polynomials

Polynomials

Introduction

• Review of algebraic expressions, including addition, subtraction, multiplication, and division.
• Recalling factorization of algebraic expressions.
• Algebraic identities:
  • $(x + y)^2 = x^2 + 2xy + y^2$
  • $(x – y)^2 = x^2 – 2xy + y^2$
  • $x^2 – y^2 = (x + y) (x – y)$
• In this chapter, focus on 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, denoted by letters like x, y, z.
• Examples of algebraic expressions: $2x, 3x, –x, –\frac{1}{2}x$ (form: 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: $4 × 3 = 12$ units.
• Perimeter of a square with side 10 units: $4 × 10 = 40$ units.
• If each side of a square is $x$ units, the perimeter is $4x$ units.
• Area of square PQRS with side $x$: $x × x = x^2$ square units.
• $x^2$ is an algebraic expression.
• Other algebraic expressions: $2x, x^2 + 2x, x^3 – x^2 + 4x + 7$
• Polynomials in one variable have whole numbers as exponents of the variable.
  • Example: $x^3 – x^2 + 4x + 7$ is a polynomial in x.
  • $3y^2 + 5y$ is a polynomial in y.
  • $t^2 + 4$ is a polynomial in t.
• Terms of a polynomial are the expressions separated by addition or subtraction.
  • Example: in $x^2 + 2x$, the terms are $x^2$ and $2x$.
  • Terms of $3y^2 + 5y + 7$ are $3y^2$, $5y$, and $7$.
  • Terms of $-x^3 + 4x^2 + 7x – 2$ are $-x^3$, $4x^2$, $7x$, and $-2$.
• Each term has a coefficient.
  • In $-x^3 + 4x^2 + 7x – 2$, coefficients are -1 (for $x^3$), 4 (for $x^2$), 7 (for $x$), and -2 (coefficient of $x^0$).
  • Coefficient of $x$ in $x^2 – x + 7$ is -1.
• Constants like 2, -5, 7 are constant polynomials.
• The constant polynomial 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.
• Polynomials can be denoted as p(x), q(x), r(x), etc.
  • Examples:
    • $p(x) = 2x^2 + 5x – 3$
    • $q(x) = x^3 – 1$
    • $r(y) = y^3 + y + 1$
    • $s(u) = 2 – u – u^2 + 6u^5$
• A polynomial can have any finite number of terms, e.g., $x^{150} + x^{149} + … + x^2 + x + 1$ (151 terms).
• Monomials have only one term (e.g., $2x, 2, 5x^3, –5x^2, y, u^4$).
• Binomials have 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 have three terms (e.g., $p(x) = x + x^2 + π, q(x) = 2 + x – x^2, r(u) = u + u^2 – 2, t(y) = y^4 + y + 5$).
• The degree of a polynomial is the highest power of the variable.
  • Example: In $3x^7 – 4x^6 + x + 9$, the term with the highest power of x is $3x^7$, so the degree is 7.
  • In $5y^6 – 4y^2 – 6$, the degree is 6.
• The degree of a non-zero constant polynomial is zero.
  *Example 1:
  Find the degree of each of the polynomials given below:
  (i) $x^5 – x^4 + 3$
  (ii) $2 – y^2 – y^3 + 2y^8$
  (iii) $2$
  Solution:
  (i) The highest power of the variable is 5. So, the degree of the polynomial is 5.
  (ii) The highest power of the variable is 8. So, the degree of the polynomial is 8.
  (iii) The only term here is 2 which can be written as $2x^0$. So the exponent of x is 0. Therefore, the degree of the polynomial is 0.
• Linear polynomials have 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 $a ≠ 0$.
• Quadratic polynomials have degree two (e.g., $2x^2 + 5, 5x^2 + 3x + π, x^2, x^2 + \frac{2}{5}x$).
  • A quadratic polynomial in x has the form $ax^2 + bx + c$, where $a ≠ 0$ and $a$, $b$, $c$ are constants.
• Cubic polynomials have 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 $a ≠ 0$ and $a$, $b$, $c$, and $d$ are constants.
• A polynomial in one variable $x$ of degree $n$ is $a<em>n x^n + a</em>{n-1}x^{n-1} + … + a<em>1x + a</em>0$, where $a<em>0, a</em>1, a<em>2, …, a</em>n$ are constants and $a_n ≠ 0$.
• The zero polynomial is denoted by 0, and its degree is not defined.
• Polynomials in more than one variable exist (e.g., $x^2 + y^2 + xyz$).

Exercise 2.1

• Identifying polynomials in one variable.
• Writing coefficients of $x^2$ in given polynomials.
• 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 a polynomial $p(x) = 5x^3 – 2x^2 + 3x – 2$ at $x = 1$ is $p(1) = 5(1)^3 – 2(1)^2 + 3(1) – 2 = 4$.
  $p(0) = 5(0)^3 – 2(0)^2 + 3(0) –2 = –2$
  *Example 2:
  Find the value of each of the following polynomials at the indicated value of variables:
  (i) $p(x) = 5x^2 – 3x + 7$ at $x = 1$.
  (ii) $q(y) = 3y^3 – 4y + \sqrt{11}$ at $y = 2$.
  (iii) $p(t) = 4t^4 + 5t^3 – t^2 + 6$ at $t = a$.
  Solution:
  (i) $p(x) = 5x^2 – 3x + 7$
  The value of the polynomial $p(x)$ at $x = 1$ is given by
  $p(1) = 5(1)^2 – 3(1) + 7 = 5 – 3 + 7 = 9$
  (ii) $q(y) = 3y^3 – 4y + \sqrt{11}$
  The value of the polynomial $q(y)$ at $y = 2$ is given by
  $q(2) = 3(2)^3 – 4(2) + \sqrt{11} = 24 – 8 + \sqrt{11} = 16 + \sqrt{11}$
  (iii) $p(t) = 4t^4 + 5t^3 – t^2 + 6$
  The value of the polynomial $p(t)$ at $t = a$ is given by
  $p(a) = 4a^4 + 5a^3 – a^2 + 6$
• If $p(x) = x – 1$, then $p(1) = 1 – 1 = 0$, so 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 root is the zero of the polynomial.
• The constant polynomial 5 has no zero because $5x^0 = 5$ for any x.
• Non-zero constant polynomials have no zeroes.
• Every real number is a zero of the zero polynomial.
  *Exampe 3:
  Check whether –2 and 2 are zeroes of the polynomial $x + 2$.
  Solution : Let $p(x) = x + 2$. Then
  $p(2) = 2 + 2 = 4$,
  $p(–2) = –2 + 2 = 0$
  Therefore, –2 is a zero of the polynomial $x + 2$, but 2 is not.
  *Example 4:
  Find a zero of the polynomial $p(x) = 2x + 1$.
  Solution : Finding a zero of $p(x)$, is the same as solving the equation $p(x) = 0$
  Now, $2x + 1 = 0$ gives us
  $x = \frac{-1}{2}$
  So, $\frac{-1}{2}$ is a zero of the polynomial $2x + 1$.
• If $p(x) = ax + b, a ≠ 0$, then its zero is $x = -\frac{b}{a}$.
  *Example 5:
  Verify whether 2 and 0 are zeroes of the polynomial $x^2 – 2x$.
  Solution : Let $p(x) = x^2 – 2x$
  Then $p(2) = 2^2 – 4 = 4 – 4 = 0$
  and $p(0) = 0 – 0 = 0$
  Hence, 2 and 0 are both zeroes of the polynomial $x^2 – 2x$.
• Observations:
  • 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) Since $x – a$ is a factor of $p(x)$, $p(x) = (x – a) g(x)$ for same polynomial $g(x)$. In this case, $p(a) = (a – a) g(a) = 0$.
      *Example 6:
      Examine whether $x + 2$ is a factor of $x^3 + 3x^2 + 5x + 6$ and of $2x + 4$.
      Solution : The zero of $x + 2$ is –2. Let $p(x) = x^3 + 3x^2 + 5x + 6$ and $s(x) = 2x + 4$
      Then, $p(–2) = (–2)^3 + 3(–2)^2 + 5(–2) + 6$
      = –8 + 12 – 10 + 6 = 0
      So, by the Factor Theorem, $x + 2$ is a factor of $x^3 + 3x^2 + 5x + 6$.
      Again, $s(–2) = 2(–2) + 4 = 0$
      So, $x + 2$ is a factor of $2x + 4$.
      In fact, you can check this without applying the Factor Theorem, since $2x + 4 = 2(x + 2)$.
      *Example 7:
      Find the value of k, if $x – 1$ is a factor of $4x^3 + 3x^2 – 4x + k$.
      Solution : As $x – 1$ is a factor of $p(x) = 4x^3 + 3x^2 – 4x + k$, $p(1) = 0$
      Now, $p(1) = 4(1)^3 + 3(1)^2 – 4(1) + k$
      So, $4 + 3 – 4 + k = 0$
      i.e., $k = –3$
• Factorizing quadratic polynomials of the form $ax^2 + bx + c$ by splitting the middle term.
  • If $ax^2 + bx + c = (px + q) (rx + s)$, then $a = pr$, $b = ps + qr$, and $c = qs$.
  • To factorize $ax^2 + bx + c$, write $b$ as the sum of two numbers whose product is $ac$.
    *Example 8:
    Factorise $6x^2 + 17x + 5$ by splitting the middle term, and by using the Factor Theorem.
    Solution 1 : (By splitting method) : If we can find two numbers p and q such that
    $p + q = 17$ and $pq = 6 × 5 = 30$, then we can get the factors.
    So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us $p + q = 17$.
    $6x^2 + 17x + 5 = 6x^2 + (2 + 15)x + 5$
    $= 6x^2 + 2x + 15x + 5$
    $= 2x(3x + 1) + 5(3x + 1)$
    $= (3x + 1) (2x + 5)$
    Solution 2 : (Using the Factor Theorem)
    $6x^2 + 17x + 5 = \frac{6}{6} (6x^2 + 17x + 5)$
    $= \frac{1}{6} (36x^2 + 102x + 30)$
    $= \frac{1}{6} (36x^2 + 12x + 90x + 30)$
    $= \frac{1}{6} (6x (6x+2) + 30 (3x +1))$
    $= \frac{1}{6} (6x + 30) (6x+2)$
    $= (x + 5) (3x + 1)$

*Example 9:
Factorise $y^2 – 5y + 6$ by using the Factor Theorem.
Solution : Let $p(y) = y^2 – 5y + 6$. Now, if $p(y) = (y – a) (y – b)$, you know that the constant term will be ab. So, $ab = 6$. So, to look for the factors of $p(y)$, we look at the factors of 6. The factors of 6 are 1, 2 and 3. Now,
$p(2) = 2^2 – (5 × 2) + 6 = 0$
So, $y – 2$ is a factor of $p(y)$.
Also, $p(3) = 3^2 – (5 × 3) + 6 = 0$
So, $y – 3$ is also a factor of $y^2 – 5y + 6$.
Therefore, $y^2 – 5y + 6 = (y – 2)(y – 3)$
Note that $y^2 – 5y + 6$ can also be factorised by splitting the middle term –5y.
*Example 10:
Factorise $x^3 – 23x^2 + 142x – 120$.
Solution : Let $p(x) = x^3 – 23x^2 + 142x – 120$
We shall now look for all the factors of –120. Some of these are$\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 60$.
By trial, we find that $p(1) = 0$. So $x – 1$ is a factor of $p(x)$.
Now we see that

$x^3 – 23x^2 + 142x – 120 = x^3 – x^2 – 22x^2 + 22x + 120x – 120$

$= x^2(x –1) – 22x(x – 1) + 120(x – 1)$

$= (x – 1) (x^2 – 22x + 120)$

Now $x^2 – 22x + 120$ can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
$x^2 – 22x + 120 = x^2 – 12x – 10x + 120$

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

Exercise 2.3

• Determining if $(x + 1)$ is a factor of given polynomials.
• Using the Factor Theorem to determine if $g(x)$ is a factor of $p(x)$.
• Finding the value of $k$ if $(x – 1)$ is a factor of $p(x)$.
• Factorizing given quadratic polynomials.
• Factorizing given cubic polynomials.

Algebraic Identities

• Algebraic identity: an algebraic equation true for all variable values.
• Previously 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: $(x + a) (x + b) = x^2 + (a + b)x + ab$
    *Example 11:
    Find the following products using appropriate identities:
    (i) $(x + 3) (x + 3)$
    (ii) $(x – 3) (x + 5)$
    Solution :
    (i) Here we can use Identity I : $(x + y)^2 = x^2 + 2xy + y^2$. Putting $y = 3$ in it, we get
    $(x + 3) (x + 3) = (x + 3)^2 = x^2 + 2(x)(3) + (3)^2$
    $= x^2 + 6x + 9$
    (ii) Using Identity IV above, i.e., $(x + a) (x + b) = x^2 + (a + b)x + ab$, we have
    $(x – 3) (x + 5) = x^2 + (–3 + 5)x + (–3)(5)$
    $= x^2 + 2x – 15$
    *Example 12:
    Evaluate $105 × 106$ without multiplying directly.
    Solution : $105 × 106 = (100 + 5) × (100 + 6)$
    $= (100)^2 + (5 + 6) (100) + (5 × 6)$, using Identity IV
    $= 10000 + 1100 + 30$
    $= 11130$
• Using identities for factorization.
  *Example 13:
  Factorise:
  (i) $49a^2 + 70ab + 25b^2$
  (ii) $\frac{25}{4} x^2 – \frac{9}{4} y^2$
  Solution :
  (i) Here you can see that
  $49a^2 = (7a)^2, 25b^2 = (5b)^2, 70ab = 2(7a) (5b)$
  Comparing the given expression with $x^2 + 2xy + y^2$, we observe that
  $x = 7a$ and $y = 5b$.
  Using Identity I, we get
  $49a^2 + 70ab + 25b^2 = (7a + 5b)^2$
  $= (7a + 5b) (7a + 5b)$
  (ii) We have
  $\frac{25}{4} x^2 – \frac{9}{4} y^2 = (\frac{5}{2} x )^2 – (\frac{3}{2} y)^2$
  Now comparing it with Identity III, we get
  $\frac{25}{4} x^2 – \frac{9}{4} y^2 = (\frac{5}{2} x + \frac{3}{2} y) (\frac{5}{2} x - \frac{3}{2} y)$
• Extending Identity I to a trinomial: $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
  *Example 14:
  Write $(3a + 4b + 5c)^2$ in expanded form.
  Solution : Comparing the given expression with $(x + y + z)^2$, we find that $x = 3a, y = 4b$ and $z = 5c$. Therefore, using Identity V, we have
  $(3a + 4b + 5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)$
  $= 9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ac$
  *Example 15:
  Expand $(4a – 2b – 3c)^2$.
  Solution : Using Identity V, we have
  $(4a – 2b – 3c)^2 = [4a + (–2b) + (–3c)]^2$
  $= (4a)^2 + (–2b)^2 + (–3c)^2 + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a)$
  $= 16a^2 + 4b^2 + 9c^2 – 16ab + 12bc – 24ac$
  *Example 16:
  Factorise $4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz$.
  Solution : We have
  $4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz = (2x)^2 + (–y)^2 + (z)^2 + 2(2x)(–y) + 2(–y)(z) + 2(2x)(z)$
  $= [2x + (–y) + z]^2$
  $= (2x – y + z)^2 = (2x – y + z)(2x – y + z)$
• Extending Identity I to compute $(x + y)^3$.
  • Identity VI: $(x + y)^3 = x^3 + y^3 + 3xy (x + y)$
  • Identity VII: $(x – y)^3 = x^3 – y^3 – 3xy(x – y) = x^3 – 3x^2y + 3xy^2 – y^3$
    *Example 17:
    Write the following cubes in the expanded form:
    (i) $(3a + 4b)^3$
    (ii) $(5p – 3q)^3$
    Solution :
    (i) Comparing the given expression with $(x + y)^3$, we find that $x = 3a$ and $y = 4b$. So, using Identity VI, we have:
    $(3a + 4b)^3 = (3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b)$
    $= 27a^3 + 64b^3 + 108a^2b + 144ab^2$
    (ii) Comparing the given expression with $(x – y)^3$, we find that $x = 5p, y = 3q$. So, using Identity VII, we have:
    $(5p – 3q)^3 = (5p)^3 – (3q)^3 – 3(5p)(3q)(5p – 3q)$
    $= 125p^3 – 27q^3 – 225p^2q + 135pq^2$
    *Example 18:
    Evaluate each of the following using suitable identities:
    (i) $(104)^3$
    (ii) $(999)^3$
    Solution :
    (i) We have $(104)^3 = (100 + 4)^3$
    $= (100)^3 + (4)^3 + 3(100)(4)(100 + 4)$
    $= 1000000 + 64 + 124800$
    $= 1124864$
    (ii) We have $(999)^3 = (1000 – 1)^3$
    $= (1000)^3 – (1)^3 – 3(1000)(1)(1000 – 1)$
    $= 1000000000 – 1 – 2997000$
    $= 997002999$
    *Example 19:
    Factorise $8x^3 + 27y^3 + 36x^2y + 54xy^2$
    Solution : The given expression can be written as
    $(2x)^3 + (3y)^3 + 3(4x^2)(3y) + 3(2x)(9y^2)$
    $= (2x)^3 + (3y)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2$
    $= (2x + 3y)^3$
    $= (2x + 3y)(2x + 3y)(2x + 3y)$
• Identity VIII: $x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)$
  *Example 20:
  Factorise : $8x^3 + y^3 + 27z^3 – 18xyz$
  Solution : Here, we have $8x^3 + y^3 + 27z^3 – 18xyz = (2x)^3 + y^3 + (3z)^3 – 3(2x)(y)(3z)$
  $= (2x + y + 3z)[(2x)^2 + y^2 + (3z)^2 – (2x)(y) – (y)(3z) – (2x)(3z)]$
  $= (2x + y + 3z) (4x^2 + y^2 + 9z^2 – 2xy – 3yz – 6xz)$

Exercise 2.4

• Using suitable identities to find products.
• Evaluating products without direct multiplication.
• Factorizing using appropriate identities.
• Expanding using suitable identities.
• Writing cubes in expanded form.
• Evaluating using suitable identities.
• Factorizing given expressions.
• Verifying given identities.
• Factorizing given expressions [Hint : See Question 9.]
  *If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
  *Without actually calculating the cubes, find the value of each of the following:
  (i) (–12)3 + (7)3 + (5)3
  (ii) (28)3 + (–15)3 + (–13)3
  *Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
  Area : 25a2 – 35a + 12 Area : 35y2 + 13y –12

*What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
Volume : 3x2 – 12x Volume : 12ky2 + 8ky – 20k

Summary

• A polynomial $p(x)$ in one variable $x$ is an algebraic expression 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 $a</em>n ≠ 0$.
• $a<em>0, a</em>1, a<em>2, …, a</em>n$ are the coefficients of $x^0, x, x^2, …, x^n$, and $n$ is the degree of the polynomial.
• Each of $a<em>n x^n, a</em>{n-1} x^{n-1}, …, a_0$, with $$