p(x) =(x+1)(x-2),x=-1, 2
Answers
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).
p(x)= (x+1)(x-2)
= (x²+x-2x-2)
= (x²-x-2)
p(-1)= (1²-1-2)
= (1-1-2)
=-2
p(2)= (2²-2-2)
= (4-2-2)
= 0
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