Dividing Polynomials & the Factor Theorem — Challenge
Dividing polynomials is the long division you already know, and a remainder of zero names a factor.
What these problems cover
These questions use 7 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Rational — Any number writable as ab with integers a and b, b ≠ 0.
- Factor — A number that divides evenly into another.
- Roots — The x values where y = 0.
- Divisor — The polynomial you are dividing by.
- Remainder — What is left when the divisor no longer fits. Always lower degree than the divisor.
- Remainder theorem — Dividing f(x) by x - a leaves remainder f(a).
- Factor theorem — If f(a) = 0 then x - a is a factor, and the other way round.
Problem 1
What is (x2+5x+6)÷(x+2)?
- x+3
- x+2
- x+6
- x-3
- x+5
Worked answer
x+3
Factorise the numerator: two numbers multiplying to 6 and adding to 5 are 2 and 3, so x2+5x+6 = (x+2)(x+3).
The factor x+2 divides out, leaving x+3.
Check by expanding: (x+2)(x+3) = x2+5x+6.
Problem 2
What is the remainder when x2-1 is divided by x-1?
- 1
- 0
- 2
- -1
- -2
Worked answer
0
By the remainder theorem, the remainder on dividing by x-1 is the value of the polynomial at x=1.
12-1 = 0.
A remainder of 0 means x-1 is a factor, and indeed x2-1 = (x-1)(x+1).
Problem 3
If f(x)=x3-2x+1, what is f(1)?
- 1
- -1
- 0
- 2
- 4
Worked answer
0
Substitute x=1 into each term: 13 = 1, -2(1) = -2 and the constant is 1.
1-2+1 = 0.
Since f(1) = 0, the factor theorem says x-1 is a factor of f(x).
Problem 4
By the factor theorem, x-a is a factor of f(x) exactly when what holds?
- f(a)=0
- f(0)=a
- f(a)=a
- f(-a)=0
- f(1)=a
Worked answer
f(a)=0
Dividing f(x) by x-a leaves a remainder equal to f(a).
A factor is a divisor leaving no remainder, so x-a is a factor precisely when f(a) = 0.
Note the sign: for the factor x+2 the relevant value is f(-2).
Problem 5
What is the remainder when x3+2x2-x+3 is divided by x-1?
- 3
- 4
- 5
- 6
- 7
Worked answer
5
The remainder theorem gives the remainder as the value at x=1.
1+2-1+3 = 5.
Long division would reach the same answer with far more work.
Problem 6
What is (2x2+7x+3)÷(x+3)?
- 2x+3
- 2x+1
- x+1
- 2x-1
- 2x+7
Worked answer
2x+1
The leading terms must give 2x2, so the answer starts with 2x.
The constants must multiply to 3, and 3· 1 = 3, suggesting (x+3)(2x+1).
Expanding confirms it: 2x2+x+6x+3 = 2x2+7x+3.
Problem 7
What is the remainder when x2+4 is divided by x+2?
- 0
- 4
- 6
- 8
- -8
Worked answer
8
Dividing by x+2 means evaluating at x=-2, since x+2 = x-(-2).
(-2)2+4 = 4+4 = 8.
The remainder is not zero, so x+2 is not a factor of x2+4.
Problem 8
Is x-2 a factor of x3-8?
- Yes
- No
- Only if x=0
- It cannot be decided
- Only for x>2
Worked answer
Yes
Evaluate the polynomial at x=2: 23-8 = 0.
Since the value is zero, the factor theorem says x-2 is a factor.
In fact x3-8 = (x-2)(x2+2x+4).
Problem 9
What is (x3-1)÷(x-1)?
- x2-x+1
- x2+x+1
- x2+1
- x2-1
- x2+x-1
Worked answer
x2+x+1
Since 13-1 = 0, the division leaves no remainder.
Dividing gives x2+x+1, the standard factorisation of a difference of cubes.
Check by expanding: (x-1)(x2+x+1) = x3+x2+x-x2-x-1 = x3-1.