Ninth Grade Mathematics: Division Algorithm & Long Division
Just as integer division partitions a dividend into a quotient and an integer remainder, polynomial division decomposes a dividend polynomial into a quotient polynomial and a remainder of strictly lower degree. Mastering polynomial long division provides the mechanical engine required to factor high-degree polynomials and locate their roots.
The Division Algorithm for Polynomials
Let P(x) and D(x) be two polynomials such that D(x) is not the zero polynomial, and the degree of P(x) is greater than or equal to the degree of D(x). Then there exist unique polynomials Q(x) (the quotient) and R(x) (the remainder) such that:
The Division Algorithm Equation:
P(x) = D(x) * Q(x) + R(x)
where:
P(x) is the Dividend
D(x) is the Divisor
Q(x) is the Quotient
R(x) is the Remainder, satisfying either R(x) = 0 or deg(R(x)) < deg(D(x))
The crucial condition is that the degree of the remainder R(x) must be strictly less than the degree of the divisor D(x). If R(x) = 0, we say that D(x) divides P(x) completely, or that D(x) is a factor of P(x).
Fraction Form:
P(x) R(x)
---- = Q(x) + ------
D(x) D(x)
Step-by-Step Long Division Method
Polynomial long division mirrors numerical long division: divide leading terms, multiply the entire divisor, subtract, and bring down the next term.
Algorithm Steps:
1. Arrange both dividend and divisor in standard form (descending exponents).
2. Insert placeholder terms with coefficient 0 for any missing powers of x.
3. Divide the leading term of the dividend by the leading term of the divisor.
4. Multiply this quotient term by the complete divisor.
5. Subtract the product from the current dividend terms (change signs and add).
6. Bring down the next term from the dividend.
7. Repeat until the degree of the remainder is strictly less than that of the divisor.
Fully Worked Example
Divide P(x) = 2x^3 - 7x^2 + 11*x - 14 by D(x) = x - 2:
2*x^2 - 3*x + 5 <--- Quotient Q(x)
+-----------------------------
x - 2 | 2*x^3 - 7*x^2 + 11*x - 14 <--- Dividend P(x)
- (2*x^3 - 4*x^2)
------------------
- 3*x^2 + 11*x
- (-3*x^2 + 6*x)
-----------------
5*x - 14
- (5*x - 10)
------------
- 4 <--- Remainder R(x)
Verification: Multiply divisor by quotient and add remainder: (x - 2) * (2x^2 - 3x + 5) + (-4) = (2x^3 - 3x^2 + 5x - 4x^2 + 6x - 10) - 4 = 2x^3 - 7x^2 + 11x - 14. The result matches the dividend exactly.
Handling Missing Intermediate Terms
If the dividend skips a power of the variable (such as x^3 - 8, which has no x^2 or x terms), you must supply placeholder terms with a zero coefficient: x^3 + 0x^2 + 0x - 8. This guarantees that like terms align vertically during subtraction.
Divide x^3 - 8 by x - 2:
x^2 + 2*x + 4
+-----------------------------
x - 2 | x^3 + 0*x^2 + 0*x - 8
- (x^3 - 2*x^2)
----------------
2*x^2 + 0*x
- (2*x^2 - 4*x)
---------------
4*x - 8
- (4*x - 8)
-----------
0 <--- Remainder is 0! (Exact factor)
Because the remainder is 0, we have factored the difference of cubes: x^3 - 8 = (x - 2)(x^2 + 2x + 4).
Chapter Exercises
Exercise 1: Divide P(x) = 3x^3 + 5x^2 - 11x - 4 by D(x) = x + 3 using polynomial long division. State the quotient Q(x) and remainder R(x), and write the division identity P(x) = D(x)Q(x) + R(x).
Exercise 2: Divide P(x) = 4x^4 - 2x^3 + 3x - 5 by D(x) = 2x^2 - x + 1. Find the quotient and remainder.
Exercise 3: What real number k must be added to P(x) = 2x^3 - 5x^2 + 4*x so that the resulting polynomial is completely divisible by x - 2?
Solutions and Explanations
Solution 1: Divide the leading term 3x^3 by x to get 3x^2. Multiply 3x^2 by (x + 3) to get 3x^3 + 9x^2. Subtract this from the dividend: (3x^3 + 5x^2) - (3x^3 + 9x^2) = -4x^2. Bring down -11x. Divide -4x^2 by x to get -4x. Multiply -4x by (x + 3) to get -4x^2 - 12x. Subtract: (-4x^2 - 11x) - (-4x^2 - 12x) = x. Bring down -4. Divide x by x to get 1. Multiply 1 by (x + 3) to get x + 3. Subtract: (x - 4) - (x + 3) = -7. The quotient is Q(x) = 3x^2 - 4x + 1 and the remainder is R(x) = -7. The division identity is 3x^3 + 5x^2 - 11x - 4 = (x + 3)(3x^2 - 4x + 1) - 7.
Solution 2: Write dividend with placeholder: 4x^4 - 2x^3 + 0x^2 + 3x - 5. Divide leading term 4x^4 by 2x^2 to get 2x^2. Multiply 2x^2 by (2x^2 - x + 1) to get 4x^4 - 2x^3 + 2x^2. Subtract: (4x^4 - 2x^3 + 0x^2) - (4x^4 - 2x^3 + 2x^2) = -2x^2. Bring down + 3x. Divide -2x^2 by 2x^2 to get -1. Multiply -1 by (2x^2 - x + 1) to get -2x^2 + x - 1. Subtract: (-2x^2 + 3x - 5) - (-2x^2 + x - 1) = 2x - 4. Since deg(2x - 4) = 1 is strictly less than deg(divisor) = 2, division terminates. The quotient is 2x^2 - 1 and the remainder is 2*x - 4.
Solution 3: Let the new polynomial be P(x) = 2x^3 - 5x^2 + 4x + k. Dividing by x - 2, we can evaluate P(2) using synthetic evaluation or direct substitution: P(2) = 2(2^3) - 5(2^2) + 4(2) + k = 2(8) - 5(4) + 8 + k = 16 - 20 + 8 + k = 4 + k. For P(x) to be completely divisible by x - 2, the remainder must equal zero: 4 + k = 0, which means k = -4. Therefore, adding -4 to the polynomial ensures exact divisibility.