Which polynomial has zeros at –3, 5 and 2?
Answers
Let p(x)=ax2+bx+c be the required polynomial whose zeroes are -2 and 5. But we know that, if we multiply/divide any polynomial by any arbitary constant. Then, the zeroes of polynomial never change. Hence, the required number of polynomials are infinite i.e., more than 3.
Step-by-step explanation:
Explanation:
Since we are given the zeroes of the polynomial function, we can write the solution in terms of factors.
In general, given 3 zeroes of a polynomial function, a, b, and c, we can write the function as the multiplication of the factors
(
x
−
a
)
,
(
x
−
b
)
,
and
(
x
−
c
)
Simply:
f
(
x
)
=
(
x
−
a
)
(
x
−
b
)
(
x
−
c
)
In this case, we can show that each of a, b, and c are zeroes of the function:
f
(
a
)
=
(
a
−
a
)
(
a
−
b
)
(
a
−
c
)
=
(
0
)
(
a
−
b
)
(
a
−
c
)
=
0
f
(
b
)
=
(
b
−
a
)
(
b
−
b
)
(
b
−
c
)
=
(
b
−
a
)
(
0
)
(
b
−
c
)
=
0
f
(
a
)
=
(
c
−
a
)
(
c
−
b
)
(
c
−
c
)
=
(
c
−
a
)
(
c
−
b
)
(
0
)
=
0
Since the value of the function at x=a, b and c is equal to 0, then the function
f
(
x
)
=
(
x
−
a
)
(
x
−
b
)
(
x
−
c
)
has zeroes at a, b, and c.
With the generalized form, we can substitute for the given zeroes,
x
=
0
,
−
2
,
and
−
3
, where
a
=
0
,
b
=
−
2
,
and
c
=
−
3
.
f
(
x
)
=
(
x
−
0
)
(
x
−
(
−
2
)
)
(
x
−
(
−
3
)
)
Simplifying gives:
f
(
x
)
=
x
(
x
+
2
)
(
x
+
3
)
From here, we can put it in standard polynomial form by foiling the right side:
f
(
x
)
=
x
(
x
2
+
5
x
+
6
)
And distributing the x yields a final answer of:
f
(
x
)
=
x
3
+
5
x
2
+
6
x
To double check the answer, just plug in the given zeroes, and ensure the value of the function at those points is equal to 0.
f
(
0
)
=
(
0
)
3
+
5
(
0
)
2
+
6
(
0
)
=
0
f
(
−
2
)
=
(
−
2
)
3
+
5
(
−
2
)
2
+
6
(
−
2
)
=
−
8
+
20
−
12
=
0
f
(
−
3
)
=
(
−
3
)
3
+
5
(
−
3
)
2
+
6
(
−
3
)
=
−
27
+
45
−
18
=
0
Thus, the function has zeroes as given by x=0, -2, and -3.