CBSE Class 10 Polynomials Sure Shot Questions Set 07

FACTORS AND COEFFICIENTS

Factor :
Each combination of the constants and variables, which form a term, is called a factor.

For examples :
(i) 7, x and 7x are factors of 7x, in which 7 is constant (numerical) factor and x is variable (literal) factor.
(ii) In \( -5x^2y \), the numerical factor is \( -5 \) and literal factors are : x, y, xy, \( x^2 \) and \( x^2y \).

Coefficient :

Any factor of a term is called the coefficient of the remaining term.

For example :
(i) In 7x ; 7 is coefficient of x
(ii) In \( -5x^2y \); 5 is coefficient of \( -x^2y \); \( -5 \) is coefficient of \( x^2y \).

Question. Write the coefficient of :
(i) \( x^2 \) in \( 3x^3 - 5x^2 + 7 \)
(ii) xy in 8xyz
(iii) –y in \( 2y^2 - 6y + 2 \)
(iv) \( x^0 \) in \( 3x + 7 \)
Answer: (i) –5
(ii) 8z
(iii) 6
(iv) Since \( x^0 = 1 \), Therefore \( 3x + 7 = 3x + 7x^0 \), coefficient of \( x^0 \) is 7.

DEGREE OF A POLYNOMIAL

The greatest power (exponent) of the terms of a polynomial is called degree of the polynomial.

For example :
(a) In polynomial \( 5x^2 - 8x^7 + 3x \):
(i) The power of term \( 5x^2 = 2 \)
(ii) The power of term \( -8x^7 = 7 \)
(iii) The power of term 3x = 1
Since, the greatest power is 7, therefore degree of the polynomial \( 5x^2 - 8x^7 + 3x \) is 7
(b) The degree of polynomial :
(i) \( 4y^3 - 3y + 8 \) is 3
(ii) \( 7p + 2 \) is 1 (\( p = p^1 \))
(iii) \( 2m - 7m^8 + m^{13} \) is 13 and so on.

Question. Find which of the following algebraic expression is a polynomial.
(i) \( 3x^2 - 5x \)
(ii) \( x + \frac{1}{x} \)
(iii) \( \sqrt{y} - 8 \)
(iv) \( z^5 - \sqrt[3]{z} + 8 \)
Answer: (i) \( 3x^2 - 5x = 3x^2 - 5x^1 \). It is a polynomial.
(ii) \( x + \frac{1}{x} = x^1 + x^{-1} \). It is not a polynomial.
(iii) \( \sqrt{y} - 8 = y^{1/2} - 8 \). Since, the power of the first term (\( \sqrt{y} \)) is \( \frac{1}{2} \), which is not a whole number.
(iv) \( z^5 - \sqrt[3]{z} + 8 = z^5 - z^{1/3} + 8 \). Since, the exponent of the second term is 1/3, which is not a whole number. Therefore, the given expression is not a polynomial.

Question. Find the degree of the polynomial :
(i) \( 5x - 6x^3 + 8x^7 + 6x^2 \)
(ii) \( 2y^{12} + 3y^{10} - y^{15} + y + 3 \)
(iii) x
(iv) 8
Answer: (i) Since the term with highest exponent (power) is \( 8x^7 \) and its power is 7. ∴ The degree of given polynomial is 7.
(ii) The highest power of the variable is 15 ⇒ degree = 15.
(iii) \( x = x^1 \) ⇒ degree is 1.
(iv) \( 8 = 8x^0 \) ⇒ degree = 0

TYPES OF POLYNOMIALS

(A) Based on degree :
If degree of polynomial is:
1. One: Linear. Examples: \( x + 3, y - x + 2, \sqrt{3}x - 3 \)
2. Two: Quadratic. Examples: \( 2x^2 - 7, \frac{1}{3}x^2 + y^2 - 2xy, x^2 + 1 + 3y \)
3. Three: Cubic. Examples: \( x^3 + 3x^2 - 7x + 8, 2x^2 + 5x^3 + 7 \)
4. Four: bi-quadratic. Examples: \( x^4 + y^4 + 2x^2y^2, x^4 + 3, \dots \)

(B) Based on Terms :
If number of terms in polynomial is:
1. One: Monomial. Examples: \( 7x, 5x^9, \frac{7}{3}x^{16}, xy, \dots \)
2. Two: Binomial. Examples: \( 2 + 7y^6, y^3 + x^{14}, 7 + 5x^9, \dots \)
3. Three: Trinomial. Examples: \( x^3 - 2x + y, x^{31} + y^{32} + z^{33}, \dots \)

Note : (1) Degree of constant polynomials (Ex. 5, 7, –3, 8/5, …) is zero.
(2) Degree of zero polynomial (zero = 0 = zero polynomial) is not defined.

POLYNOMIAL IN ONE VARIABLE

If a polynomial has only one variable then it is called polynomial in one variable.

Examples:
\( P(x) = 2x^3 + 5x - 3 \) Cubic trinomial
\( Q(x) = 7x^7 - 5x^5 - 3x^3 + x + 3 \) polynomial of degree 7
\( R(y) = y \) Linear, monomial
\( S(t) = t^2 + 3 \) Quadratic Binomial

Note : General form of a polynomial in one variable x of degree 'n' is \( a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \dots + a_2x^2 + a_1x + a_0, a_n \neq 0 \), where \( a_n, a_{n-1}, \dots a_2, a_1, a_0 \) all are constants.
∴ for linear \( ax + b, a \neq 0 \)
for quadratic \( ax^2 + bx + c, a \neq 0 \)
for cubic \( ax^3 + bx^2 + cx + d, a \neq 0 \)

REMAINDER THEOREM

• (i) Remainder obtained on dividing polynomial p(x) by x – a is equal to p(a).
• (ii) If a polynomial p(x) is divided by (x + a) the remainder is the value of p(x) at x = –a.
• (iii) (x – a) is a factor of polynomial p(x) if p(a) = 0
• (iv) (x + a) is a factor of polynomial p(x) if p(–a) = 0
• (v) (x – a)(x – b) is a factor of polynomial p(x), if p(a) = 0 and p(b) = 0.

Question. Find the remainder when \( 4x^3 - 3x^2 + 2x - 4 \) is divided by
(a) \( x - 1 \)
(b) \( x + 2 \)
(c) \( x + \frac{1}{2} \)
Answer: Let \( p(x) = 4x^3 - 3x^2 + 2x - 4 \)
(a) When p(x) is divided by (x – 1), then by remainder theorem, the required remainder will be p(1)
\( p(1) = 4(1)^3 - 3(1)^2 + 2(1) - 4 \)
\( = 4 \times 1 - 3 \times 1 + 2 \times 1 - 4 \)
\( = 4 - 3 + 2 - 4 = -1 \)
(b) When p(x) is divided by (x + 2), then by remainder theorem, the required remainder will be p(–2).
\( p(-2) = 4(-2)^3 - 3(-2)^2 + 2(-2) - 4 \)
\( = 4 \times (-8) - 3 \times 4 - 4 - 4 \)
\( = -32 - 12 - 8 = -52 \)
(c) When p(x) is divided by \( x + \frac{1}{2} \), then by remainder theorem, the required remainder will be \( p(-\frac{1}{2}) \)
\( p(-\frac{1}{2}) = 4(-\frac{1}{2})^3 - 3(-\frac{1}{2})^2 + 2(-\frac{1}{2}) - 4 \)
\( = 4 \times (-\frac{1}{8}) - 3 \times \frac{1}{4} - 2 \times \frac{1}{2} - 4 \)
\( = -\frac{1}{2} - \frac{3}{4} - 1 - 4 = -\frac{1}{2} - \frac{3}{4} - 5 \)
\( = \frac{-2 - 3 - 20}{4} = \frac{-25}{4} \)

VALUES OF A POLYNOMIAL

For a polynomial \( f(x) = 3x^2 - 4x + 2 \).
To find its value at x = 3; replace x by 3 everywhere.
So, the value of \( f(x) = 3x^2 - 4x + 2 \) at x = 3 is \( f(3) = 3 \times 3^2 - 4 \times 3 + 2 = 27 - 12 + 2 = 17 \).
Similarly, the value of polynomial \( f(x) = 3x^2 - 4x + 2 \),
(i) at x = –2 is \( f(-2) = 3(-2)^2 - 4(-2) + 2 = 12 + 8 + 2 = 22 \)
(ii) at x = 0 is \( f(0) = 3(0)^2 - 4(0) + 2 = 0 - 0 + 2 = 2 \)
(iii) at x = \( \frac{1}{2} \) is \( f(\frac{1}{2}) = 3(\frac{1}{2})^2 - 4(\frac{1}{2}) + 2 = \frac{3}{4} - 2 + 2 = \frac{3}{4} \)

Question. Find the value of the polynomial \( 5x - 4x^2 + 3 \) at:
(i) x = 0
(ii) x = –1
Answer: Let \( p(x) = 5x - 4x^2 + 3 \).
(i) At x = 0, \( p(0) = 5 \times 0 - 4 \times (0)^2 + 3 = 0 - 0 + 3 = 3 \)
(ii) At x = –1, \( p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4 + 3 = - 6 \)

ZEROES OF A POLYNOMIAL

If for x = a, the value of the polynomial p(x) is 0 i.e., p(a) = 0; then x = a is a zero of the polynomial p(x).
For example :
(i) For polynomial \( p(x) = x - 2; p(2) = 2 - 2 = 0 \)
∴ x = 2 or simply 2 is a zero of the polynomial \( p(x) = x - 2 \).
(ii) For the polynomial \( g(u) = u^2 - 5u + 6; g(3) = (3)^2 - 5 \times 3 + 6 = 9 - 15 + 6 = 0 \)
∴ 3 is a zero of the polynomial \( g(u) = u^2 - 5u + 6 \).
Also, \( g(2) = (2)^2 - 5 \times 2 + 6 = 4 - 10 + 6 = 0 \)
∴ 2 is also a zero of the polynomial \( g(u) = u^2 - 5u + 6 \)

(a) Every linear polynomial has one and only one zero.

(b) A given polynomial may have more than one zeroes.

(c) If the degree of a polynomial is n; the largest number of zeroes it can have is also n.
For example : If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes; if the degree of a polynomial is 8; largest number of zeroes it can have is 8.

(d) A zero of a polynomial need not be 0.
For example : If \( f(x) = x^2 - 4 \), then \( f(2) = (2)^2 - 4 = 4 - 4 = 0 \). Here, zero of the polynomial \( f(x) = x^2 - 4 \) is 2 which itself is not 0.

(e) 0 may be a zero of a polynomial.
For example : If \( f(x) = x^2 - x \), then \( f(0) = 0^2 - 0 = 0 \). Here 0 is the zero of polynomial \( f(x) = x^2 - x \).

Question. Verify whether the indicated numbers are zeroes of the polynomial corresponding to them in the following cases :
(i) \( p(x) = 3x + 1, x = -\frac{1}{3} \)
(ii) \( p(x) = (x + 1)(x - 2), x = -1, 2 \)
(iii) \( p(x) = x^2, x = 0 \)
(iv) \( p(x) = \lambda x + m, x = -\frac{m}{\lambda} \)
(v) \( p(x) = 2x + 1, x = \frac{1}{2} \)
Answer: (i) \( p(x) = 3x + 1 \Rightarrow p(-\frac{1}{3}) = 3 \times (-\frac{1}{3}) + 1 = -1 + 1 = 0 \). ∴ \( x = -\frac{1}{3} \) is a zero of \( p(x) = 3x + 1 \).
(ii) \( p(x) = (x + 1)(x - 2) \Rightarrow p(-1) = (-1 + 1)(-1 - 2) = 0 \times -3 = 0 \) and, \( p(2) = (2 + 1)(2 - 2) = 3 \times 0 = 0 \). ∴ x = –1 and x = 2 are zeroes of the given polynomial.
(iii) \( p(x) = x^2 \Rightarrow p(0) = 0^2 = 0 \). ∴ x = 0 is a zero of the given polynomial.
(iv) \( p(x) = \lambda x + m \Rightarrow p(-\frac{m}{\lambda}) = \lambda(-\frac{m}{\lambda}) + m = -m + m = 0 \). ∴ \( x = -\frac{m}{\lambda} \) is a zero of the given polynomial.
(v) \( p(x) = 2x + 1 \Rightarrow p(\frac{1}{2}) = 2 \times \frac{1}{2} + 1 = 1 + 1 = 2 \neq 0 \). ∴ \( x = \frac{1}{2} \) is not a zero of the given polynomial.

Question. Find the zero of the polynomial in each of the following cases :
(i) \( p(x) = x + 5 \)
(ii) \( p(x) = 2x + 5 \)
(iii) \( p(x) = 3x - 2 \)
Answer: To find the zero of a polynomial p(x) means to solve the polynomial equation p(x) = 0.
(i) For the zero of polynomial \( p(x) = x + 5 \): \( p(x) = 0 \Rightarrow x + 5 = 0 \Rightarrow x = -5 \). ∴ x = –5 is a zero of the polynomial \( p(x) = x + 5 \).
(ii) \( p(x) = 0 \Rightarrow 2x + 5 = 0 \Rightarrow 2x = -5 \) and \( x = -\frac{5}{2} \). ∴ \( x = -\frac{5}{2} \) is a zero of \( p(x) = 2x + 5 \).
(iii) \( p(x) = 0 \Rightarrow 3x - 2 = 0 \Rightarrow 3x = 2 \) and \( x = \frac{2}{3} \). ∴ \( x = \frac{2}{3} \) is zero of \( p(x) = 3x - 2 \).

RELATIONSHIP BETWEEN THE ZEROES AND THE COEFFICIENTS OF A POLYNOMIAL

Consider quadratic polynomial \( P(x)=2x^2–16x+ 30 \).
Now, \( 2x^2 – 16x + 30 = (2x – 6) (x – 3) = 2 (x – 3) (x – 5) \)
The zeroes of \( P(x) \) are 3 and 5.
Sum of the zeroes \( = 3 + 5 = 8 = \frac{-(-16)}{2} = - \left[ \frac{\text{coefficient of } x}{\text{coefficient of } x^2} \right] \)
Product of the zeroes \( = 3 \times 5 = 15 = \frac{30}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2} \)
So if \( ax^2 + bx + c, a \neq 0 \) is a quadratic polynomial and \( \alpha, \beta \) are two zeroes of polynomial then \( \alpha + \beta = -\frac{b}{a} , \alpha\beta = \frac{c}{a} \)

Question. Find the zeroes of the quadratic polynomial \( 6x^2 – 13x + 6 \) and verify the relation between the zeroes and its coefficients.
Answer: We have, \( 6x^2 – 13x + 6 = 6x^2 – 4x – 9x + 6 \)
\( = 2x (3x – 2) – 3 (3x – 2) = (3x – 2) (2x – 3) \)
So, the value of \( 6x^2 – 13x + 6 \) is 0, when \( (3x – 2) = 0 \) or \( (2x – 3) = 0 \) i.e.,
When \( x = \frac{2}{3} \) or \( \frac{3}{2} \)
Therefore, the zeroes of \( 6x^2 – 13x + 6 \) are \( \frac{2}{3} \) and \( \frac{3}{2} \).
Sum of the zeroes \( = \frac{2}{3} + \frac{3}{2} = \frac{13}{6} = \frac{-(-13)}{6} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
Product of the zeroes \( = \frac{2}{3} \times \frac{3}{2} = \frac{6}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2} \)

Question. Find the zeroes of the quadratic polynomial \( 4x^2 – 9 \) and verify the relation between the zeroes and its coefficients.
Answer: We have, \( 4x^2 – 9 = (2x)^2 – 3^2 = (2x – 3) (2x + 3) \)
So, the value of \( 4x^2 – 9 \) is 0, when \( 2x – 3 = 0 \) or \( 2x + 3 = 0 \)
i.e., when \( x = \frac{3}{2} \) or \( x = -\frac{3}{2} \).
Therefore, the zeroes of \( 4x^2 – 9 \) are \( \frac{3}{2} \) & \( -\frac{3}{2} \).
Sum of the zeroes \( = \frac{3}{2} – \frac{3}{2} = 0 = \frac{-(0)}{4} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
Product of the zeroes \( = \left( \frac{3}{2} \right) \left( -\frac{3}{2} \right) = \frac{-9}{4} = \frac{\text{constant term}}{\text{coefficient of } x^2} \)

Question. Find the zeroes of the quadratic polynomial \( 9x^2 – 5 \) and verify the relation between the zeroes and its coefficients.
Answer: We have, \( 9x^2 – 5 = (3x)^2 – (\sqrt{5})^2 = (3x – \sqrt{5}) (3x + \sqrt{5}) \)
So, the value of \( 9x^2 – 5 \) is 0, when \( 3x – \sqrt{5} = 0 \) or \( 3x + \sqrt{5} = 0 \)
i.e., when \( x = \frac{\sqrt{5}}{3} \) or \( x = \frac{-\sqrt{5}}{3} \).
Sum of the zeroes \( = \frac{\sqrt{5}}{3} - \frac{\sqrt{5}}{3} = 0 = \frac{-(0)}{9} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
Product of the zeroes \( = \left( \frac{\sqrt{5}}{3} \right) \left( \frac{-\sqrt{5}}{3} \right) = \frac{-5}{9} = \frac{\text{constant term}}{\text{coefficient of } x^2} \)

Question. If \( \alpha \) and \( \beta \) are the zeroes of \( ax^2 + bx + c, a \neq 0 \) then verify the relation between the zeroes and its coefficients.
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of polynomial \( ax^2 + bx + c \).
Therefore, \( (x – \alpha), (x – \beta) \) are the factors of the polynomial \( ax^2 + bx + c \).
\( \Rightarrow ax^2 + bx + c = k (x – \alpha) (x – \beta) \)
\( \Rightarrow ax^2 + bx + c = k \{x^2 – (\alpha + \beta) x + \alpha\beta\} \)
\( \Rightarrow ax^2 + bx + c = kx^2 – k (\alpha + \beta) x + k\alpha\beta \dots (1) \)
Comparing the coefficients of \( x^2, x \) and constant terms of (1) on both sides, we get
\( a = k, b = – k (\alpha + \beta) \) and \( c = k\alpha\beta \)
\( \Rightarrow \alpha + \beta = -\frac{b}{k} \) and \( \alpha\beta = \frac{c}{k} \)
\( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \) [∵ k = a]
Sum of the zeroes \( = \frac{-b}{a} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
Product of the zeroes \( = \frac{c}{a} = \frac{\text{constant term}}{\text{coefficient of } x^2} \)

Question. Prove relation between the zeroes and the coefficient of the quadratic polynomial \( ax^2 + bx + c \).
Answer: Let \( \alpha \) and \( \beta \) be the zeroes of the polynomial \( ax^2 + bx + c \)
\( \therefore \alpha = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \dots (1) \)
\( \beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \dots (2) \)
By adding (1) and (2), we get
\( \alpha + \beta = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-2b}{2a} = -\frac{b}{a} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
Hence, sum of the zeroes of the polynomial \( ax^2 + bx + c \) is \( -\frac{b}{a} \)
By multiplying (1) and (2), we get
\( \alpha\beta = \left( \frac{-b + \sqrt{b^2 - 4ac}}{2a} \right) \left( \frac{-b - \sqrt{b^2 - 4ac}}{2a} \right) \)
\( = \frac{(-b)^2 - (\sqrt{b^2 - 4ac})^2}{4a^2} = \frac{b^2 - b^2 + 4ac}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a} \)
\( = \frac{\text{constant term}}{\text{coefficient of } x^2} \)
Hence, product of zeroes \( = \frac{c}{a} \)

In general, it can be proved that if \( \alpha, \beta, \gamma \) are the zeroes of a cubic polynomial \( ax^3 + bx^2 + cx + d \), then
\( \alpha + \beta + \gamma = \frac{-b}{a} \)
\( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \)
\( \alpha\beta\gamma = \frac{-d}{a} \)
Note, \( \frac{b}{a}, \frac{c}{a} \) and \( \frac{d}{a} \) are meaningful because \( a \neq 0 \).

Question. find the zeroes of the quadratic polynomial \( x^2 – 2x – 8 \) and verify a relationship between zeroes and its coefficients.
Answer: \( x^2 – 2x – 8 = x^2 – 4x + 2x – 8 = x (x – 4) + 2 (x – 4) = (x – 4) (x + 2) \)
So, the value of \( x^2 – 2x – 8 \) is zero when \( x – 4 = 0 \) or \( x + 2 = 0 \) i.e., when \( x = 4 \) or \( x = – 2 \).
So, the zeroes of \( x^2 – 2x – 8 \) are 4, – 2.
Sum of the zeroes \( = 4 – 2 = 2 = \frac{-(-2)}{1} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
Product of the zeroes \( = 4 (-2) = –8 = \frac{-8}{1} = \frac{\text{constant term}}{\text{coefficient of } x^2} \)

Question. Verify that the numbers given along side of the cubic polynomials are their zeroes. Also verify the relationship between the zeroes and the coefficients. \( 2x^3 + x^2 – 5x + 2 ; \frac{1}{2}, 1, – 2 \)
Answer: Here, the polynomial \( p(x) \) is \( 2x^3 + x^2 – 5x + 2 \)
Value of the polynomial \( 2x^3 + x^2 – 5x + 2 \) when x = 1/2
\( = 2 \left( \frac{1}{2} \right)^3 + \left( \frac{1}{2} \right)^2 - 5 \left( \frac{1}{2} \right) + 2 = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = 0 \)
So, 1/2 is a zero of \( p(x) \).
On putting x = 1 in the cubic polynomial \( 2x^3 + x^2 – 5x + 2 = 2(1)^3 + (1)^2 – 5(1) + 2 = 2 + 1 – 5 + 2 = 0 \)
On putting x = – 2 in the cubic polynomial \( 2x^3 + x^2 – 5x + 2 = 2(–2)^3 + (–2)^2 – 5 (–2) + 2 = – 16 + 4 + 10 + 2 = 0 \)
Hence, \( \frac{1}{2}, 1, – 2 \) are the zeroes of the given polynomial.
Sum of the zeroes of \( p(x) = \frac{1}{2} + 1 - 2 = -\frac{1}{2} = \frac{-\text{coefficient of } x^2}{\text{coefficient of } x^3} \)
Sum of the products of two zeroes taken at a time \( = \frac{1}{2} \times 1 + \frac{1}{2} \times (-2) + 1 \times (-2) = \frac{1}{2} - 1 - 2 = -\frac{5}{2} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} \)
Product of all the three zeroes \( = \left( \frac{1}{2} \right) \times (1) \times (-2) = -1 = \frac{-(2)}{2} = \frac{-\text{constant term}}{\text{coefficient of } x^3} \)

SYMMETRIC FUNCTIONS OF ZEROS OF A QUADRATIC POLYNOMIAL

Symmetric function :
An algebraic expression in \( \alpha \) and \( \beta \), which remains unchanged, when \( \alpha \) and \( \beta \) are interchanged is known as symmetric function in \( \alpha \) and \( \beta \).
For example, \( \alpha^2 + \beta^2 \) and \( \alpha^3 + \beta^3 \) etc. are symmetric functions. Symmetric function is to be expressed in terms of \( (\alpha + \beta) \) and \( \alpha\beta \). So, this can be evaluated for a given quadratic equation.

Some useful relations involving \( \alpha \) and \( \beta \) :

1. \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta \)

2. \( (\alpha – \beta)^2 = (\alpha + \beta)^2 – 4\alpha\beta \)

3. \( \alpha^2 – \beta^2 = (\alpha + \beta) (\alpha – \beta) = (\alpha + \beta) \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} \)

4. \( \alpha^3 + \beta^3 = (\alpha + \beta)^3 – 3\alpha\beta (\alpha + \beta) \)

5. \( \alpha^3 – \beta^3 = (\alpha – \beta)^3 + 3\alpha\beta (\alpha – \beta) \)

6. \( \alpha^4 + \beta^4 = [(\alpha + \beta)^2 – 2\alpha\beta]^2 – 2(\alpha\beta)^2 \)

7. \( \alpha^4 – \beta^4 = (\alpha^2 + \beta^2) (\alpha^2 – \beta^2) \) then use (1) and (3)

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( ax^2 + bx + c \). Find the value of (i) \( \alpha – \beta \) (ii) \( \alpha^2 + \beta^2 \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( ax^2 + bx + c \).
\( \therefore \alpha + \beta = -\frac{b}{a} ; \alpha\beta = \frac{c}{a} \)
(i) \( (\alpha – \beta)^2 = (\alpha + \beta)^2 – 4\alpha\beta = \left( -\frac{b}{a} \right)^2 - \frac{4c}{a} = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2} \)
\( \alpha – \beta = \frac{\sqrt{b^2 - 4ac}}{a} \)
(ii) \( \alpha^2 + \beta^2 = \alpha^2 + \beta^2 + 2\alpha\beta – 2\alpha\beta = (\alpha + \beta)^2 – 2\alpha\beta = \left( -\frac{b}{a} \right)^2 - 2 \left( \frac{c}{a} \right) = \frac{b^2 - 2ac}{a^2} \)

Question. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( ax^2 + bx + c \). Find the value of (i) \( \alpha^2 – \beta^2 \) (ii) \( \alpha^3 + \beta^3 \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of \( ax^2 + bx + c \)
\( \therefore \alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a} \)
(i) \( \alpha^2 – \beta^2 = (\alpha + \beta) (\alpha – \beta) = -\frac{b}{a} \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = -\frac{b}{a} \sqrt{\left(-\frac{b}{a}\right)^2 - 4\frac{c}{a}} = -\frac{b}{a} \frac{\sqrt{b^2 - 4ac}}{a} = -\frac{b \sqrt{b^2 - 4ac}}{a^2} \)
(ii) \( \alpha^3 + \beta^3 = (\alpha + \beta) (\alpha^2 + \beta^2 – \alpha\beta) = (\alpha + \beta) [(\alpha^2 + \beta^2 + 2\alpha\beta) – 3\alpha\beta] = (\alpha + \beta) [(\alpha + \beta)^2 – 3\alpha\beta] \)
\( = -\frac{b}{a} \left[ \left(-\frac{b}{a}\right)^2 - \frac{3c}{a} \right] = -\frac{b}{a} \left[ \frac{b^2 - 3ac}{a^2} \right] = \frac{-b^3 + 3abc}{a^3} \)

TO FORM A QUADRATIC POLYNOMIAL WITH THE GIVEN ZEROES

Let zeroes of a quadratic polynomial be \( \alpha \) and \( \beta \).
\( \therefore x = \alpha, x = \beta \)
\( x – \alpha = 0, x – \beta = 0 \)
The obviously the quadratic polynomial is \( (x – \alpha) (x – \beta) \) i.e., \( x^2 – (\alpha + \beta) x + \alpha\beta \)
\( x^2 - (\text{Sum of the zeroes}) x + \text{Product of the zeroes} \)

Question. Form the quadratic polynomial whose zeroes are 4 and 6.
Answer: Sum of the zeroes \( = 4 + 6 = 10 \)
Product of the zeroes \( = 4 \times 6 = 24 \)
Hence the polynomial formed \( = x^2 – (\text{sum of zeroes}) x + \text{Product of zeroes} = x^2 – 10x + 24 \)

Question. Form the quadratic polynomial whose zeroes are –3, 5.
Answer: Here, zeroes are – 3 and 5.
Sum of the zeroes \( = – 3 + 5 = 2 \)
Product of the zeroes \( = (–3) \times 5 = – 15 \)
Hence the polynomial formed \( = x^2 – (\text{sum of zeroes}) x + \text{Product of zeroes} = x^2 – 2x – 15 \)

Question. Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively- (i) \( \frac{1}{4}, – 1 \) (ii) \( \sqrt{2}, \frac{1}{3} \) (iii) \( 0, \sqrt{5} \)
Answer: Let the polynomial be \( ax^2 + bx + c \) and its zeroes be \( \alpha \) and \( \beta \).
(i) Here, \( \alpha + \beta = \frac{1}{4} \) and \( \alpha . \beta = – 1 \)
Thus the polynomial formed \( = x^2 – (\text{Sum of zeroes})x + \text{Product of zeroes} = x^2 – \left( \frac{1}{4} \right) x – 1 = x^2 – \frac{x}{4} – 1 \)
The other polynomial are \( k \left( x^2 - \frac{x}{4} - 1 \right) \). If k = 4, then the polynomial is \( 4x^2 – x – 4 \).
(ii) Here, \( \alpha + \beta = \sqrt{2}, \alpha\beta = \frac{1}{3} \)
Thus the polynomial formed \( = x^2 – (\sqrt{2}) x + \frac{1}{3} \) or \( x^2 – \sqrt{2}x + \frac{1}{3} \). Other polynomial are \( k \left( x^2 - \sqrt{2}x + \frac{1}{3} \right) \). If k = 3, then the polynomial is \( 3x^2 – 3\sqrt{2}x + 1 \).
(iii) Here, \( \alpha + \beta = 0 \) and \( \alpha\beta = \sqrt{5} \)
Thus the polynomial formed \( = x^2 – (0) x + \sqrt{5} = x^2 + \sqrt{5} \)

Question. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time, and product of its zeroes as 2, – 7 and –14, respectively.
Answer: Let the cubic polynomial be \( ax^3 + bx^2 + cx + d \Rightarrow x^3 + \frac{b}{a} x^2 + \frac{c}{a} x + \frac{d}{a} \dots (1) \)
and its zeroes are \( \alpha, \beta \) and \( \gamma \), then
\( \alpha + \beta + \gamma = 2 = -\frac{b}{a} \)
\( \alpha\beta + \beta\gamma + \gamma\alpha = -7 = \frac{c}{a} \)
\( \alpha\beta\gamma = -14 = -\frac{d}{a} \)
Putting the values of \( \frac{b}{a}, \frac{c}{a} \) and \( \frac{d}{a} \) in (1), we get
\( x^3 + (–2) x^2 + (–7)x + 14 \Rightarrow x^3 – 2x^2 – 7x + 14 \)

Question. Find the cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and product of its zeroes as 0, –7 and –6 respectively.
Answer: Let the cubic polynomial be \( ax^3 + bx^2 + cx + d \Rightarrow x^3 + \frac{b}{a} x^2 + \frac{c}{a} x + \frac{d}{a} \dots (1) \)
and its zeroes are \( \alpha, \beta, \gamma \). Then \( \alpha + \beta + \gamma = 0 = -\frac{b}{a} \)
\( \alpha\beta + \beta\gamma + \gamma\alpha = -7 = \frac{c}{a} \)
\( \alpha\beta\gamma = -6 = \frac{-d}{a} \)
Putting the values of \( \frac{b}{a}, \frac{c}{a} \) and \( \frac{d}{a} \) in (1), we get
\( x^3 – (0) x^2 + (–7) x + (–6) \) or \( x^3 – 7x + 6 \)

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomials \( ax^2 + bx + c \) then form the polynomial whose zeroes are \( 1/\alpha \) and \( 1/\beta \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of \( ax^2 + bx + c \). So \( \alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a} \)
Sum of the zeroes \( = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c} \)
Product of the zeroes \( = \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c} \)
But required polynomial is \( x^2 – (\text{sum of zeroes}) x + \text{Product of zeroes} \)
\( \Rightarrow x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c} \) or \( x^2 + \frac{b}{c}x + \frac{a}{c} \) or \( c(x^2 + \frac{b}{c}x + \frac{a}{c}) \Rightarrow cx^2 + bx + a \)

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( x^2 + 4x + 3 \), form the polynomial whose zeroes are \( 1 + \beta/\alpha \) and \( 1 + \alpha/\beta \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( x^2 + 4x + 3 \). Then, \( \alpha + \beta = – 4, \alpha\beta = 3 \)
Sum of the zeroes \( = 1 + \frac{\beta}{\alpha} + 1 + \frac{\alpha}{\beta} = \frac{\alpha\beta + \beta^2 + \alpha\beta + \alpha^2}{\alpha\beta} = \frac{\alpha^2 + \beta^2 + 2\alpha\beta}{\alpha\beta} = \frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{(-4)^2}{3} = \frac{16}{3} \)
Product of the zeroes \( = (1 + \frac{\beta}{\alpha})(1 + \frac{\alpha}{\beta}) = 1 + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \frac{\alpha\beta}{\alpha\beta} = 2 + \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{2\alpha\beta + \alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{(-4)^2}{3} = \frac{16}{3} \)
But required polynomial is \( x^2 – (\text{sum of zeroes}) x + \text{product of zeroes} \)
or \( x^2 – \frac{16}{3}x + \frac{16}{3} \) or \( k(x^2 - \frac{16}{3}x + \frac{16}{3}) \) or \( 3(x^2 - \frac{16}{3}x + \frac{16}{3}) \) (if k = 3) \( \Rightarrow 3x^2 – 16x + 16 \)

WORKING RULE TO DIVIDE A POLYNOMIAL BY ANOTHER POLYNOMIAL

Step 1: First arrange the term of dividend and the divisor in the decreasing order of their degrees.
Step 2: To obtain the first term of quotient divide the highest degree term of the dividend by the highest degree term of the divisor.
Step 3: To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor.
Step 4: Continue this process till the degree of remainder is less than the degree of divisor.

Division Algorithm for Polynomial

If \( p(x) \) and \( g(x) \) are any two polynomials with \( g(x) \neq 0 \), then we can find polynomials \( q(x) \) and \( r(x) \) such that \( p(x) = q(x) \times g(x) + r(x) \)
where \( r(x) = 0 \) or \( \text{degree of } r(x) < \text{degree of } g(x) \).
The result is called Division Algorithm for polynomials.
Dividend = Quotient × Divisor + Remainder

Question. Apply the division algorithm to find the quotient and remainder on dividing \( p(x) \) by \( g(x) \) as given below \( p(x) = x^4 – 3x^2 + 4x + 5, g(x) = x^2 + 1 – x \)
Answer: We have, \( p(x) = x^4 – 3x^2 + 4x + 5, g(x) = x^2 + 1 – x \)
Quotient \( = x^2 + x – 3 \), remainder = 8.
We stop here since degree of (8) < degree of (\( x^2 – x + 1 \)). So, quotient \( = x^2 + x – 3 \), remainder = 8. Therefore, Quotient × Divisor + Remainder \( = (x^2 + x – 3) (x^2 – x + 1) + 8 = x^4 – x^3 + x^2 + x^3 – x^2 + x – 3x^2 + 3x – 3 + 8 = x^4 – 3x^2 + 4x + 5 = \text{Dividend} \). Therefore the Division Algorithm is verified.

Question. On dividing \( x^3 – 3x^2 + x + 2 \) by a polynomial \( g(x) \), the quotient and remainder were \( x – 2 \) and \( –2x + 4 \), respectively. Find \( g(x) \).
Answer: \( p(x) = x^3 – 3x^2 + x + 2, q(x) = x – 2 \) and \( r(x) = –2x + 4 \)
By Division Algorithm, we know that \( p(x) = q(x) \times g(x) + r(x) \). Therefore,
\( x^3 – 3x^2 + x + 2 = (x – 2) \times g(x) + (–2x + 4) \)
\( \Rightarrow x^3 – 3x^2 + x + 2 + 2x – 4 = (x – 2) \times g(x) \)
\( \Rightarrow g(x) = \frac{x^3 - 3x^2 + 3x - 2}{x - 2} \)
On dividing \( x^3 – 3x^2 + 3x – 2 \) by \( x – 2 \), we get \( g(x) = x^2 – x + 1 \).

Question. Give examples of polynomials \( p(x), q(x) \) and \( r(x) \), which satisfy the division algorithm and (i) \( \text{deg } p(x) = \text{deg } q(x) \) (ii) \( \text{deg } q(x) = \text{deg } r(x) \) (iii) \( \text{deg } q(x) = 0 \).
Answer: (i) Let \( q(x) = 3x^2 + 2x + 6, \text{degree of } q(x) = 2 \); \( p(x) = 12x^2 + 8x + 24, \text{degree of } p(x) = 2 \). Here, \( \text{deg } p(x) = \text{deg } q(x) \)
(ii) \( p(x) = x^5 + 2x^4 + 3x^3 + 5x^2 + 2 \); \( q(x) = x^2 + x + 1, \text{degree of } q(x) = 2 \); \( g(x) = x^3 + x^2 + x + 1 \); \( r(x) = 2x^2 – 2x + 1, \text{degree of } r(x) = 2 \). Here, \( \text{deg } q(x) = \text{deg } r(x) \)
(iii) Let \( p(x) = 2x^4 + 8x^3 + 6x^2 + 4x + 12 \); \( q(x) = 2, \text{degree of } q(x) = 0 \); \( g(x) = x^4 + 4x^3 + 3x^2 + 2x + 6 \); \( r(x) = 0 \). Here, \( \text{deg } q(x) = 0 \)

Question. If the zeroes of polynomial \( x^3 – 3x^2 + x + 1 \) are \( a – b, a, a + b \). Find \( a \) and \( b \).
Answer: \( \because a – b, a, a + b \) are zeros. \( \therefore \text{product } (a – b) a(a + b) = –1 \Rightarrow (a^2 – b^2) a = –1 \dots (1) \)
and sum of zeroes is \( (a – b) + a + (a + b) = 3 \Rightarrow 3a = 3 \Rightarrow a = 1 \dots (2) \)
by (1) and (2) \( (1 – b^2)1 = –1 \Rightarrow 2 = b^2 \Rightarrow b = \pm \sqrt{2} \). \( \therefore a = 1 \) & \( b = \pm \sqrt{2} \) Ans.

Question. If two zeroes of the polynomial \( x^4 – 6x^3 – 26x^2 + 138x – 35 \) are \( 2 \pm \sqrt{3} \), find other zeroes.
Answer: \( \because 2 \pm \sqrt{3} \) are zeroes. \( \therefore x = 2 \pm \sqrt{3} \Rightarrow x – 2 = \pm \sqrt{3} \) (squaring both sides) \( \Rightarrow (x – 2)^2 = 3 \Rightarrow x^2 + 4 – 4x – 3 = 0 \Rightarrow x^2 – 4x + 1 = 0 \), is a factor of given polynomial.
\( \therefore \text{other factors} = \frac{x^4 - 6x^3 - 26x^2 + 138x - 35}{x^2 - 4x + 1} = x^2 – 2x – 35 \)
\( \therefore \text{other factors} = x^2 – 7x + 5x – 35 = x(x – 7) + 5(x – 7) = (x – 7) (x + 5) \)
\( \therefore \text{other zeroes are } (x – 7) = 0 \Rightarrow x = 7; x + 5 = 0 \Rightarrow x = – 5 \). Ans.

Question. If the polynomial \( x^4 – 6x^3 + 16x^2 – 25x + 10 \) is divided by another polynomial \( x^2 – 2x + k \), the remainder comes out to be \( x + a \), find \( k \) & \( a \).
Answer: On dividing \( x^4 – 6x^3 + 16x^2 – 25x + 10 \) by \( x^2 – 2x + k \), the remainder is \( x [4k – 25 + 16 – 2k] + 10 – 8k + k^2 \).
According to questions, remainder is \( x + a \). \( \therefore \text{coefficient of } x = 1 \Rightarrow 2k – 9 = 1 \Rightarrow k = (10/2) = 5 \).
Also constant term = a \( \Rightarrow k^2 – 8k + 10 = a \Rightarrow (5)^2 – 8(5) + 10 = a \Rightarrow a = 25 – 40 + 10 \Rightarrow a = – 5 \).
\( \therefore k = 5, a = –5 \). Ans.