Polynomials Notes

Polynomials

Introduction

• Review of algebraic expressions and operations (addition, subtraction, multiplication, division).

• Recalling factorization of algebraic expressions.

• Algebraic identities:

  • $(x + y)^2 = x^2 + 2xy + y^2$

  • $(x - y)^2 = x^2 - 2xy + y^2$

  • $x^2 - y^2 = (x + y)(x - y)$

• Chapter focus: polynomials, terminology, Remainder Theorem, Factor Theorem, algebraic identities, factorization, and evaluating expressions.

Polynomials in One Variable

• A variable is a symbol that can take any real value (e.g., x, y, z).

• Examples of algebraic expressions: $2x, 3x, -x, -\frac{1}{2}x$ (a constant × x).

• Constants are denoted by letters like a, b, c.

• Constants have fixed values in a problem, while variables can change.

• Perimeter of a square with side 3 units: $4 \times 3 = 12$ units.

• Perimeter of a square with side 10 units: $4 \times 10 = 40$ units.

• Perimeter of a square with side x units: $4x$ units.

• Area of a square with side x units: $x \times x = x^2$ square units.

• Polynomials in one variable have whole numbers as exponents of the variable.

• Examples: $x^3 - x^2 + 4x + 7, 3y^2 + 5y, t^2 + 4$

• Terms of a polynomial are the expressions being added or subtracted.

• Example: $x^2 + 2x$ has terms $x^2$ and $2x$.

• Example: $3y^2 + 5y + 7$ has terms $3y^2$, $5y$, and $7$.

• Example: $-x^3 + 4x^2 + 7x - 2$ has terms $-x^3$, $4x^2$, $7x$, and $-2$.

• Each term has a coefficient.

• In $-x^3 + 4x^2 + 7x - 2$:

  • Coefficient of $x^3$ is $-1$.

  • Coefficient of $x^2$ is $4$.

  • Coefficient of $x$ is $7$.

  • Coefficient of $x^0$ is $-2$.

• Coefficient of $x$ in $x^2 - x + 7$ is $-1$.

• Constants (e.g., 2, -5, 7) are constant polynomials.

• $0$ is the zero polynomial.

• Expressions like $x + \frac{1}{x} = x + x^{-1}$ are not polynomials because the exponent $-1$ is not a whole number.

• $\sqrt{x} + 3 = x^{\frac{1}{2}} + 3$ is not a polynomial because the exponent $\frac{1}{2}$ is not a whole number.

• $\frac{3}{y} + y^2$ is not a polynomial because it can be written as $3y^{-1} + y^2$ which has -1 as an exponent.

Polynomial Notation and Types

• Polynomials in variable x can be denoted as p(x), q(x), r(x), etc.

  • Example: $p(x) = 2x^2 + 5x - 3$

  • Example: $q(x) = x^3 - 1$

  • Example: $r(y) = y^3 + y + 1$

  • Example: $s(u) = 2 - u - u^2 + 6u^5$

• Polynomials can have any finite number of terms (e.g., $x^{150} + x^{149} + … + x^2 + x + 1$ has 151 terms).

• Monomials: Polynomials with one term (e.g., $2x, 2, 5x^3, -5x^2, y, u^4$).

• Binomials: Polynomials with two terms (e.g., $p(x) = x + 1, q(x) = x^2 - x, r(y) = y^9 + 1, t(u) = u^{15} - u^2$).

• Trinomials: Polynomials with three terms (e.g., $p(x) = x + x^2 + \pi, q(x) = 2 + x - x^2, r(u) = u + u^2 - 2, t(y) = y^4 + y + 5$).

• Degree of a polynomial: The highest power of the variable in the polynomial.

  • Example: In $p(x) = 3x^7 - 4x^6 + x + 9$, the degree is 7.

  • Example: In $q(y) = 5y^6 - 4y^2 - 6$, the degree is 6.

• The degree of a non-zero constant polynomial is zero.

Degree of Polynomials

• Example (i): $x^5 - x^4 + 3$ has degree 5.

• Example (ii): $2 - y^2 - y^3 + 2y^8$ has degree 8.

• Example (iii): $2$ has degree 0 (since $2 = 2x^0$).

Linear, Quadratic, and Cubic Polynomials

• Linear polynomial: A polynomial of degree one (e.g., $4x + 5, 2y, t + \sqrt{2}, 3 - u$).

• A linear polynomial in x has the form $ax + b$, where $a$ and $b$ are constants and $a \neq 0$.

• Quadratic polynomial: A polynomial of degree two (e.g., $2x^2 + 5, 5x^2 + 3x + \pi, x^2, x^2 + \frac{2}{5}x$).

• A quadratic polynomial in x has the form $ax^2 + bx + c$, where $a \neq 0$ and a, b, c are constants.

• Cubic polynomial: A polynomial of degree three (e.g., $4x^3, 2x^3 + 1, 5x^3 + x^2, 6x^3 - x, 6 - x^3, 2x^3 + 4x^2 + 6x + 7$).

• A cubic polynomial in x has the form $ax^3 + bx^2 + cx + d$, where $a \neq 0$ and a, b, c, and d are constants.

• A polynomial in one variable x of degree n has the form $a<em>n x^n + a</em>{n-1}x^{n-1} + … + a<em>1x + a</em>0$, where $a<em>0, a</em>1, a<em>2, …, a</em>n$ are constants and $a_n \neq 0$.

• Zero polynomial: All coefficients are zero. Denoted by 0. The degree of the zero polynomial is not defined.

• Polynomials in more than one variable (e.g., $x^2 + y^2 + xyz$).

Exercise 2.1

• Identifying polynomials in one variable.

• Writing coefficients of $x^2$ in given expressions.

• Giving examples of a binomial of degree 35 and a monomial of degree 100.

• Writing the degree of given polynomials.

• Classifying polynomials as linear, quadratic, and cubic.

Zeroes of a Polynomial

• Value of polynomial $p(x) = 5x^3 - 2x^2 + 3x - 2$ at $x = 1$ is $p(1) = 5(1)^3 - 2(1)^2 + 3(1) - 2 = 4$.

• Value of polynomial $p(x) = 5x^3 - 2x^2 + 3x - 2$ at $x = 0$ is $p(0) = 5(0)^3 - 2(0)^2 + 3(0) - 2 = -2$.

• If $p(1) = 0$, then 1 is a zero of the polynomial p(x).

• A zero of a polynomial p(x) is a number c such that $p(c) = 0$.

• $p(x) = 0$ is a polynomial equation, and its solutions are called roots.

• A non-zero constant polynomial has no zero.

• Every real number is a zero of the zero polynomial.

Examples of Finding Zeroes

• Example: Check if -2 and 2 are zeroes of $p(x) = x + 2$.

  • $p(2) = 2 + 2 = 4$

  • $p(-2) = -2 + 2 = 0$

  • Therefore, -2 is a zero, but 2 is not.

• Example: Find a zero of $p(x) = 2x + 1$.

  • $2x + 1 = 0$ implies $x = -\frac{1}{2}$.

• If $p(x) = ax + b$, a ≠ 0, then $x = -\frac{b}{a}$ is the only zero of p(x).

• Example: Verify if 2 and 0 are zeroes of $p(x) = x^2 - 2x$.

  • $p(2) = 2^2 - 2(2) = 0$

  • $p(0) = 0^2 - 2(0) = 0$

  • Hence, 2 and 0 are zeroes of the polynomial.

Observations about Zeroes

• A zero of a polynomial need not be 0.

• 0 may be a zero of a polynomial.

• Every linear polynomial has one and only one zero.

• A polynomial can have more than one zero.

Exercise 2.2

• Finding the value of a polynomial at a given value.

• Finding p(0), p(1), and p(2) for given polynomials.

• Verifying whether given values are zeroes of the polynomial.

• Finding the zero of the polynomial in each case.

Factorization of Polynomials

• Factor Theorem: If p(x) is a polynomial of degree n > 1 and a is any real number, then

  • (i) x – a is a factor of p(x), if p(a) = 0, and

  • (ii) p(a) = 0, if x – a is a factor of p(x).

• Proof by the Remainder Theorem: $p(x)=(x – a) q(x) + p(a)$.

  • (i) If $p(a) = 0$, then $p(x) = (x – a) q(x)$, which shows that x – a is a factor of p(x).

  • (ii) If x – a is a factor of p(x), $p(x) = (x – a) g(x)$ for some polynomial g(x). In this case, $p(a) = (a – a) g(a) = 0$.

Examples of Applying Factor Theorem

• Example: Examine whether x + 2 is a factor of $x^3 + 3x^2 + 5x + 6$ and of $2x + 4$.

  • The zero of x + 2 is –2.

  • Let $p(x) = x^3 + 3x^2 + 5x + 6$ and $s(x) = 2x + 4$.

  • $p(–2) = (–2)^3 + 3(–2)^2 + 5(–2) + 6 = –8 + 12 – 10 + 6 = 0$

  • So, by the Factor Theorem, x + 2 is a factor of $x^3 + 3x^2 + 5x + 6$.

  • $s(–2) = 2(–2) + 4 = 0$

  • So, x + 2 is a factor of 2x + 4.

• Example: Find the value of k, if x – 1 is a factor of $4x^3 + 3x^2 – 4x + k$.

  • As x – 1 is a factor of $p(x) = 4x^3 + 3x^2 – 4x + k$, $p(1) = 0$.

  • $p(1) = 4(1)^3 + 3(1)^2 – 4(1) + k$

  • $4 + 3 – 4 + k = 0$

  • $k = –3$

Factorization Techniques

• Factorizing quadratic polynomials of the type $ax^2 + bx + c$ by splitting the middle term.

• If $ax^2 + bx + c = (px + q) (rx + s)$, then comparing coefficients:

  • $a = pr$

  • $b = ps + qr$

  • $c = qs$

• To factorize $ax^2 + bx + c$, write b as the sum of two numbers whose product is ac.

Examples of Factorization

• Example: Factorise $6x^2 + 17x + 5$ by splitting the middle term, and by using the Factor Theorem.

  • Solution 1 (By splitting method): Find p and q such that $p + q = 17$ and $pq = 6 \times 5 = 30$. The pair 2 and 15 satisfies these conditions.

    • $6x^2 + 17x + 5 = 6x^2 + (2 + 15)x + 5 = 6x^2 + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1) (2x + 5)$

  • Solution 2 (Using the Factor Theorem): $6x^2 + 17x + 5 = \frac{6}{6}(6x^2 + 17x + 5) =6 p(x)$, where $p(x) = x^2 + \frac{17}{6}x + \frac{5}{6}$

    • If a and b are the zeroes of p(x), then $6x^2 + 17x + 5 = 6(x – a) (x – b)$. So, $ab = \frac{5}{6}$.

    • By trial, find that $p(-\frac{1}{3}) = 0$. So, $(x + \frac{1}{3})$ is a factor of p(x). Similarly, by trial, find that $(x + \frac{5}{2})$ is a factor of p(x).

    • $6x^2 + 17x + 5 = (3x + 1) (2x + 5)$

• Example: Factorise $y^2 – 5y + 6$ by using the Factor Theorem.

  • If $p(y) = (y – a) (y – b)$, then $ab = 6$. The factors of 6 are 1, 2, and 3.

  • $p(2) = 2^2 – (5 \times 2) + 6 = 0$

  • So, y – 2 is a factor of p(y).

  • $p(3) = 3^2 – (5 \times 3) + 6 = 0$

  • So, y – 3 is also a factor of $y^2 – 5y + 6$.

  • $y^2 – 5y + 6 = (y – 2)(y – 3)$

Factorizing Cubic Polynomials

• The splitting method is not appropriate to start with.

• Need to find at least one factor first.

• Example: Factorise $x^3 – 23x^2 + 142x – 120$.

  • Let $p(x) = x^3 – 23x^2 + 142x – 120$.

  • Factors of –120 are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.

  • By trial, find that $p(1) = 0$. So x – 1 is a factor of p(x).

  • $x^3 – 23x^2 + 142x – 120 = x^3 – x^2 – 22x^2 + 22x + 120x – 120 = x^2(x –1) – 22x(x – 1) + 120(x – 1) = (x – 1) (x^2 – 22x + 120)$

  • $x^2 – 22x + 120 = x^2 – 12x – 10x + 120 = x(x – 12) – 10(x – 12) = (x – 12) (x – 10)$

  • $x^3 – 23x^2 – 142x – 120 = (x – 1)(x – 10)(x – 12)$

Exercise 2.3

• Determining whether (x + 1) is a factor of given polynomials.

• Using the Factor Theorem to determine whether g(x) is a factor of p(x) in each case.

• Finding the value of k, if x – 1 is a factor of p(x) in each case.

• Factorising given quadratic expressions.

• Factorising given cubic expressions.

Algebraic Identities

• Algebraic identity: An algebraic equation that is true for all values of the variables occurring in it.

• Studied identities:

  • Identity I: $(x + y)^2 = x^2 + 2xy + y^2$

  • Identity II: $(x – y)^2 = x^2 – 2xy + y^2$

  • Identity III: $x^2 – y^2 = (x + y) (x – y)$

  • Identity IV: $(x + a) (x + b) = x^2 + (a + b)x + ab$

Examples of Using Identities

• Example: Find the following products using appropriate identities:

  • (i) $(x + 3) (x + 3) = (x + 3)^2 = x^2 + 2(x)(3) + (3)^2 = x^2 + 6x + 9$

  • (ii) $(x – 3) (x + 5) = x^2 + (–3 + 5)x + (–3)(5) = x^2 + 2x – 15$

• Example: Evaluate 105 × 106 without multiplying directly.

  • $105 × 106 = (100 + 5) × (100 + 6) = (100)^2 + (5 + 6) (100) + (5 × 6) = 10000 + 1100 + 30 = 11130$

• Example: Factorise:

  • (i) $49a^2 + 70ab + 25b^2 = (7a)^2 + 2(7a)(5b) + (5b)^2 = (7a + 5b)^2 = (7a + 5b) (7a + 5b)$

  • (ii) $\frac{25}{4}x^2 – \frac{9}{4}y^2 = (\frac{5}{2}x)^2 – (\frac{3}{2}y)^2 = (\frac{5}{2}x + \frac{3}{2}y )( \frac{5}{2}x - \frac{3}{2}y)$

Extension to Trinomials

• Identity V: $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

• Expanded form: The right-hand side expression.

• The expansion of $(x + y + z)^2$ consists of three square terms and three product terms.

Examples of Trinomial Expansion

• Example: Write $(3a + 4b + 5c)^2$ in expanded form.

  • $(3a + 4b + 5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a) = 9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ac$

• Example: Expand $(4a – 2b – 3c)^2$.

  • $(4a – 2b – 3c)^2 = [4a + (–2b) + (–3c)]^2 = (4a)^2 + (–2b)^2 + (–3c)^2 + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a) = 16a^2 + 4b^2 + 9c^2 – 16ab + 12bc – 24ac$

• Example: Factorise $4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz$.

  • $4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz = (2x)^2 + (–y)^2 + (z)^2 + 2(2x)(–y) + 2(–y)(z) + 2(2x)(z) = [2x + (–y) + z]^2 = (2x – y + z)^2 = (2x – y + z)(2x – y + z)$

Cubing Binomials

• Identity VI: $(x + y)^3 = x^3 + y^3 + 3xy (x + y)$

• Identity VII: $(x – y)^3 = x^3 – y^3 – 3xy(x – y) = x^3 – 3x^2y + 3xy^2 – y^3$

Examples of Cubing Binomials

• Example: Write the following cubes in the expanded form:

  • (i) $(3a + 4b)^3 = (3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b) = 27a^3 + 64b^3 + 108a^2b + 144ab^2$

  • (ii) $(5p – 3q)^3 = (5p)^3 – (3q)^3 – 3(5p)(3q)(5p – 3q) = 125p^3 – 27q^3 – 225p^2q + 135pq^2$

• Example: Evaluate each of the following using suitable identities:

  • (i) $(104)^3 = (100 + 4)^3 = (100)^3 + (4)^3 + 3(100)(4)(100 + 4) = 1000000 + 64 + 124800 = 1124864$

  • (ii) $(999)^3 = (1000 – 1)^3 = (1000)^3 – (1)^3 – 3(1000)(1)(1000 – 1) = 1000000000 – 1 – 2997000 = 997002999$

• Example: Factorise $8x^3 + 27y^3 + 36x^2y + 54xy^2$.

  • $8x^3 + 27y^3 + 36x^2y + 54xy^2 = (2x)^3 + (3y)^3 + 3(4x^2)(3y) + 3(2x)(9y^2) = (2x)^3 + (3y)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2 = (2x + 3y)^3 = (2x + 3y)(2x + 3y)(2x + 3y)$

Identity VIII

• Identity VIII: $x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)$

• Example: Factorise: $8x^3 + y^3 + 27z^3 – 18xyz$.

  • $8x^3 + y^3 + 27z^3 – 18xyz = (2x)^3 + y^3 + (3z)^3 – 3(2x)(y)(3z) = (2x + y + 3z)[(2x)^2 + y^2 + (3z)^2 – (2x)(y) – (y)(3z) – (2x)(3z)] = (2x + y + 3z) (4x^2 + y^2 + 9z^2 – 2xy – 3yz – 6xz)$

Exercise 2.4

• Using suitable identities to find products.

• Evaluating products without multiplying directly.

• Factorising using appropriate identities.

• Expanding using suitable identities.

• Factorising given expressions.

• Writing cubes in expanded form.

• Evaluating using suitable identities.

• Factorising given expressions.

• Verifying given identities.

• Factorising given expressions.

Summary

• A polynomial p(x) in one variable x is an algebraic expression in x of the form
  $p(x) = a<em>n x^n + a</em>{n–1}x^{n – 1} + . . . + a<em>2x^2 + a</em>1x + a<em>0$ where $a</em>0, a<em>1, a</em>2, . . ., a<em>n$ are constants and $a</em>n ≠ 0$.

• Polynomials are classified by the number of terms (monomial, binomial, trinomial) and by degree (linear, quadratic, cubic).

• A real number ‘a’ is a zero of a polynomial p(x) if $p(a) = 0$.

• Factor Theorem: x – a is a factor of the polynomial p(x) if $p(a) = 0$.

• Key Identities:

  • $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

  • $(x + y)^3 = x^3 + y^3 + 3xy(x + y)$

  • $(x – y)^3 = x^3 – y^3 – 3xy(x – y)$

  • $x^3 + y^3 + z^3 – 3xyz = (x + y + z) (x^2 + y^2 + z^2 – xy – yz – zx)$