به انگلیسی

Polynomial Division and the Factor Theorem

Mathematics · Algebra · ages 16-17

Name ______________________ Date ____________

A cubic f(x) has factor x + 3. Which equation must hold?
- f(3) = 0
- f(-3) = 0
- f(0) = 3
- f(-3) = 3
Begin dividing x³ + 5x² - 2x + 8 by x + 1. Which term starts the quotient?
- x³
- x
- x²
- 1
Before long division, write x³ - 6x + 5 with all powers in order. What coefficient belongs to its missing x² term?
Answer: ______________
Let f(x) = x³ - 4x² + x + 6. A calculation gives f(2) = 0. Select a factor of f(x).
- (x - 2)
- (x + 2)
- (x - 6)
- (2x - 1)
The remainder from dividing a polynomial by x - 1 must be zero or positive.
Circle one: True False
The cubic x³ - 3x² + 4 has factor x - 2. How many different real roots does it have?
Answer: ______________
What quotient results from dividing x³ + x² - 10x + 8 by x - 2?
- x² - 3x - 4
- x² + 3x + 4
- x² + 3x - 4
- x² - 10x + 8
Division of f(x) by x - 5 leaves remainder 12. Choose the value of f(5).
- 0
- 12
- 5
- -12
Test the proposed factors of f(x) = x³ - 2x² - 11x + 12. Which one divides f(x) exactly?
- (x - 3)
- (x + 1)
- (x + 4)
- (x + 3)
Sam obtains x³ + 2x² - 9x - 18 = (x + 2)(x² - 9). Why is this not yet fully factorised into linear factors?
- x² - 9 splits into (x - 3)(x + 3).
- The factor x + 2 must be changed to x - 2.
- A cubic cannot be written using three linear factors.
- The calculation must begin again with divisor x + 3.

Answer key

For grown-ups. Fold this page away before handing over the rest.

f(-3) = 0 · At x = -3, the factor x + 3 equals zero. Therefore the polynomial also equals zero: f(-3) = 0.
x² · Choose the term that multiplies the divisor's leading x to produce x³. Since x² times x is x³, the quotient starts with x².
0 · There is no x² term, so its coefficient is 0. Write x³ + 0x² - 6x + 5 to keep the powers aligned.
(x - 2) · The zero at input 2 confirms x - 2. The other proposed factors correspond to inputs -2, 6 and 1/2, whose values are not zero.
False · Polynomial remainders can be negative. For x³ - 5 divided by x - 1, the remainder is f(1) = 1 - 5 = -4.
2 · Division gives x² - x - 2 = (x - 2)(x + 1). The cubic is (x - 2)²(x + 1), so its different roots are 2 and -1.
x² + 3x - 4 · The successive quotient terms are x², 3x and -4. Their products with x - 2 remove the dividend completely, giving x² + 3x - 4.
12 · In f(x) = (x - 5)Q(x) + 12, substituting 5 removes the product. Only 12 remains, so f(5) = 12.
(x + 3) · The test inputs are 3, -1, -4 and -3. Their polynomial values are -12, 20, -40 and 0 respectively, so x + 3 is the factor.
x² - 9 splits into (x - 3)(x + 3). · The displayed product is valid, but its quadratic factor has the pair -3 and 3. The full linear factorisation is (x + 2)(x - 3)(x + 3).