

Polynomial and/or Polynomial
Functions and Equations 
Polynomial functions 
Roots or zeros of
polynomial function 
Vieta's
formulas 





Polynomial functions 
Roots or zeros of
polynomial function 
The
zeros of a polynomial function are the values of x
for which the function equals zero. 
That
is, the solutions of the equation f
(x)
= 0,
that are called roots of the polynomial, are the zeros of the
polynomial function or the xintercepts of its
graph in a coordinate plane. 
At
these points the graph of the polynomial function cuts or
touches the xaxis. 
If
the graph of a polynomial intersects with the xaxis
at (r,
0), or x
= r
is a root or zero of a polynomial, then 
(x

r)
is a factor of that polynomial.

Every
polynomial of degree n
has exactly n
real and/or complex zeros. 
An
nth
degree polynomial has at most n
real zeros. 
Some
of the roots may be repeated. The number of times a root is
repeated is called multiplicity or
order of the root. 
The
number r_{i}
is a root of the
polynomial f
(x)
if and only if
f (x)
is divisible by
(x

r_{i}).

Therefore,
a polynomial and/or polynomial
function with
real coefficients can be expressed as a product of its leading
coefficient a_{n
}and
n
linear factors of the form (x
 r_{i}),
where r_{i}_{
}denotes its real root and/or complex root, 
f
(x)
= a_{n}x^{n}
+ a_{n}_{}_{1}x^{n}^{}^{1}
+
.
. . +
a_{1}x
+ a_{0}
= a_{n}(x
 r_{1})(x
 r_{2})
.
. . (x
 r_{n}).

Thus, finding the roots of a
polynomial f(x)
is equivalent to finding its
linear divisors or is equivalent to polynomial
factorization into
linear factors. 

Vieta's
formulas 
Suppose,
r_{1}, r_{2}, r_{3},
· · · ,
r_{n},
are roots of a polynomial f
(x)
of degree n
(counting multiplicities) with
leading coefficient a_{n}
= 1,
that is, 
f
(x)
= x^{n}
+ a_{n}_{}_{1}x^{n}^{}^{1}^{}
+ a_{n}_{}_{2}x^{n}^{}^{2}
+
.
. . +
a_{2}x^{2}
+ a_{1}x
+ a_{0} 
then we can write f
(x)
= (x
 r_{1})(x
 r_{2})(x
 r_{3})
· · · (x
 r_{n}).

By
expanding the above expression and collecting like terms, and
after comparing corresponding coefficients of
both expressions, we get 

Vieta's
formulas that show coefficients and roots relations. 
Note
that the coefficients of a polynomial are expressed as the sum
of corresponding combinations of
roots. 
Let
write coefficients of the quadratic polynomial by roots 
f (x)
= x^{2}
+ a_{1}x
+ a_{0}
= (x
 r_{1})(x
 r_{2})
=
x^{2}

(r_{1}
+ r_{2})x
+ r_{1}r_{2}, 
the
cubic polynomial 
f (x)
= x^{3}
+
a_{2}x^{2}
+ a_{1}x
+ a_{0}
= (x
 r_{1})(x
 r_{2})(x
 r_{3})
or 
f (x) =
x^{3}

(r_{1}
+ r_{2}
+ r_{3})x^{2}
+ (r_{1}r_{2}
+ r_{1}r_{3}
+ r_{2}r_{3})x

r_{1}r_{2}r_{3}, 
and
the quartic polynomial 
f (x)
= x^{4}
+
a_{3}x^{3}
+
a_{2}x^{2}
+ a_{1}x
+ a_{0}
= (x
 r_{1})(x
 r_{2})(x
 r_{3})(x
 r_{4})
or 
f (x) =
x^{4}

(r_{1}
+ r_{2}
+ r_{3}
+ r_{4})x^{3}
+ (r_{1}r_{2}
+ r_{1}r_{3}
+ r_{1}r_{4}
+ r_{2}r_{3}
+ r_{2}r_{4}
+ r_{3}r_{4})x^{2}
 

(r_{1}r_{2}r_{3}
+ r_{1}r_{2}r_{4}
+ r_{1}r_{3}r_{4}
+ r_{2}r_{3}r_{4})x
+ r_{1}r_{2}r_{3}r_{4}. 

Example:
Write the cubic
polynomial whose roots are r_{1}
= 1 and r_{2,3}
= 1 ±
i, assuming its
leading coefficient a_{3}
= 1. Find its source function
and draw their graphs. 
Solution:
Let write the cubic polynomial by roots 
f (x)
= x^{3}

(r_{1}
+ r_{2}
+ r_{3})x^{2}
+ (r_{1}r_{2}
+ r_{1}r_{3}
+ r_{2}r_{3})x

r_{1}r_{2}r_{3}, 
and
calculate coefficients 
a_{2}
=

(r_{1}
+ r_{2}
+ r_{3})
=

[1
+ (1 + i)
+ (1 
i)]
=

3, 
a_{1}
=
r_{1}r_{2}
+ r_{1}r_{3}
+ r_{2}r_{3}
=
1 ·
(1 + i)
+ 1
·
(1 
i)
+ (1 + i)
·
(1 
i)
=
4, 
a_{0}
=

r_{1}r_{2}r_{3}
=

1 ·
(1 + i)
·
(1 
i)
=

2, 
therefore
f (x)
= x^{3}

3x^{2}
+ 4x

2
or f
(x)
= (x

1)[x
 (1 + i)][x
 (1

i)] 
To
find the source form, f_{s}(x)
=
a_{3}x^{3}
+ a_{1}x
of the general cubic polynomial
we should calculate 
coordinates
of translations x_{0}
and y_{0}, 

and plug into y
+ y_{0}
= a_{3}(x
+ x_{0})^{3}
+
a_{2}(x
+ x_{0})^{2}
+
a_{1}(x
+ x_{0})
+
a_{0} 
thus,
y
= (x
+
1)^{3}
 3(x
+ 1)^{2}
+ 4(x
+ 1)
 2
= x^{3}
+ x
or f_{s}(x)
= x^{3}
+ x. 
Alternatively,
coefficients of the source polynomial can be calculated like
coefficients of the Taylor polynomial 

since,
a_{n}
= a_{n},
a_{n
}_{}_{1}
= 0 and a_{0}
= f
(x_{0}),
and where f ^{(n

k)}^{
}(x_{0})
denotes (n

k)th derivative at
x_{0}. 
Hence,
f (x)
= x^{3}
 3x^{2}
+ 4x
 2, f '(x)
= 3x^{2}
 6x
+ 4
and f '(x_{0})
or f '(1)
= 3x^{2}
 6x
+ 4
= 1
then 


Therefore,
the given cubic polynomial is translated only in the direction
of the xaxis
by x_{0}
= 1, as shows the above
picture. Note that the point of inflection of a cubic polynomial
I (x_{0},
y_{0}), since f
'' (x_{0}) = 0. 








Precalculus contents
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