NCERT Solution Class 9th Maths Chapter – 2 Polynomials
Chapter – 2
Polynomials
Examples
Example 1 : Find the degree of each of the polynomials given below: (i) x5 – x4 + 3 (ii) 2 – y2 – y3 + 2y8 (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 2x0 . So the exponent of x is 0. Therefore, the degree of the polynomial is 0.
Example 2 : Find the value of each of the following polynomials at the indicated value of variables: (i) p(x) = 5x2 – 3x + 7 at x = 1. (ii) q(y) = 3y3 – 4y + √11 at y = 2. (iii) p(t) = 4t4 + 5t3 – t2 + 6 at t = a.
Solution – (i) p(x) = 5x2 – 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) = 3y3 – 4y + √11 The value of the polynomial q(y) at y = 2 is given by q(2) = 3(2)3 – 4(2) + √11 = 24 – 8 + √11 = 16 + √11
(iii) p(t) = 4t4 + 5t3 – t2 + 6 The value of the polynomial p(t) at t = a is given by p(a) = 4a4 + 5a3 – a2 + 6
Example 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 = -1/2 So, -1/2 is a zero of the polynomial 2x + 1. Now, if p(x) = ax + b, a ≠ 0, is a linear polynomial, how can we find a zero of p(x)? Example 4 may have given you some idea. Finding a zero of the polynomial p(x), amounts to solving the polynomial equation p(x) = 0. Now, p(x) = 0 means ax + b = 0, a ≠ 0 So, ax = –b i.e., x = – b/a So, x = -b/a is the only zero of p(x), i.e., a linear polynomial has one and only one zero. Now we can say that 1 is the zero of x – 1, and –2 is the zero of x + 2.
Example 5 : Verify whether 2 and 0 are zeroes of the polynomial x2 – 2x.
Solution – Let p(x) = x2 – 2x Then p(2) = 22– 4 = 4 – 4 = 0 and p(0) = 0 – 0 = 0
Hence, 2 and 0 are both zeroes of the polynomial x2 – 2x. Let us now list our observations: (i) A zero of a polynomial need not be 0. (ii) 0 may be a zero of a polynomial. (iii) Every linear polynomial has one and only one zero. (iv) A polynomial can have more than one zero.
Example 6 : Examine whether x + 2 is a factor of x3 + 3×2 + 5x + 6 and of 2x + 4.
Solution – The zero of x + 2 is –2. Let p(x) = x3 + 3x2 + 5x + 6 and s(x) = 2x + 4 Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 = –8 + 12 – 10 + 6 = 0
Example 7 : Find the value of k, if x – 1 is a factor of 4x3 + 3x2 – 4x + k.
Solution – As x – 1 is a factor of p(x) = 4x3 + 3x2 – 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
Example 8 : Factorise 6x2 + 17x + 5 by splitting the middle term, and by using the Factor Theorem.
Solution – (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 So, 6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5 = 6x2 + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1) (2x + 5)
Example 9 : Factorise y2 – 5y + 6 by using the Factor Theorem.
Solution – Let p(y) = y2 – 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) = 22 – (5 × 2) + 6 = 0 So, y – 2 is a factor of p(y). Also, p(3) = 32 – (5 × 3) + 6 = 0 So, y – 3 is also a factor of y2 – 5y + 6. Therefore, y2 – 5y + 6 = (y – 2)(y – 3) Note that y2 – 5y + 6 can also be factorised by splitting the middle term –5y. Now, let us consider factorising cubic polynomials. Here, the splitting method will not be appropriate to start with. We need to find at least one factor first, as you will see in the following example
