- Constants and Variable
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- Polynomial expression
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- how to find the degree of a polynomial
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- Value of the polynomial
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- Zeros or roots of the polynomial
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- Adding Polynomials
- |
- subtracing Polynomials
- |
- Multiplying Polynomials
- |
- Dividing Polynomails
- |
- How to factor polynomials
- |
- Solved Examples Polynomials
- |

In this page we have ** NCERT book Solutions for Class 9th Maths:Polynomial** for
EXERCISE 2 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

**Question 1:**

Find the value of the polynomial P(x)= 5x-4x^{2} +3 at

(i) x = 0 (ii) x = −1 (iii) x = 2

**Solution:**

(i) P(x)= 5x-4x^{2} +3

P(0)= 0-0+3=3

(ii) P(x)= 5x-4x^{2} +3

P(-1) =-5-4+3=-6

(iii) P(x)= 5x-4x^{2} +3

P(2)= 10-16+3=-3

**Question 2**

Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y^{2} − y + 1 (ii) p(t) = 2 + t + 2t^{2} − t^{3}

(iii) p(x) = x^{3} (iv) p(x) = (x − 1) (x + 1)

**Solution:**

(i) p(y) = y^{2} − y + 1

p(0) = (0)^{2} − (0) + 1 = 1

p(1) = (1)^{2} − (1) + 1 = 1

p(2) = (2)^{2} − (2) + 1 = 3

(ii) p(t) = 2 + t + 2t^{2} − t^{3}

p(0) = 2 + 0 + 2 (0)^{2} − (0)3 = 2

p(1) = 2 + (1) + 2(1)^{2} − (1)^{3}

= 2 + 1 + 2 − 1 = 4

p(2) = 2 + 2 + 2(2)^{2} − (2)^{3}

= 2 + 2 + 8 − 8 = 4

(iii) p(x) = x^{3}

p(0) = (0)^{3} = 0

p(1) = (1)^{3} = 1

p(2) = (2)^{3} = 8

(iv) p(x) = (x − 1) (x + 1)

p(0) = (0 − 1) (0 + 1) = (− 1) (1) = − 1

p(1) = (1 − 1) (1 + 1) = 0 (2) = 0

p(2) = (2 − 1 ) (2 + 1) = 1(3) = 3

**Question 3**

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) *p*(*x*) = 3*x *+ 1, *x *=– 1/3

(ii) *p*(*x*) = 5*x *– π, *x *=4/5

(iii) *p*(*x*) = *x ^{2}* – 1,

(iv) *p*(*x*) = (*x *+ 1) (*x *– 2), *x *= – 1, 2

(v) *p*(*x*) = *x ^{2} ,*

(vi) *p*(*x*) = *lx *+ *m*, *x *= – *m/l*

(vii) *p*(*x*) = 3*x ^{2}* – 1,

(viii) *p*(*x*) = 2*x *+ 1, *x *=1/2

**Solution:**

(i) p(x) = 3x + 1,x= -1/3

p(-1/3) = 3 (-1/3)+1=−1+1=0

p(-1/3) = 0 which means that -1 /3is zero of the polynomial p(x) = 3x+1.

(ii) p(x) = 5x−π, x = 4/5

p(4/5) = 5(4/5) –π = 4−π

p(4/5) ≠ 0 which means that 4/5 is not zero of the polynomial p(x) = 5x − π.

(iii) p(x) = x^{2}−1,x = 1,− 1

p(1)=1^{2}−1=1−1=0

p(−1)=(−1)^{2}−1=1−1=0

Both p(1) and p(−1) are equal to 0. It means that 1 and -1 are zeroes of the polynomial p(x) = x^{2 }− 1.

(iv) p(x) = (x+1)(x−2),x=−1,2

p(−1) = (−1+1)(−1−2) = 0×−3 = 0

p(2) = (2+1)(2−2) = 3×0 = 0

Both p(−1) and p(2) are equal to 0. It means that -1 and 2 are zeroes of the polynomial p(x) = (x+1)(x−2).

(v) p(x) = x^{2},x = 0

p(0) = 0^{2} = 0

p(0) = 0 which means that 0 is the zero of the polynomial p(x) = x^{2}.

(vi) p(x) =* l*x + m,x = =-m/

p(-m/l) = *l*(-m/l) + m = −m + m = 0

p(-m/l) = 0 which means that (–m/l) is zero of the polynomial p(x)=*l*x + m

(vii) p(x) = 3x^{2} − 1,* x *=-1/√3 and 2/√3

p(-1/√3) = 3(-1/√3)^{2}−1=3(1/3)−1=1−1=0

p(2/√3) = 3(2/√3)^{2 }−1 = 3 × (4/3)−1= 4−1 = 3

p(-1/√3) = 0 which means that -1/√3 is zero of the polynomial p(x) = 3x^{2 }− 1.

p(2/√3) ≠ 0 which means that 2/√3 is not zero of the polynomial p(x)=3x^{2 }− 1.

(viii) p(x) = 2x + 1 ,x =1/2

p(1/2)=2 × (1/2) + 1=1+1=2

p(1/2) ≠ 0. It means that 1/2 is not zero of the polynomial p(x)=2x+1.

**Question 4. **

Find the zero of the polynomial in each of the following cases:

(i) *p*(*x*) = *x *+ 5 (ii) *p*(*x*) = *x *– 5 (iii) *p*(*x*) = 2*x *+ 5

(iv) *p*(*x*) = 3*x *– 2 (v) *p*(*x*) = 3*x *(vi) *p*(*x*) = *ax*, *a *≠ 0

(vii) *p*(*x*) = *cx *+ *d*, *c *≠ 0, *c*, *d *are real numbers.

**Solution:**

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

(i) p(x) = x + 5

p(x) = 0

x + 5 = 0

x = − 5

Therefore, for x = −5, the value of the polynomial is 0 and hence, x = −5 is a zero of the given polynomial.

(ii) p(x) = x − 5

p(x) = 0

x − 5 = 0

x = 5

Therefore, for x = 5, the value of the polynomial is0 and hence, x = 5 is a zero of the given polynomial.

(iii) p(x) = 2x + 5

p(x) = 0

2x + 5 = 0

2x = − 5

x = - 5/2

Therefore, for x = - 5/2 , the value of the polynomial is 0 and hence, x = - 5/2 is a zero of the given polynomial.

(iv) p(x) = 3x − 2

p(x) = 0

3x − 2 = 0

x=2/3, so 2/3 is the zero of the polynomial

(v) p(x) = 3x

p(x) = 0

3x = 0

x = 0

So x=0 is the zero of the polynomial

(vi) p(x) = ax

p(x) = 0

ax = 0

x = 0

Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vii) p(x) = cx + d

p(x) = 0

cx+ d = 0

x=-d/c

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