Chapter 2 - Polynomials, Solved Examples, Class 9, Maths PDF Download

Ex 1. : Consider a polynomial f(x) = 3x² - 4x + 2, find the value at x = 3.

Sol. replace x by 3 everywhere

So, the value of f(x) = 3x² - 4x + 2, at x = 3 is

f(3) = 3 x 3² - 4 x 3 + 2 = 27 -12 + 2 = 17

Similarly the value of polynomiak f(x) = 3x² - 4x + 2

at x = -2 is, 

at x = 0 is,

at x = 1/2 is,

DO YOUR SELF

Find the value of polynomial 5x - 4x² + 3 at :

(i) x = 0

(ii) x = -1

(iii) x = 2

(iv) ZEROES OF A POLYNOMIAL

A real number α is a zero of a polynomial p(x) if the value of the polynomial p(x) is zero at x = α. i.e. p(α) = 0

OR

The value of the variable x, for which the polynomial p(x) becomes zero is called zero of the polynomial.

Ex : consider, a polynomial p(x) = x² - 5x + 6 ; replace x by 2 and 3.
p(2) = (2)² - 5 × 2 + 6 = 4 - 10 + 6 = 0,

p(3) = (3)2 - 5 × 3 + 6 = 9 - 15 + 6 = 0

∴ 2 and 3 are the zeroes of the polynomial p(x).

REMARK 

1. The constant polynomial has no zero.

2. Every linear polynomial has one and only one zero or root.
consider a linear polynomial p(x) = ax + b, a ≠ 0 ⇒ p(x) = 0 ⇒ ax + b  = 0 ⇒ ax = -b ⇒ x = -b/a  is a zero of the polynomial.

3. A given polynomial can have more than one zero or root.

4. If the degree of a polynomial is n, the maximum number of zeroes it can have is also n.
Ex. : If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes, if the degree of
polynomial is 8, maximum number of zeroes it can have is 8.

5. A zero of a polynomial need not be 0

Ex . : Consider a polynomial f(x) = x² - 4, then f(2) = (2)² - 4 = 4 - 4 = 0 here, zero of the polynomial
f(x) = x² - 4 is 2 which itself is not 0.

6. 0 may be zero of the polynomial.

Ex. Consider a polynomial f(x) = x² - x, then f(0) = 0² - 0 = 0 here, 0 is the zero of the polynomial

f(x) = x² - x

Ex 2. Find which of the following algebraic expression is a polynomial.
 (i)  

(iv) 
 

Sol. (i)  Is a polynomial in one variable.

(ii)  = x¹ + x^-1 Is not a polynomial as the second term 1/x has degree (-1).

(iii) = y^1/2 - 8 Is not a polynomial since, the power of the first term √y is 1/2, which is not a whole number.

(iv)
Since, the exponent of the second term is 1/3, which in not a whole number. Therefore, the given expression is not a polynomial.

Ex 3. Find the degree of the polynomial :
 (i) 5x² - 6x³ + 8x⁷ + 6x² 

(ii) 2y¹² + 3y¹⁰ - y¹⁵ + y + 3 

(iii) x 

(iv) 8

Sol. (i) Since the term with highest exponent (power) is 8x⁷ 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 = x¹ ⇒ degree is 1.
(iv) 8 = 8x⁰ ⇒ degree is 0.

Ex 4. Write the coefficient of x² in each of the following :
 (i)  

(ii) 

(iii) 

(iv) 

Sol. (i) In the polynomial  the coefficient of x² is 1.
(ii) In the polynomial  the coefficient of x² is - 1.
(iii) In the polynomial  the coefficient of x² is 1/2.

(iv) In the polynomial  the coefficient of x² = 0. [We may write  = ax² +. So the
coefficient of x² is 0]

Ex 5. Classify the following as linear, quadratic and cubic polynomials :
 (i) x² + x 

(ii) x - x³ 

(iii) 1 + x 

(iv) 3t
 (v) r² 

(vi) 7x³

(vii) y + y² + 5 

(viii) 3xyz

Sol. (i) The polynomial x² + x is a quadratic polynomial as its degree is 2.

(ii) Degree of the polynomial x - x³ is 3. It is a cubic polynomial

(iii) Degree of the polynomial 1 + x is 1. It is a linear polynomial.

(iv) Degree of the polynomial 3t is 1. It is a linear polynomial.

(v) Degree of the polynomial r² is 2. It is a quadratic polynomial.

(vi) Degree of the polynomial 7x³ is 3. It is a cubic polynomial.

(vii)Degree of the polynomial y + y² + 5 is 2. It is a quadratic polynomial.

(viii) 3xyz is a polynomial in 3 variables x, y and z. Its degree is 1 + 1 + 1 = 3. It is a cubic polynomial.

Ex 6. Find q(0), q(1) and q(2) for each of the following polynomials :
 (i) q(x) = x² + 3x 

(ii) q(y) = 2 + y + 2y² - 5y³ 

(iii) q(t) = t³

Sol.

q (x) = x² + 3x
q(0) - (0)² + 3 x 0 = 0
q(l) = (l)² + 3 x 1 = 4

(i) q(2) = (2)²+ 3 x 2 = 4 + 6 = 10
q(y) = 2 + y + 2y² - 5y³ 
q(0) = 2 + 0 + 2 (0)² - 5 (0)³ = 2

(ii) q(1) = 2 + 1 + 2 (1)² - 5 (1)³ = 2 + 1 + 2 - 5 = 0

q(t) = t³ 
q(0) =0
q(1) = (1)³ = 1

(iii) q(2) = (2)³ = 8.

E7. Check whether 0 and 3 are zeroes of the polynomial x² - 3x.
Sol. 

let p (x) = x1 - 3x

Then p(0) = (0)" - 3*0 = 0

∴ 0 is a zero of the given polymcinial
Again,    p(3) = (3)² -3 x 3 = 9 - 9 = 0
∴ 3 is alari a zero cf the given polynomial.
Hence 0 and 3 are both zeroes of the polynomial x² - 3x

Ex 8. Show that 3 is a zero of the polynomial x³ - 8x² + 8x + 21.

Sol . Let p(x) = x³ - 8x² + 8x + 21.

Now p(3) = (3)3 - 8 (3)2 + 8.3 + 21 = 27 - 72 + 24 + 21 = 0

∴ 3 is zero of the polynomial - x³ + 8x² + 8x + 21.

Ex 9. Which of the number 1, -1, and -3 are zeroes of the polynomial 2x⁴ + 9x³ + 11x² + 4x - 6.
 Sol. 
Let f(x) = 2x⁴ + 9x³ + 11x² + 4x - 6
f(1) = 2(1)⁴ + 9(1)³ + 11(1)² + 4(1) - 6 = 2 + 9 + 11 + 4 - 6 = 20  0
∴ 1 is not a zero of the polynomial f(x)

Again f(-1) = 2(-1)⁴ + 9(-1)³ + 11(-1)² + 4(-1) - 6 = 2 - 9 + 11 - 4 - 6 = - 6  0
∴ -1 is not a zero the polynomial f(x)

Also f(-3) = 2(-3)⁴ + 9(-3)³ + 11(-3)² + 4(-3) - 6 = 162 - 243 + 99 - 12 - 6 = 0
∴ -3 is a zero of the polynomial f(x).
Thus 1 and -1 are not zeroes of f(x) whereas -3 is a zero of f(x).

Ex 10. Verify whether the indicated numbers are zeroes (roots) of the polynomial corresponding to them in the following cases:
 (i)

(ii) p(x) = (x + 1) (x - 2), x = -1, 2

(iii) p(x) = x², x = 0 

(iv)

(v)

Sol. 

(i) p(x) = 3x + 1

∴  is a zero of the polynomial.

(ii)  p(x) - (x + 1) (x- 2)
⇒ P(-1) = (-1 + 1) (-1- 2) = 0 x (-3) = 0 and, p (2) = (2 + 1) [2 - 2) = 3 x 0 = 0
∴' x = -1 and x = 2 are zeroes of the given polynomial

(iii) p(x) = x2   ⇒    p(0) = (0)² = 0

∴  x = 0 is a zero of the given polynomial

(iv)  p (x) - ℓx + m

  is a zero of the given polynomial.

(v) p(x) = 2x + 1     

∴ x = 1/2 is not a zero of the given polynomial.