Comprehensive Study Guide: Polynomials, Divisibility, and Root Analysis

Polynomial Long Division and Remainder Theorem

• Polynomial Long Division Process:

  • Long division continues iteratively until the maximum power (degree) of the remaining inner terms (remainder) is strictly smaller than the maximum power (degree) of the outer terms (divisor).

  • Example problem: Find the remainder when $p(x) = 2x^4 - 6x^2 + 7x - 5$ is divided by $x - 3$

  • Step-by-step long division breakdown:

    • First term division: Divide highest power term $2x^4$ by $x$ to obtain quotient term $2x^3$. Multiply $2x^3(x - 3) = 2x^4 - 6x^3$.

    • Subtract from dividend $2x^4 + 0x^3 - 6x^2 + 7x - 5$ to get intermediate term $6x^3 - 6x^2 + 7x - 5$

    • Second term division: Divide $6x^3$ by $x$ to obtain quotient term $6x^2$. Multiply $6x^2(x - 3) = 6x^3 - 18x^2$.

    • Subtract to obtain intermediate term $12x^2 + 7x - 5$

    • Third term division: Divide $12x^2$ by $x$ to obtain quotient term $12x$. Multiply $12x(x - 3) = 12x^2 - 36x$.

    • Subtract to obtain intermediate term $43x - 5$

    • Fourth term division: Divide $43x$ by $x$ to obtain quotient term $43$. Multiply $43(x - 3) = 43x - 129$.

    • Subtract to obtain final numerical remainder $113$

• The Remainder Theorem:

  • Definition: If a polynomial $p(x)$ is divided by a linear expression $(x - a)$, then the remainder $R$ resulting from the division is equal to $p(a)$.

  • Application to Example:

    • To evaluate the remainder of $p(x) = 2x^4 - 6x^2 + 7x - 5$ divided by $x - 3$ using the theorem:

    • Set divisor equal to zero: x - 3 = 0 \n\nimplies x = 3

    • Substitute $x = 3$ into $p(x)$:             $p(3) = 2(3)^4 - 6(3)^2 + 7(3) - 5$             $p(3) = 2(81) - 6(9) + 21 - 5$             $p(3) = 162 - 54 + 21 - 5$             $p(3) = 113$

    • The calculated value $113$ represents the remainder $R$.

The Factor Theorem

• Definition and Core Principle:

  • If $(x - a)$ is a factor of a polynomial $p(x)$, then the remainder $p(a) = 0$

  • Conversely, if $p(a) = 0$, then $(x - a)$ is an exact linear factor of $p(x)$.

  • Numerical analogy: $4$ is a factor of $12$ because 12 \n\ndiv 4 = 3 with a remainder of $0$

• Solving for Unknown Polynomial Parameters:

  • Problem: Determine the unknown constant $a$ if $(x - 2)$ is a factor of $p(x) = 2x^3 - ax - 3$

  • Step 1: Apply the Factor Theorem by setting the factor equal to zero: x - 2 = 0 \n\nimplies x = 2

  • Step 2: Since $(x - 2)$ is a factor, set the remainder equal to zero: $p(2) = 0$

  • Step 3: Substitute $x = 2$ into $p(x)$:         $p(2) = 2(2)^3 - a(2) + 3 = 0$         $2(8) - 2a + 3 = 0$         $16 - 2a + 3 = 0$         $19 - 2a = 0$         $2a = 19$         $a = 9.5$

Divisor Types and Factorization Strategies

• Classification of Divisors:

  • Linear Divisors: Polynomial expressions where the maximum power of $x$ is $1$ (e.g., $x - 2$ or $3x - 1$). Remainder Theorem and Factor Theorem apply directly without requiring division steps.

  • Non-Linear Divisors: Polynomial expressions where the maximum power of $x$ is greater than or equal to $2$ (e.g., quadratic expressions $x^2 - 4$). These require polynomial long division or algebraic structural matching.

• Degree Reduction Rule for Factorization:

  • When factoring a polynomial of maximum degree $N$ using a known factor of degree $M$, the remaining factor will have a maximum degree equal to $N - M$:

    • If polynomial degree $N = 3$ and known factor degree $M = 1$, the remaining factor is quadratic: $(ax^2 + bx + c)$.

    • If polynomial degree $N = 2$ and known factor degree $M = 1$, the remaining factor is linear: $(ax + b)$.

    • If polynomial degree $N = 4$ and known factor degree $M = 1$, the remaining factor is cubic: $(ax^3 + bx^2 + cx + d)$.

Complete Factorization of Cubic Polynomials

• Complete Factorization Procedure:

  • Problem: Factorize $p(x) = 8x^3 - 6x^2 - 3x + 1$ completely, given that $(x + 1)$ is a known factor.

  • Step 1: Set up the polynomial as a product of the known linear factor and an unknown quadratic factor:         $p(x) = (x + 1)(ax^2 + bx + c)$

  • Step 2: Determine the highest power coefficient $a$ by matching the $x^3$ terms:         x \n\ntimes ax^2 = 8x^3 \n\nimplies a = 8

  • Step 3: Determine the constant term $c$ by matching the constant terms:         1 \n\ntimes c = 1 \n\nimplies c = 1

  • Step 4: Substitute $a = 8$ and $c = 1$ into the expression:         $p(x) = (x + 1)(8x^2 + bx + 1)$

  • Step 5: Determine middle coefficient $b$ by equating either the $x^2$ coefficients or $x$ coefficients:

    • Expanding $(x + 1)(8x^2 + bx + 1) = 8x^3 + (b + 8)x^2 + (b + 1)x + 1$

    • Equating $x^2$ terms: b + 8 = -6 \n\nimplies b = -14

    • Quadratic expression becomes: $(8x^2 - 6x + 1)$ or $(8x^2 - 14x + 1)$

  • Step 6: Factorize the quadratic term into linear factors:         $(8x^2 - 6x + 1) = (4x - 1)(2x - 1)$

  • Final factored form:         $p(x) = (x + 1)(4x - 1)(2x - 1)$

• Second Factorization Example:

  • Polynomial: $p(x) = 2x^3 - 3x^2 - 5x + 2$ with known factor $(2x + 1)$

  • Set up structural form:         $2x^3 - 3x^2 - 5x + 2 = (2x + 1)(ax^2 + bx + c)$

  • Match outer terms:

    • 2x \n\ntimes ax^2 = 2x^3 \n\nimplies a = 1

    • 1 \n\ntimes c = 2 \n\nimplies c = 2

  • Match middle term $b$:

    • $(2x + 1)(x^2 + bx + 2) = 2x^3 + (2b + 1)x^2 + (b + 4)x + 2$

    • Equating $x^2$ coefficient: 2b + 1 = -3 \n\nimplies 2b = -4 \n\nimplies b = -2

  • Resulting factored polynomial form:         $p(x) = (2x + 1)(x^2 - 2x + 2)$

Solving Simultaneous Linear Equations in Polynomials

• System Formulation:

  • Given polynomial $p(x) = ax^3 + bx^2 + 5x + 11$

  • Condition 1: Divided by $(x - 2)$, the remainder is $12$

    • Apply Remainder Theorem: $p(2) = 12$

    • $a(2)^3 + b(2)^2 + 5(2) + 11 = 12$

    • $8a + 4b + 10 + 11 = 12$

    • $8a + 4b + 21 = 12$

    • $8a + 4b = -9$

  • Condition 2: Given a second linear factor equation yielding:         $a + 2b = -4$

  • Step-by-step solution of simultaneous equations:

    • Equation 1: $8a + 4b = -9$

    • Equation 2: a + 2b = -4 \n\nimplies a = -4 - 2b

    • Substitute Equation 2 into Equation 1:             $8(-4 - 2b) + 4b = -9$             $-32 - 16b + 4b = -9$             $-32 - 12b = -9$             -12b = 23 \n\nimplies b = -\n\nfrac{23}{12}

    • Alternatively, using whole integer coefficients derived from direct evaluation:             $a = 2$             $b = -3$

Polynomials and Complex/Non-Real Roots Analysis

• Root Determination from Factorized Form:

  • Given polynomial equation: $(x + 1)(x^2 + bx + 3) = 0$

  • Setting each factor to zero yields potential roots:

    • Linear factor root: x + 1 = 0 \n\nimplies x = -1

    • Quadratic factor equation: $x^2 + bx + 3 = 0$

• Discriminant Analysis for Real vs Non-Real Roots:

  • For the quadratic equation $ax^2 + bx + c = 0$, the discriminant is defined as \n\nDelta = b^2 - 4ac

  • Applying coefficients $a = 1$ and $c = 3$:         \n\nDelta = b^2 - 4(1)(3) = b^2 - 12

  • Condition for No Real Roots (Complex Roots):

    • If \n\nDelta < 0, the quadratic factor has no real roots.

    • b^2 - 12 < 0 \n\nimplies b^2 < 12

  • Demonstration with $b = 2$:         \n\nDelta = 2^2 - 4(1)(3) = 4 - 12 = -8

  • Since $-8 < 0$, the factor $x^2 + 2x + 3 = 0$ yields No Real Roots, demonstrating that the polynomial has complex conjugate roots.

Worksheet Assignments

• Worksheet Details:

  • Worksheet Reference: Worksheet A 12 Polynomials

  • Required Problems: Complete all questions in Worksheet A 12 except Questions 8, 15, and 16.