointing out is that as the problem is stated, there is no numeric value of r that will make that true. You appear to be assuming that is a number, not a function of x.

There is no number "r" such $\displaystyle (x+ 3)(x+ 5)+ r= x^2+ 9x- 7$ because $\displaystyle (x+ 3)(x+ 5)= x^2+ 8x+ 15$. In order that this be equal to $\displaystyle x^2+ 9x- 7$, rather than $\displaystyle x^2+ 8x- 7$ we need r to be [tex](x^2+ 9x- 7)- (x^2+ 8x+ 15)= x- 22. That is what r must be.

7. Re: Remainder theorem

Thank you for responding.

Can you please read the below paragraph to further understand my confusion?

"The theorem says that when you divide a polynomial f(x) by another one g(x), there is a quotient q(x) and a remainder r(x). These are related by the equation: f(x) = g(x)q(x) + r(x)

The remainder r(x) either is zero, or has a degree lower than g(x).

For example, if g(x) is a linear polynomial, then r(x) should be a constant, and if g(x) is quadratic, then r(x) should be a linear polynomial of the form Ax+B."

Since g(x) in this case is of a degree of 1, I was assuming that the reminder then must have a degree of 0 or in other words, a constant value. Why is 8b an exception to the rule explained in the paragraph above?

8. Re: Remainder theorem

Originally Posted by otownsend
Thank you for responding.

Can you please read the below paragraph to further understand my confusion?

"The theorem says that when you divide a polynomial f(x) by another one g(x), there is a quotient q(x) and a remainder r(x). These are related by the equation: f(x) = g(x)q(x) + r(x)

The remainder r(x) either is zero, or has a degree lower than g(x).

For example, if g(x) is a linear polynomial, then r(x) should be a constant, and if g(x) is quadratic, then r(x) should be a linear polynomial of the form Ax+B."

Since g(x) in this case is of a degree of 1, I was assuming that the reminder then must have a degree of 0 or in other words, a constant value. Why is 8b an exception to the rule explained in the paragraph above?
This problem simply has nothing to do with that theorem. There is no division involved here. What "g(x)" are you talking about?

9. Re: Remainder theorem

The following might help

Let $\displaystyle f(x)=x^2+9x-7$

When we divide f(x) by (x+3) the quotient is (x+6) and the remainder is (-25)

When we divide f(x) by (x+5) the quotient is (x+4) and the remainder is (-27)

When we divide f(x) by (x+3)(x+5) the quotient is (1) and the remainder is (ax+b)

10. Re: Remainder theorem

Ok thanks for helping. I'll think about this some more and get back if I have any more questions.