Question

verify whether the following are zeros of the polynomial indicated against them:- p(y)=q y+r,y=-r/q

Answer 1

$\Large\mathtt\green{ }\huge\underline\mathtt\red{Answer : }$

In 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=−31, we get,

p(x)=p(3−1)=3(−31)+1=−1+1=0.

So, x=−31 is the zero of the given polynomial p(x).

(ii) p(x)=5x−π:

Put x=54, we get,

p(x)=p(54)=5(54)−π=4−π=0.

So, x=54 is not the zero of the given polynomial p(x).

(iii) p(x)=x2−1:

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)=x2:

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=−lm, we get,

p(x)=p(−lm)=l(−lm)+m=−m+m=0.

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

(vii) p(x)=3x2−1:

Put x=−31, we get,

p(x)=p(−31)=3(−31)2−1=(3×31)−1=1−1=0.

So, x=−31 is the zero of the given polynomial p(x).  

Now put x=32, we get,

p(x)=p(32)=3(32)2−1=(3×34)−1=4−1=3=0.

So, x=32 is not the zero of the given polynomial p(x).  

(viii) p(x)=2x+1:

Put x=21, we get,

p(x)=p(21)=2(21)+1=1+1=2=0.

So, x=21 is not the zero of the given polynomial p(x).  

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Answer 2

Answer:

\Large\mathtt\green{ }\huge\underline\mathtt\red{Answer : }

Answer:

In 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=−31, we get,

p(x)=p(3−1)=3(−31)+1=−1+1=0.

So, x=−31 is the zero of the given polynomial p(x).

(ii) p(x)=5x−π:

Put x=54, we get,

p(x)=p(54)=5(54)−π=4−π=0.

So, x=54 is not the zero of the given polynomial p(x).

(iii) p(x)=x2−1:

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)=x2:

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=−lm, we get,

p(x)=p(−lm)=l(−lm)+m=−m+m=0.

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

(vii) p(x)=3x2−1:

Put x=−31, we get,

p(x)=p(−31)=3(−31)2−1=(3×31)−1=1−1=0.

So, x=−31 is the zero of the given polynomial p(x).

Now put x=32, we get,

p(x)=p(32)=3(32)2−1=(3×34)−1=4−1=3=0.

So, x=32 is not the zero of the given polynomial p(x).

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

Put x=21, we get,

p(x)=p(21)=2(21)+1=1+1=2=0.

So, x=21 is not the zero of the given polynomial p(x).

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