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 Step 2 through Step 4 treating the remainder as the dividend. Stop when the degree of the remainder is less than the degree of the divisor.
In our case both the dividend and the divisor are written in standard form and the dividend has all terms present. We can go straight to dividing!
l r d + 1 & |l d^2-d+1
â–Ľ
Divide
d^2/d= d
r d r d+1 & |l d^2-d+1
Multiply term by divisor
r d rl d+1 & |l d^2-d+1 & d^2+d
Subtract down
r d r d+1 & |l - 2d+1
â–Ľ
Divide
- 2d/d= - 2
r d-2 r d+1 & |l - 2d+1
Multiply term by divisor
rd - 2 rl d+1 & |l - 2d+1 & - 2d-2
Subtract down
r d-2 r d+1 & |l 3
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 d-2 with a remainder of 3.
( d^2-d+1) Ă· (d+1) = d-2 + 3/d+1
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.
We are asked to divide a polynomial by a binomial.
(d^3-d^2+d-1)Ă·(d+1)
As in Part A, we will use polynomial long division to find the quotient.
l r d + 1 & |l d^3-d^2+d-1
â–Ľ
Divide
d^3/d= d^2
r d^2 r d+1 & |l d^3-d^2+d-1
Multiply term by divisor
r d^2 rl d+1 & |l d^3-d^2+d-1 & d^3+d^2
Subtract down
r d^2 r d+1 & |l - 2d^2+d-1
â–Ľ
Divide
- 2d^2/d= - 2d
r d^2-2d r d+1 & |l - 2d^2+d-1
Multiply term by divisor
rd^2 - 2d rl d+1 & |l - 2d^2+d-1 & - 2d^2-2d
Subtract down
r d^2-2d r d+1 & |l 3d-1
â–Ľ
Divide
3d/d= 3
r d^2-2d+3 r d+1 & |l 3d-1
Multiply term by divisor
rd^2-2d+ 3 rl d+1 & |l 3d-1 & 3d+3
Subtract down
r d^2-2d+3 r d+1 & |l - 4
The quotient is d^2-2d+3 with a remainder of - 4.
(d^3-d^2+d-&1)Ă·(d+1)
&=d^2-2d+3+- 4/d+1
Note that we can rewrite our result by putting the minus sign in front of the fraction.
d^2-2d+3+- 4/d+1
⇕
d^2-2d+3-4/d+1
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.
This time we are asked to divide a polynomial with degree 4 by the same binomial as in Parts A and B.
(d^4-d^3+d^2-d+1)Ă·(d+1)Again, we will use polynomial long division.
The quotient is d^3-2d^2+3d-4 with a remainder of 5.
(d^4-d^3&+d^2-d+1)Ă·(d+1)
&=d^3-2d^2+3d-4+5/d+1
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.
We are asked to predict the result of the division based on our results from Part A, Part B, and Part C.
( d^5-d^4+d^3-d^2+d-1) Ă· (d+1)
The divisor is the same as in previous parts but the dividend is different. Let's combine our previous results in a table and look for a pattern.
Dividend
Quotient
Remainder
d^2-d+1
d-2
3
d^3-d^2+d-1
d^2-2d+3
- 4
d^4-d^3+d^2-d+1
d^3-2d^2+3d-4
5
d^5-d^4+d^3-d^2+d-1
?
?
We can see that all of the dividends follow the same pattern.
Each dividend has all terms present.
The term with the highest degree has a coefficient of 1, the next term has a coefficient of - 1, the next term after that has a coefficient of 1, and so on.
Now we can observe how the pattern in the dividends is reflected in the quotients and remainders.
Each quotient has all terms present.
The degree of the quotient is one less than the degree of the dividend.
The term with the highest degree has a positive coefficient, the next term has a negative coefficient, and so on. In other words, the signs of the coefficients are alternating between positive and negative.
The sign of the remainder is opposite to the sign of the coefficient of the term with the lowest degree, the constant.
Finally, we will also look at the absolute values of the coefficients of each quotient and remainder.
Coefficients of Quotient
Remainder
Absolute Values
1, - 2
3
1, 2, 3
1, - 2, 3
- 4
1, 2, 3, 4
1, - 2, 3, - 4
5
1, 2, 3, 4, 5
We can see that the absolute values of the coefficients of the quotient and the remainder are consecutive integers, starting from 1. Finally, we will use our knowledge about the pattern to predict the result of our division.
( d^5-d^4+d^3-d^2+d-1) Ă· (d+1)
Let's list what we know about the quotient and the remainder.
Since the degree of the dividend is 5, the degree of the quotient will be 4.
The quotient will have all terms present. Therefore, it will have five terms.
The absolute values of the coefficients of the quotient and the remainder are consecutive integers, starting from one.
The signs of the coefficients and the remainder are alternating between positive and negative.
Knowing all the above patterns and conclusions, let's write our predictions of the quotient and the remainder.
Quotient: & d^4-2d^3+3d^2-4d+5
Remainder: & - 6
We will verify our predictions in Part E.
We will verify our prediction from Part D.
(&d^5-d^4+d^3-d^2+d-1)Ă·(d+1)
&? =d^4-2d^3+3d^2-4d+5+- 6/d+1
Let's divide the polynomials!
We have that the quotient is d^4-2d^3+3d^2-4d+5 with a remainder of - 6. This result confirms our predictions from Part D.
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.
