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Intermediate Algebra / LEVEL 2 · DIFFICULTY 2/5

Polynomial Structure

Use roots, remainders and factor conditions.

3 stages · 24 practice problems · two 6-question assessment forms

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  1. MINI QUEST IA 2.1Roots and CoefficientsRead the lesson
  2. MINI QUEST IA 2.2Polynomial RemaindersRead the lesson
  3. MINI QUEST IA 2.3Factors and Unknown CoefficientsRead the lesson
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STAGE IA 2.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Roots and Coefficients

Useful preparation: Discriminants

Goal: Understand and apply roots and coefficients.

Before you begin: Discriminants

Understand the idea

Expanding a product of three linear factors groups terms by how many roots are selected. This connects polynomial coefficients to sums, pairwise products and the triple product of the roots.

(x−r)(x−s)(x−t)=x³−(r+s+t)x²+(rs+rt+st)x−rst

Choose and carry out a method

For a monic cubic, the x coefficient is the sum of all three pairwise root products. List the pairs systematically to avoid omissions.

Check the reasoning

Check the alternating signs: the x² coefficient is the negative root sum, while the x coefficient is the positive pairwise sum.

WORKED EXAMPLE 1

A monic cubic has roots 2, 4, and 6. What is the coefficient of x in the polynomial?

  1. In a monic cubic, the x coefficient is the sum of pairwise root products.
  2. 2·4+2·6+4·6=8+12+24.
  3. The coefficient is 44; the constant is the negative of the three-root product.

44

WORKED EXAMPLE 2

A monic cubic has roots 3, 5, and 7. What is the coefficient of x in the polynomial?

  1. In a monic cubic, the x coefficient is the sum of pairwise root products.
  2. 3·5+3·7+5·7=15+21+35.
  3. The coefficient is 71; the constant is the negative of the three-root product.

71

Common pitfalls

Possible mix-up: The x coefficient is the product of all roots.

It is the sum of pairwise products; the triple product determines the constant.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Explain where each pairwise product appears during expansion.

Preview the eight practice prompts
  1. A monic cubic has roots 5, 7, and 9. What is the coefficient of x in the polynomial?
  2. A monic cubic has roots 6, 8, and 10. What is the coefficient of x in the polynomial?
  3. A monic cubic has roots 7, 9, and 11. What is the coefficient of x in the polynomial?
  4. A monic cubic has roots 8, 10, and 12. What is the coefficient of x in the polynomial?
  5. A monic cubic has roots 9, 11, and 13. What is the coefficient of x in the polynomial?
  6. A monic cubic has roots 10, 12, and 14. What is the coefficient of x in the polynomial?
  7. A calibration polynomial is monic and vanishes at 11, 13, and 15. It has degree three. What is the coefficient of its linear term? New context
  8. A calibration polynomial is monic and vanishes at 12, 14, and 16. It has degree three. What is the coefficient of its linear term? New context
Open stage IA 2.1 in the student workspace →

STAGE IA 2.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Polynomial Remainders

Useful preparation: Roots and Coefficients

Goal: Understand and apply polynomial remainders.

Before you begin: Roots and Coefficients

Understand the idea

Dividing P(x) by x−r leaves a constant remainder. Substituting x=r eliminates the divisor multiple, so the remainder is P(r).

P(x)=(x−r)Q(x)+P(r)

Choose and carry out a method

Evaluate the polynomial at r, respecting powers and signs. Long division is optional when only the remainder is requested.

Check the reasoning

The remainder has degree below the divisor’s degree. A linear divisor therefore leaves a number, not a polynomial of positive degree.

WORKED EXAMPLE 1

Find the remainder when x³-2x+4 is divided by x-3.

  1. The remainder theorem says division by x-r leaves remainder P(r).
  2. P(3)=3³-2·3+4.
  3. The remainder is 25; its degree is smaller than the divisor’s degree.

25

WORKED EXAMPLE 2

Find the remainder when x³-3x+5 is divided by x-4.

  1. The remainder theorem says division by x-r leaves remainder P(r).
  2. P(4)=4³-3·4+5.
  3. The remainder is 57; its degree is smaller than the divisor’s degree.

57

Common pitfalls

Possible mix-up: Evaluate at −r for divisor x−r.

The divisor becomes zero at x=r.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why does substitution identify the remainder even without finding the quotient?

Preview the eight practice prompts
  1. Find the remainder when x³-5x+7 is divided by x-6.
  2. Find the remainder when x³-6x+8 is divided by x-2.
  3. Find the remainder when x³-7x+9 is divided by x-3.
  4. Find the remainder when x³-8x+10 is divided by x-4.
  5. Find the remainder when x³-9x+11 is divided by x-5.
  6. Find the remainder when x³-10x+12 is divided by x-6.
  7. A cubic rule P(x)=x³-11x+13 is to be split into a multiple of (x-2) plus a constant adjustment. What adjustment is needed? New context
  8. A cubic rule P(x)=x³-12x+14 is to be split into a multiple of (x-3) plus a constant adjustment. What adjustment is needed? New context
Open stage IA 2.2 in the student workspace →

STAGE IA 2.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Factors and Unknown Coefficients

Useful preparation: Polynomial Remainders

Goal: Understand and apply factors and unknown coefficients.

Before you begin: Polynomial Remainders

Understand the idea

A polynomial has factor x−r exactly when its remainder on division by x−r is zero. This turns a divisibility condition into an equation for an unknown coefficient.

(x−r) divides P(x) ⇔ P(r)=0

Choose and carry out a method

Substitute r into the polynomial, set the value equal to zero and solve the resulting coefficient equation.

Check the reasoning

Substitute the recovered coefficient back into P(r). A root condition proves the factor but does not identify every other root.

WORKED EXAMPLE 1

If x-3 is a factor of x³+kx+21, find k.

  1. A factor x-r requires P(r)=0.
  2. 3³+k·3+21=0. Divide by nonzero 3.
  3. k=-(3²+7)=-16.

-16

WORKED EXAMPLE 2

If x-4 is a factor of x³+kx+32, find k.

  1. A factor x-r requires P(r)=0.
  2. 4³+k·4+32=0. Divide by nonzero 4.
  3. k=-(4²+8)=-24.

-24

Common pitfalls

Possible mix-up: Set the coefficient of x equal to zero.

Set the entire polynomial value at the proposed root equal to zero.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

How would two specified roots create two conditions on unknown coefficients?

Preview the eight practice prompts
  1. If x-6 is a factor of x³+kx+60, find k.
  2. If x-7 is a factor of x³+kx+77, find k.
  3. If x-8 is a factor of x³+kx+96, find k.
  4. If x-9 is a factor of x³+kx+117, find k.
  5. If x-10 is a factor of x³+kx+140, find k.
  6. If x-11 is a factor of x³+kx+165, find k.
  7. A design polynomial P(x)=x³+kx+192 must have zero output at x=12. What value of k enforces that condition? New context
  8. A design polynomial P(x)=x³+kx+221 must have zero output at x=13. What value of k enforces that condition? New context
Open stage IA 2.3 in the student workspace →
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