ANSWER:

order to verify the values are zeros of polynomial p(x), we must replace the variable x with the given values.

If p(x)=0, then that given value is zero of polynomial p(x).

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

Put x=−

3

1

, we get,

p(x)=p(

3

−1

)=3(−

3

1

)+1=−1+1=0.

So, x=−

3

1

is the zero of the given polynomial p(x).

(ii) p(x)=5x−π:

Put x=

5

4

, we get,

p(x)=p(

5

4

)=5(

5

4

)−π=4−π



=0.

So, x=

5

4

is not the zero of the given polynomial p(x).

(iii) p(x)=x

2

−1:

Put x=1, we get,

p(x)=p(1)=(1)

2

−1=1−1=0.

So, x=1 is the zero of the given polynomial p(x).

Now put x=−1, we get,

p(x)p(−1)=(−1)

2

−1=1−1=0.

So, x=−1 is the zero of the given polynomial p(x).

(iv) p(x)=(x+1)(x−2):

Put x=−1, we get,

p(x)=(−1+1)(−1−2)=0(−3)=0.

So, x=−1 is the zero of the given polynomial p(x).

Now put x=2, we get,

p(x)=(2+1)(2−2)=3(0)=0.

So, x=2 is the zero of the given polynomial p(x).

(v) p(x)=x

2

:

Put x=0, we get,

p(x)=p(0)=(0)

2

=0.

So, x=0 is the zero of the given polynomial p(x).

(VI) p(x)=lx+m:

Put x=−

L

m

, we get,

p(x)=p(−

l

m

)=l(−

l

m

)+m=−m+m=0.

So, x=−

l

m

is the zero of the given polynomial p(x).

(vii) p(x)=3x

2

−1:

Put x=−

3

1

, we get,

p(x)=p(−

3

1

)=3(−

3

1

)

2

−1=(3×

3

1

)−1=1−1=0.

So, x=−

3

1

is the zero of the given polynomial p(x).

Now put x=

3

2

, we get,

p(x)=p(

3

2

)=3(

3

2

)

2

−1=(3×

3

4

)−1=4−1=3



=0.

So, x=

3

2

is not the zero of the given polynomial p(x).

(VIII) p(x)=2x+1:

Put x=

2

1

, we get,

p(x)=p(

2

1

)=2(

2

1

)+1=1+1=2



=0.

So, x=

2

1

is not the zero of the given polynomial p(x). order to verify the values are zeros of polynomial p(x), we must replace the variable x with the given values.

If p(x)=0, then that given value is zero of polynomial p(x).

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

Put x=−

3

1

, we get,

p(x)=p(

3

−1

)=3(−

3

1

)+1=−1+1=0.

So, x=−

3

1

is the zero of the given polynomial p(x).

(ii) p(x)=5x−π:

Put x=

5

4

, we get,

p(x)=p(

5

4

)=5(

5

4

)−π=4−π



=0.

So, x=

5

4

is not the zero of the given polynomial p(x).

(iii) p(x)=x

2

−1:

Put x=1, we get,

p(x)=p(1)=(1)

2

−1=1−1=0.

So, x=1 is the zero of the given polynomial p(x).

Now put x=−1, we get,

p(x)p(−1)=(−1)

2

−1=1−1=0.

So, x=−1 is the zero of the given polynomial p(x).

(iv) p(x)=(x+1)(x−2):

Put x=−1, we get,

p(x)=(−1+1)(−1−2)=0(−3)=0.

So, x=−1 is the zero of the given polynomial p(x).

Now put x=2, we get,

p(x)=(2+1)(2−2)=3(0)=0.

So, x=2 is the zero of the given polynomial p(x).

(v) p(x)=x

2

:

Put x=0, we get,

p(x)=p(0)=(0)

2

=0.

So, x=0 is the zero of the given polynomial p(x).

(vi) p(x)=lx+m:

Put x=−

l

m

, we get,

p(x)=p(−

l

m

)=l(−

l

m

)+m=−m+m=0.

So, x=−

l

m

is the zero of the given polynomial p(x).

(vii) p(x)=3x

2

−1:

Put x=−

3

1

, we get,

p(x)=p(−

3

1

)=3(−

3

1

)

2

−1=(3×

3

1

)−1=1−1=0.

So, x=−

3

1

is the zero of the given polynomial p(x).

Now put x=

3

2

, we get,

p(x)=p(

3

2

)=3(

3

2

)

2

−1=(3×

3

4

)−1=4−1=3



=0.

So, x=

3

2

is not the zero of the given polynomial p(x).

(viii) p(x)=2x+1:

Put x=

2

1

, we get,

p(x)=p(

2

1

)=2(

2

1

)+1=1+1=2



=0.

So, x=

2

1

is not the zero of the given polynomial p(x).