put p(2), p(-2),p(1)
=x⁴-3x²+2x-4
plzz do fast it's urgent
Answers
Step-by-step explanation:
Verify whether the following are zeroes of the polynomial, indicated against them.
1
(1) p(x) = 3x + 1, x =
3
(iii) p(x)= x2 – 1, x=1,-1
4
(ü) p(x) = 5x – , x=
5
(iv) p(x) = (x + 1) (x - 2), r=-1,2
m
(v) p(x)= x?, x=0
(vi) p(x) = lx + m, x= -
1
1
(vii) p(x) = 3x2 - 1, x=
1 2.
(viii) p(x) = 2x + 1, x =
13 13
in each of the following cases:
2
(i) The given polynomial is: p(x) = 3x + 1 , x = \frac{-1}{3}x=
3
−1
Now, put the given value of x in the polynomial;
p(\frac{-1}{3}) = 3(\frac{-1}{3})+1p(
3
−1
)=3(
3
−1
)+1
= -1 + 1 = 0
Since the result comes out to be 0, then x = \frac{-1}{3}x=
3
−1
is the zero of the given polynomial.
(ii) The given polynomial is: p(x) = 5x-\pi5x−π , x = \frac{4}{5}x=
5
4
Now, put the given value of x in the polynomial;
p(\frac{4}{5}) = 5(\frac{4}{5})-\pip(
5
4
)=5(
5
4
)−π
= 4-\pi4−π \neq
= 0
Since the result doesn't come out to be 0, then x = \frac{4}{5}x=
5
4
is not the zero of the given polynomial.
(iii) The given polynomial is: p(x) = x^{2} -1 , \ x = 1,-1x
2
−1, x=1,−1
Now, first put the given value of x = 1 in the polynomial;
p(1) = (1)^{2} -1p(1)=(1)
2
−1
= 1 - 1 = 0
Also, put x = -1 in the given polynomial;
p(-1) = (-1)^{2} -1p(−1)=(−1)
2
−1
= 1 - 1 = 0
Since the result comes out to be 0, then x =1,-1x=1,−1 are the zeroes of the given polynomial.
(iv) The given polynomial is: p(x) = (x+1)(x-2), \ x =-1,2p(x)=(x+1)(x−2), x=−1,2
Now, first put the given value of x = -1 in the polynomial;
p(-1) = (-1+1)(-1-2)p(−1)=(−1+1)(−1−2)
= 0 \times (-3)0×(−3) = 0
Also, put x = 2 in the given polynomial;
p(2) = (2+1)(2-2)p(2)=(2+1)(2−2)
= 3 \times 03×0 = 0
Since the result comes out to be 0, then x =-1,2x=−1,2 are the zeroes of the given polynomial.
(v) The given polynomial is: p(x)=x^{2} , \ x = 0p(x)=x
2
, x=0
Now, put the given value of x = 0 in the polynomial;
p(0)=(0)^{2}p(0)=(0)
2
= 0
Since the result comes out to be 0, then x =0x=0 is the zero of the given polynomial.
(vi) The given polynomial is: p(x)=lx+m, \ x = \frac{-m}{l}p(x)=lx+m, x=
l
−m
Now, put the given value of x=\frac{-m}{l}x=
l
−m
in the polynomial;
p(\frac{-m}{l})=l(\frac{-m}{l})+mp(
l
−m
)=l(
l
−m
)+m
= -m + m = 0
Since the result comes out to be 0, then x =\frac{-m}{l}x=
l
−m
is the zero of the given polynomial.
(vii) The given polynomial is: p(x) = 3x^{2} -1, \ x = \frac{-1}{\sqrt{3} },\frac{2}{\sqrt{3} }p(x)=3x
2
−1, x=
3
−1
,
3
2
Now, first put the given value of x = \frac{-1}{\sqrt{3} }x=
3
−1
in the polynomial;
p(\frac{-1}{\sqrt{3} }) = 3(\frac{-1}{\sqrt{3} })^{2} -1p(
3
−1
)=3(
3
−1
)
2
−1
= 3 \times (\frac{1}{3})-13×(
3
1
)−1
= 1 - 1 = 0
Since the result comes out to be 0, then x = \frac{-1}{\sqrt{3} }x=
3
−1
is the zero of the given polynomial.
Also, put x = \frac{2}{\sqrt{3} }x=
3
2
in the given polynomial;
p(\frac{2}{\sqrt{3} }) = 3(\frac{2}{\sqrt{3} })^{2} -1p(
3
2
)=3(
3
2
)
2
−1
= 3 \times (\frac{4}{3})-13×(
3
4
)−1
= 4 - 1 = 3
Since the result doesn't come out to be 0, then x = \frac{2}{\sqrt{3} }x=
3
2
is not the zero of the given polynomial.
(viii) The given polynomial is: p(x)= 2x+1, \ x = \frac{1}{2}p(x)=2x+1, x=
2
1
Now, put the given value of x=\frac{1}{2}x=
2
1
in the polynomial;
p(x)= 2(\frac{1}{2})+1p(x)=2(
2
1
)+1
= 1 + 1 = 2
Since the result doesn't come out to be 0, then x=\frac{1}{2}x=
2
1