
A
revealing insight into the polynomial function 
continue 





The principle provides full control over the polynomial 
Hence, by
substituting coordinates of translations into the source
polynomial function we can move its graph to any desired
position on the coordinate plane. 
Since an n^{th}
degree polynomial 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}), 
the above expression shows
following coefficients and roots relations called Vieta's
formulas. 
Therefore, if the leading coefficient a_{n}
= 1
we can write for example, 
the quadratic
function shown by its 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 function expressed by roots 
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}, 
the quartic
function wrtten by roots 
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}, 
and so on. 
Note that the coefficients of a polynomial are expressed as the
alternating sums
of corresponding combinations of
products of roots. 

Example: Let
write the cubic
polynomial whose roots are r_{1}
= 1 and r_{2,3}
= 1 ±
i assuming its
leading coefficient a_{3}
= 1,
and find its source function
and draw their graphs. 
Then, find the cubic
function obtained by moving the source function by x_{t}
= 
1 and y_{t}
= 2, and draw the graph of
the translated cubic function. 
Solution:
We write the cubic polynomial
given by its roots to calculate its coefficients, since 
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}, 
then
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 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
 the source cubic function of the type 2/1 since, a_{3}a_{1}
> 0. 
The obtained source
cubic we then move to new position by
plugging the given coordinates of translations, 
x_{t}
= 1 and y_{t}
= 2,
into y

y_{t}
=
a_{3}(x

x_{t })^{3}
+
a_{1}(x

x_{t
}) 
thus,
y

2
=
(x
+
1)^{3}
+
(x
+
1)
or y
=
x^{3}
+
3x^{2}
+ 4x
+ 4 
as shows the below figure. 






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