We are asked to write a binomial and a trinomial using the same variable. Then we want to divide the trinomial by the binomial. Let's start by recalling the definitions of binomial and trinomial.
Knowing this, let's write an arbitrary binomial and an arbitrary trinomial. Let x be our variable.
ccc
Binomial & Trinomial
x+5 & 4x^2-x+1
Now we will divide 4x^2-x+1 by x+5. Let's recall the steps of polynomial long division.
Write the dividend and divisor in standard form. If a term is not present in the dividend, add the term with a coefficient of 0.
Divide the first term of the dividend by the first term of the divisor. The result is the first term of the quotient.
Multiply the first term of the quotient by the divisor and write the product under the dividend.
Subtract the product from the dividend. We will call the result of this operation the remainder.
Repeat Steps 2 through 4 treating the remainder as the dividend. Stop when the degree of the remainder is less than the degree of the divisor.
We will follow these steps when dividing our trinomial by the binomial. Since the terms of each polynomial are in descending degree order, they are already written in standard form.
Polynomial
Standard Form?
4x^2-x+1
Yes âś“
x+5
Yes âś“
Additionally, the dividend has all terms present. Therefore, we can go straight to dividing!
l r x + 5 & |l 4x^2-x+1
â–Ľ
Divide
4x^2/x= 4x
r 4x r x+5 & |l 4x^2-x+1
Multiply term by divisor
r 4x rl x+5 & |l 4x^2-x+1 & 4x^2+20x
Subtract down
r 4x r x+5 & |l - 21x+1
â–Ľ
Divide
- 21x/x= - 21
r 4x-21 r x+5 & |l - 21x+1
Multiply term by divisor
r4x - 21 rl x+5 & |l - 21x+1 & - 21x-105
Subtract down
r 4x-21 r x+5 & |l 106
The degree of the remainder is 0 and the degree of the divisor is 1. Therefore, the degree of the remainder is less than the degree of the divisor. We must stop the division right here. We have that the quotient is 4x-21 with a remainder of 106.
( 4x^2-x+1)Ă·(x+5)= 4x-21+106/x+5
Showing Our Work
Long division by hand...
When we are doing long division by hand, it looks a bit different than how we have it in this solution. Here is how yours should look when you are writing it in your notebook.
Extra
Additional note
In the text, a fifth step is introduced for bringing down the next terms. For numeric long division this makes sense but it is a bit more complicated with polynomial long division. Yes, we bring down the terms but, because there are operations between each term, we can include this as part of the subtraction step.
