2.1 Polynomials
Detailed Theory: Polynomials
1. Basic Concepts and Definitions
1.1 What is a Polynomial?
A polynomial is an algebraic expression consisting of variables (indeterminates) and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
1.2 General Form of a Polynomial
A polynomial in one variable x of degree n is written as:
P(x)=anxn+an−1xn−1+⋯+a2x2+a1x+a0
Where:
an,an−1,…,a1,a0 are constants called coefficients
an=0 (leading coefficient)
n is a non-negative integer called the degree of the polynomial
a0 is the constant term
Each akxk is called a term
1.3 Standard Form
A polynomial is in standard form when terms are arranged in descending powers of the variable.
Example:
P(x)=3x4−2x3+5x2−x+7
is in standard form.
1.4 Degree of a Polynomial
The degree of a polynomial is the highest power of the variable with a non-zero coefficient.
Notation: deg(P) or deg(P)
Examples:
P(x)=5x3−2x2+x−1 → Degree = 3
Q(x)=7 → Degree = 0 (constant polynomial)
R(x)=0 → Degree is undefined (zero polynomial)
1.5 Classification by Degree
Constant polynomial: Degree 0 (e.g., P(x)=5)
Linear polynomial: Degree 1 (e.g., P(x)=2x+3)
Quadratic polynomial: Degree 2 (e.g., P(x)=x2−4x+4)
Cubic polynomial: Degree 3 (e.g., P(x)=x3−6x2+11x−6)
Quartic polynomial: Degree 4 (e.g., P(x)=x4−10x2+9)
Quintic polynomial: Degree 5
n-th degree polynomial: Degree n
1.6 Classification by Number of Terms
Monomial: 1 term (e.g., 7x3)
Binomial: 2 terms (e.g., x2−4)
Trinomial: 3 terms (e.g., x2+5x+6)
Multinomial: More than 3 terms
1.7 Zero Polynomial
The polynomial with all coefficients zero: P(x)=0
Special Properties:
Its degree is undefined (some say −∞ or not defined)
It's the additive identity in polynomial algebra
P(x)+0=P(x) for any polynomial P(x)
2. Operations on Polynomials
2.1 Addition and Subtraction
Add/subtract corresponding coefficients of like terms (same power).
Example 1: Add (3x3−2x2+5x−1) and (2x3+4x2−3x+2)
=(3x3−2x2+5x−1)+(2x3+4x2−3x+2)
=(3+2)x3+(−2+4)x2+(5−3)x+(−1+2)
=5x3+2x2+2x+1
Example 2: Subtract (4x2−3x+2) from (2x3+5x2−x+1)
=(2x3+5x2−x+1)−(4x2−3x+2)
=2x3+5x2−x+1−4x2+3x−2
=2x3+(5−4)x2+(−1+3)x+(1−2)
=2x3+x2+2x−1
2.2 Multiplication
a) Multiplying by a Constant
Multiply each coefficient by the constant.
Example:
=3(2x3−x2+4x−1)
=6x3−3x2+12x−3
b) Multiplying Two Polynomials
Use distributive property: Multiply each term of first polynomial by each term of second polynomial, then combine like terms.
Example: Multiply (2x+3) by (x2−2x+1)
=(2x+3)(x2−2x+1)
=2x(x2−2x+1)+3(x2−2x+1)
=2x3−4x2+2x+3x2−6x+3
=2x3+(−4+3)x2+(2−6)x+3
=2x3−x2−4x+3
Degree Property:
deg(P⋅Q)=deg(P)+deg(Q)
2.3 Division of Polynomials
a) Division by a Monomial
Divide each term by the monomial.
Example: Divide (6x4−9x3+3x2) by 3x2
=3x26x4−9x3+3x2
=3x26x4−3x29x3+3x23x2
=2x2−3x+1
b) Long Division of Polynomials
Similar to numerical long division.
Example: Divide (x3−6x2+11x−6) by (x−2)
Step 1: Divide first term: =xx3=x2
Step 2: Multiply: =x2(x−2)=x3−2x2
Step 3: Subtract: =(x3−6x2)−(x3−2x2)=−4x2
Step 4: Bring down next term: =−4x2+11x
Step 5: Divide: =x−4x2=−4x
Step 6: Multiply: =−4x(x−2)=−4x2+8x
Step 7: Subtract: =(−4x2+11x)−(−4x2+8x)=3x
Step 8: Bring down: =3x−6
Step 9: Divide: =x3x=3
Step 10: Multiply: =3(x−2)=3x−6
Step 11: Subtract: =(3x−6)−(3x−6)=0
Result: Quotient = x2−4x+3, Remainder = 0
So:
=x3−6x2+11x−6=(x−2)(x2−4x+3)
c) Synthetic Division
A shortcut for division by linear factors of form (x−c).
Example: Divide (2x3−7x2+5x−1) by (x−3)
Steps:
Step 1: Write coefficients: 2,−7,5,−1
Step 2: Write c=3 (from x−3)
Step 3: Bring down first coefficient: 2
Step 4: Multiply by c and add to next coefficient:
=2×3=6
=−7+6=−1
Step 5:
=−1×3=−3
=5−3=2
Step 6:
=2×3=6
=−1+6=5
Step 7: Last number is remainder (5), others are coefficients of quotient
Quotient: 2x2−x+2, Remainder: 5
Verification:
=(2x3−7x2+5x−1)
=(x−3)(2x2−x+2)+5
3. Polynomial Equations and Roots
3.1 Polynomial Equation
An equation of the form P(x)=0 where P(x) is a polynomial.
Example:
x3−6x2+11x−6=0
3.2 Roots/Zeros of a Polynomial
A number α is a root (or zero) of polynomial P(x) if P(α)=0.
Equivalent Statements:
α is a root of P(x)
P(α)=0
(x−α) is a factor of P(x)
α is a solution of equation P(x)=0
3.3 Fundamental Theorem of Algebra
Every non-constant polynomial with complex coefficients has at least one complex root.
Consequences:
A polynomial of degree n has exactly n roots (counting multiplicity)
If coefficients are real, complex roots occur in conjugate pairs
3.4 Multiplicity of a Root
If (x−α)k is a factor of P(x) but (x−α)k+1 is not, then α is a root of multiplicity k.
Behavior at roots:
Odd multiplicity: Graph crosses x-axis at root
Even multiplicity: Graph touches x-axis and turns around at root
Example:
P(x)=(x−1)3(x+2)2(x−3)
Root x=1 has multiplicity 3 (odd, crosses axis)
Root x=−2 has multiplicity 2 (even, touches axis)
Root x=3 has multiplicity 1 (odd, crosses axis)
3.5 Finding Roots
a) For Linear Polynomials (ax+b=0)
Solution: x=−ab
b) For Quadratic Polynomials (ax2+bx+c=0)
Quadratic Formula: x=2a−b±b2−4ac
Discriminant (Δ): Δ=b2−4ac
If Δ>0: Two distinct real roots
If Δ=0: One real root (double root)
If Δ<0: Two complex conjugate roots
c) For Cubic Polynomials
General cubic: ax3+bx2+cx+d=0
Special Cases:
Factorable: Try integer factors of constant term
Sum/difference of cubes: a3±b3=(a±b)(a2∓ab+b2)
Cardano's formula (general solution, but complex)
Example: Solve x3−6x2+11x−6=0
Try factors of 6: ±1,±2,±3,±6
Check: P(1)=1−6+11−6=0✓
So (x−1) is a factor.
Divide: =x−1x3−6x2+11x−6=x2−5x+6
Solve quadratic: x2−5x+6=0→(x−2)(x−3)=0
Roots: x=1,2,3
d) For Quartic Polynomials
General quartic: ax4+bx3+cx2+dx+e=0
Special Cases:
Biquadratic: ax4+bx2+c=0 (substitute y=x2)
Factorable: Try rational roots
Ferrari's method (general solution, complex)
Example: Solve x4−5x2+4=0
Let y=x2: y2−5y+4=0
Solve: (y−1)(y−4)=0→y=1 or y=4
So x2=1 or x2=4
Roots: x=±1,±2
3.6 Rational Root Theorem
For polynomial P(x)=anxn+an−1xn−1+⋯+a0 with integer coefficients:
If p/q (in lowest terms) is a rational root, then:
p divides constant term a0
q divides leading coefficient an
Example: Find rational roots of P(x)=2x3−3x2−8x+12
Possible p: factors of 12: ±1,±2,±3,±4,±6,±12
Possible q: factors of 2: ±1,±2
Possible p/q: ±1,±2,±3,±4,±6,±12,±21,±23
Test: P(2)=2(8)−3(4)−8(2)+12=16−12−16+12=0✓
So x=2 is a root.
3.7 Relationship Between Roots and Coefficients
For Quadratic: ax2+bx+c=0 with roots α,β
Sum of roots: α+β=−ab
Product of roots: αβ=ac
Quadratic with given roots: x2−(α+β)x+αβ=0
For Cubic: ax3+bx2+cx+d=0 with roots α,β,γ
Sum: α+β+γ=−ab
Sum of products taken two at a time: αβ+βγ+γα=ac
Product: αβγ=−ad
For Quartic: ax4+bx3+cx2+dx+e=0 with roots α,β,γ,δ
Sum: α+β+γ+δ=−ab
Sum of products taken two at a time: αβ+αγ+αδ+βγ+βδ+γδ=ac
Sum of products taken three at a time: αβγ+αβδ+αγδ+βγδ=−ad
Product: αβγδ=ae
Example: For cubic x3−6x2+11x−6=0 with roots 1,2,3:
Sum: 1+2+3=6=−(−6)/1✓
Sum of products: 1⋅2+2⋅3+3⋅1=2+6+3=11=11/1✓
Product: 1⋅2⋅3=6=−(−6)/1=6✓
3.8 Formation of Polynomial from Roots
If α1,α2,…,αn are roots, then:
P(x)=a(x−α1)(x−α2)⋯(x−αn)
where a is leading coefficient.
Example: Form cubic with roots 1,−2,3 and leading coefficient 2:
=P(x)=2(x−1)(x+2)(x−3)
Expanding:
=2[(x−1)(x+2)](x−3)
=2(x2+x−2)(x−3)
=2(x3−3x2+x2−3x−2x+6)
=2(x3−2x2−5x+6)
=2x3−4x2−10x+12
4. Special Polynomials and Identities
4.1 Difference of Squares
a2−b2=(a−b)(a+b)
Examples:
x2−9=(x−3)(x+3)
4x2−25=(2x−5)(2x+5)
x4−16=(x2−4)(x2+4)=(x−2)(x+2)(x2+4)
4.2 Perfect Square Trinomials
(a+b)2=a2+2ab+b2
(a−b)2=a2−2ab+b2
Examples:
x2+6x+9=(x+3)2
4x2−12x+9=(2x−3)2
4.3 Sum and Difference of Cubes
a3+b3=(a+b)(a2−ab+b2)
a3−b3=(a−b)(a2+ab+b2)
Examples:
x3+8=(x+2)(x2−2x+4)
27x3−1=(3x−1)(9x2+3x+1)
4.4 Square of a Trinomial
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
Example:
(x+y+1)2=x2+y2+1+2xy+2y+2x
4.5 Binomial Theorem for Polynomial Expansion
(a+b)n=∑k=0n(kn)an−kbk
where (kn)=k!(n−k)!n!
Examples:
For (x+2)3:
=(x+2)3
=(03)x3+(13)x2(2)+(23)x(22)+(33)(23)
=1⋅x3+3⋅2x2+3⋅4x+1⋅8
=x3+6x2+12x+8
For (2x−1)4:
=(2x−1)4
=(04)(2x)4+(14)(2x)3(−1)+(24)(2x)2(−1)2+(34)(2x)(−1)3+(44)(−1)4
=16x4−32x3+24x2−8x+1
5. Remainder Theorem and Factor Theorem
5.1 Remainder Theorem
If polynomial P(x) is divided by (x−c), then the remainder is P(c).
Proof:
By division algorithm: P(x)=(x−c)Q(x)+R
Put x=c: P(c)=(c−c)Q(c)+R=R
Example: Find remainder when P(x)=x3−2x2+3x−1 is divided by (x−2)
Remainder =P(2)=8−8+6−1=5
5.2 Factor Theorem
(x−c) is a factor of P(x) if and only if P(c)=0.
Proof:
From Remainder Theorem: P(c)= remainder when divided by (x−c)
If P(c)=0, remainder =0, so (x−c) is a factor.
Example: Check if (x−3) is factor of P(x)=x3−6x2+11x−6
P(3)=27−54+33−6=0✓
So (x−3) is a factor.
5.3 Applications
a) Finding Roots
Use Factor Theorem to find factors, then find roots.
Example: Find roots of P(x)=x3−3x2−x+3
Try factors of 3: ±1,±3
Check: P(1)=1−3−1+3=0✓
So (x−1) is factor.
Divide: P(x)=(x−1)(x2−2x−3)
Solve quadratic: x2−2x−3=0→(x−3)(x+1)=0
Roots: x=1,3,−1
b) Finding Polynomial with Given Roots
Example: Find quadratic with roots 2 and −3
Factors: (x−2) and (x+3)
Polynomial: (x−2)(x+3)=x2+x−6
c) Determining Unknown Coefficients
Example: Find k if (x−2) is factor of P(x)=x3+kx2−4x−8
Since (x−2) is factor, P(2)=0
P(2)=8+4k−8−8=4k−8=0
So k=2
6. Graphs of Polynomial Functions
6.1 General Shape
A polynomial of degree n:
Has at most (n−1) turning points
Has at most n x-intercepts (real roots)
End behavior depends on degree and leading coefficient
6.2 End Behavior
For P(x)=anxn+⋯:
n odd, an>0: As x→−∞, P(x)→−∞; As x→∞, P(x)→∞
n odd, an<0: As x→−∞, P(x)→∞; As x→∞, P(x)→−∞
n even, an>0: As x→±∞, P(x)→∞
n even, an<0: As x→±∞, P(x)→−∞
Mnemonic: "Positive leading coefficient: Right end goes up"
6.3 Key Features
a) Linear (n=1)
Graph: Straight line
Equation: y=mx+c
Slope =m
y-intercept =c
b) Quadratic (n=2)
Graph: Parabola
Equation: y=ax2+bx+c
Vertex form: y=a(x−h)2+k where vertex =(h,k)
Vertex coordinates: h=−2ab,k=f(h)
Properties:
Opens upward if a>0, downward if a<0
Axis of symmetry: x=h
y-intercept: c
x-intercepts: roots of equation
c) Cubic (n=3)
Graph: S-shaped curve
General shape depends on sign of leading coefficient.
Special Case: Cubic with three real distinct roots
Passes through x-axis at each root
Has two turning points (local max and min)
Example:
y=x3−6x2+11x−6=(x−1)(x−2)(x−3)
Roots at x=1,2,3
Two turning points between roots
7. Inequalities Involving Polynomials
7.1 Polynomial Inequalities
To solve P(x)>0 or P(x)<0:
Method:
Step 1: Find all real roots of P(x)=0
Step 2: Plot roots on number line
Step 3: Test sign in each interval
Step 4: Include/exclude endpoints based on inequality (>,≥,<,≤)
Example: Solve x3−6x2+11x−6>0
Factor: (x−1)(x−2)(x−3)>0
Roots: x=1,2,3
Test intervals:
Interval 1: x<1 Test x=0: (−)(−)(−)=− (negative)
Interval 2: 1<x<2 Test x=1.5: (+)(−)(−)=+ (positive)
Interval 3: 2<x<3 Test x=2.5: (+)(+)(−)=− (negative)
Interval 4: x>3 Test x=4: (+)(+)(+)=+ (positive)
Solution: x∈(1,2)∪(3,∞)
7.2 Rational Inequalities
Solve Q(x)P(x)>0 or similar:
Method:
Step 1: Find roots of P(x)=0 and Q(x)=0
Step 2: Plot all on number line (roots of Q(x) are excluded)
Step 3: Test sign in each interval
Example: Solve x−1x2−4>0
Factor: x−1(x−2)(x+2)>0
Critical points: x=−2,1,2
Test intervals:
Interval 1: x<−2 Test x=−3: (−)(−)/(−)=(−) (negative)
Interval 2: −2<x<1 Test x=0: (−)(+)/(−)=(+) (positive)
Interval 3: 1<x<2 Test x=1.5: (−)(+)/(+)=(−) (negative)
Interval 4: x>2 Test x=3: (+)(+)/(+)=(+) (positive)
Solution: x∈(−2,1)∪(2,∞)
8. Solved Examples
Example 1: Complete Factorization
Factor completely: P(x)=x4−5x2+4
Solution:
Step 1: Let y=x2: =y2−5y+4=(y−1)(y−4)
Step 2: So P(x)=(x2−1)(x2−4)
Step 3: =P(x)=(x−1)(x+1)(x−2)(x+2)
Example 2: Find Polynomial with Given Conditions
Find cubic polynomial with roots 1,2,3 and P(0)=12.
Solution:
Step 1: General form: P(x)=a(x−1)(x−2)(x−3)
Step 2: Given P(0)=12: a(−1)(−2)(−3)=12
Step 3: −6a=12→a=−2
Step 4: So P(x)=−2(x−1)(x−2)(x−3)=−2(x3−6x2+11x−6)
Step 5: =P(x)=−2x3+12x2−22x+12
Example 3: Synthetic Division Application
Divide P(x)=3x4−4x3+2x2−5x+1 by (x−2)
Solution using synthetic division:
Step 1: Coefficients: 3,−4,2,−5,1 c=2
Step 2: Bring down 3
Step 3: =3×2=6 =−4+6=2
Step 4: =2×2=4 =2+4=6
Step 5: =6×2=12 =−5+12=7
Step 6: =7×2=14 =1+14=15
Result:
Quotient: 3x3+2x2+6x+7
Remainder: 15
So: =3x4−4x3+2x2−5x+1
=(x−2)(3x3+2x2+6x+7)+15
Example 4: Roots and Coefficients Relationship
If α,β are roots of x2−3x+2=0, find α3+β3.
Solution:
Step 1: From equation: α+β=3, αβ=2
Step 2: We know: α3+β3=(α+β)3−3αβ(α+β)
Step 3: =33−3×2×3=27−18=9
Example 5: Inequality Solution
Solve: (x−1)(x−2)2(x−3)3>0
Solution:
Step 1: Roots:
x=1 (multiplicity 1)
x=2 (multiplicity 2)
x=3 (multiplicity 3)
Step 2: Sign chart:
Interval 1: x<1 Test x=0: (−)(+)(−)=+ (positive)
Interval 2: 1<x<2 Test x=1.5: (+)(+)(−)=− (negative)
Interval 3: 2<x<3 Test x=2.5: (+)(+)(−)=− (negative)
Interval 4: x>3 Test x=4: (+)(+)(+)=+ (positive)
Step 3: Since inequality is >0, and:
At x=1: sign changes (odd multiplicity)
At x=2: sign doesn't change (even multiplicity)
At x=3: sign changes (odd multiplicity)
Solution: x∈(−∞,1)∪(3,∞)
9. Practice Tips for Exams
Factorization First: Always try to factor polynomials first
Rational Root Theorem: Use for finding possible rational roots
Synthetic Division: Faster than long division for (x−c)
Sum/Product of Roots: Quick way to check answers
End Behavior: Helps sketch graphs quickly
Sign Charts: Essential for solving inequalities
Multiplicity: Determines graph behavior at roots
Practice Identities: Memorize common expansions/factorizations
This comprehensive theory covers all aspects of polynomials with detailed explanations and examples, providing complete preparation for the entrance examination.
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