How do you divide a polynomial by (x+1)

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In summary, the conversation discusses dividing a polynomial p(x) by (x+1) and determining the value of b in the expression. The first person is revising mathematics and has forgotten how to divide polynomials. They ask if dividing (x+1) by p(x) would result in a remainder of a+b=1, and if this means that a=1-b. The second person points out that b should also be divided by (x+1). Additionally, they question how the remainder can always be 1 as x varies.
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laura_a
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Homework Statement


Let p(x)= (x+1)(x-3)Q(x) + a(x+1)+b

Where q(x) is a polynomial and a and b are real numbers

When p(x) is divided by (x+1) the remainder is 1, what is the value of b


I'm revising year 12 mathematics and I've forgotten how to divide polynomials
Is it

If I divided (x+1) by p(x) is it just = (x-3)q(x) + a + b so does that mean a+b is the remainder so a+b=1 therefore a = 1-b ?

Is that even close? :)
 
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  • #2
You seem to have divided OK for the first two terms, but doesn't b get divided by x + 1 as well?

As x varies then how can the remainder *always* be 1?
 

Related to How do you divide a polynomial by (x+1)

1. What is the general method for dividing a polynomial by (x+1)?

The general method for dividing a polynomial by (x+1) is called synthetic division. This method involves using the coefficients of the polynomial and the root, in this case (x+1), to simplify the division process.

2. Can you provide an example of dividing a polynomial by (x+1)?

Yes, for example, if we want to divide the polynomial 3x^3 + 5x^2 + 2x + 1 by (x+1), we would write out the coefficients as 3, 5, 2, and 1. Then, we would use synthetic division to simplify the process and get the quotient of 3x^2 + 2x + 1 with a remainder of 0.

3. Why do we use synthetic division to divide polynomials by (x+1)?

Synthetic division is a more efficient and simplified method for dividing polynomials by (x+1). It involves less steps and is easier to follow compared to traditional long division methods.

4. Are there any limitations to using synthetic division for dividing polynomials by (x+1)?

Yes, synthetic division can only be used when dividing by a linear binomial, such as (x+1). It cannot be used for dividing by polynomials with more complex factors such as quadratic or cubic expressions.

5. How is dividing a polynomial by (x+1) related to finding the roots of the polynomial?

When dividing a polynomial by (x+1), the remainder will always be 0 if (x+1) is a root of the polynomial. This means that finding the roots of a polynomial can be simplified by dividing it by (x+1) and solving for the quotient. Conversely, if (x+1) is not a root of the polynomial, then there will be a non-zero remainder and (x+1) is not a root.

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