find the zeros of the cubic polynomial f(x)=x^3-x also verify the relationship between zeros and the coefficient
Answers
Answer:
We have,
p(x)=3x
3
−5x
2
−11x−3
And the zeros are
3,−1,
3
−1
Verifying the zeros,
x=3,p(3)=3(3)
3
−5(3)
2
−11(3)−3
=81−45−33−3
=0
=x−1,p(−1)=3(−1)
3
−5(−1)
2
−11(−1)−3
=−3−5+11−3
=0
x=
3
−1
,
p(
3
−1
)=3(
3
−1
)
3
−5(
3
−1
)
2
−11(
3
−1
)−3
9
−1
−
9
5
+
3
11
−3
9
−1−5−33−27
0
Now verifying the relation between zeros and coefficients is:
for,
p(x)=3x
3
−5x
2
−11x−3
a=3,b=−1,c=−11,d=−1
and zeros α=3,β=−1γ=
3
−1
Now,
α+β+γ
=3+(−1)+
3
−1
=
3
9−3−1
3
5
=
a
−b
αβ+βγ+γα
=(3)(−1)+(−1)(
3
−1
)+(
3
−1
)(3)
3
−9+1−3
=
3
−11
=
a
c
αβγ
=(3)(−1)(
3
−1
)
1=
a
d
Thus the relation are verified.
Step-by-step explanation:
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Answer:
x=0,±1
Step-by-step explanation:
x³-x=0
x(x²-1)=0
we get x=0 and x²-1=o
x²-1=0 then we get x=±1
verification
for x=0
0³-0=0
true
for x=1
1³-1=0
true
for x=-1
-1³-(-1)=0
true