James Gray
University of Edinburgh
Applicable Mathematics (Foundation) af0
7: Polynomials
Here are some examples of
polynomials:
`3x^4 + 2x^3 - 6x^2 + 11`
`m^6-4m^3 + m^2 - 2`
`(5x-4)(x+3)`=`5x^2 + 11x
-12`
In general,
`a_nx^n + a_(n-1)n^(n-1) +
cdots + a_1 x + a_0`
is a polynomial of degree `n` with coefficients `a_n, ldots
,a_0`, where `n` is a whole number and `a_n != 0`.
Note
Polynomials of low degree have names
| Degree |
1 |
2 |
3 |
4 |
5 |
| Name |
linear |
quadratic |
cubic |
quartic |
quintic |
7.1 Synthetic Division
We have seen before that
`14 -: 5 = 2` remainder
`4`.
We say this because we know
`14 = 5 times 2 +4`.
In this situation we call 5 the
divisor, 2 the quotient and 4
the remainder.
We can do the same with polynomials
`(3x^2 -5x + 1)-:(x-2) =
3x+1` remainder 3
because
`3x^2 - 5x + 1 = (x-2)(3x+1)
+ 3`.
As with numbers we call
`(x-2)` the divisor,
`(3x+1)` the quotient and `3` the remainder.
How do we find the quotient and the remainder?
We use a technique called synthetic division which is
explained on p128 of Heinemann. To find the quotient and remainder in
the problem above (`3x^2 - 5x + 1` divided by `x-2`) we first set up a
table:
Then we follow the rules on p128 of Heinemann to
complete the table as follows
From the bottom row of the completed table we can read off
that the quotient is `3x+1` and the remainder is `3`.
Therefore `3x^2 - 5x + 1 = (x-2)(3x+1)
+ 3`.
Example
Q: Use synthetic division to find the quotient and remainder
for `4x^3 + 2x - 7` divided by `x+3`.
A: We set up the table as follows
Notice the zero in the top row which is there because in `4x^3
+ 2x - 7` the coefficient of `x^2` is `0`.
Now complete the table to get
| -3 |
4 |
0 |
2 |
-7 |
|
0 |
-12 |
36 |
-114 |
|
4 |
-12 |
38 |
-121 |
Therefore the quotient is `4x^2 -12x +38` and the remainder is
`-121`. So
`4x^3 + 2x - 7 = (x+3)(4x^2
-12x +38) -121`
7.2 The Remainder Theorem
In the previous example we found that`4x^3
+ 2x - 7 = (x+3)(4x^2 -12x +38) -121`. If we wanted to
calculate the value of this polynomial at `x=-3` we could stick `-3` in
either the left or the right side of this formula. We will
stick it in the right side to get
`(-3+3)(4(-3)^2 -12(-3) +38)
-121 = 0 times (4(-3)^2 -12(-3) +38) -121 = -121`
So the remainder when we divide by `x+3` gives the value of
the polynomial at `x=-3`. This works in all cases and is
summed up in the remainder theorem.
The
Remainder Theorem
If a polynomial
`f(x)` is divied by `(x-h)` the remainder is `f(h)`.
7.3 The Factor Theorem
The Factor Theorem
If `f(h)=0` then `(x-h)` is
a factor of `f(x)`.
If `(x-h)` is a factor of
`f(x)` then `f(h)=0`.
Example
Q: Show that `x-4` is a factor of `f(x) = 2x^4 - 9x^3 + 5x^2
-3x -4`.
A:
| 4 |
2 |
-9 |
5 |
-3 |
-4 |
|
0 |
8 |
-4 |
4 |
4 |
|
2 |
-1 |
1 |
1 |
0 |
Since `f(4) = 0`, we know that, by the factor theorem that
`(x-4)` is a factor of `f(x)`. Hence `f(x) = (x-4)(2x^3 - x^2
+ x+1).
Example: Finding a polynomial's coeffiecients
Q: If `(x+5)` is a factor of `2x^3 + px^2 - 9x + 30`, find `p`.
A:
| -5 |
2 |
p |
-9 |
30 |
|
0 |
-10 |
-5p+50 |
25p-205 |
|
2 |
p-10 |
-5p+41 |
25p-175 |
Because `x+5` is a factor, the remainder must be `0`, so `25p
-175 =0`. Therefore `p-7`.
7.4 Roots of Polynomials
The number `a` is a root
of a polynomial `f(x)` if `f(a) = 0` (and so by the factor theorem
`(x-a)` is a factor of `f(x)`).
Roots are also called the zeros
of a polynomial.
Roots are where the graph crosses the `x`-axis.
Example
Q: From the sketch of `f(x)` find its equation.
A: The roots are at `-4`, `3` and `10`. So `(x+4)`, `(x-3)`
and `(x-10)` are factors of `f(x)`. Hence
`f(x) = k(x+4)(x-3)(x-10)`
for some `k`. We can find `k` by substituting the
coordinates `(0,240)` of the `y`-intercept into the equation
above. This gives
`240 = k(0 +4)(0-3)(0-10)`
`240 = 120k`
Therefore `k=2`. Hence `f(x) = 2(x+4)(x-3)(x-10)`,
and by expanding the brackets we find that
`f(x) = 2x^3-18x^2-44x+240.`
7.5 Approximate Roots
If `f(a) <0` and `f(b)>0` then we know `f(x)`
has a root between `a` and `b`.
We can use this fact to find approximate values for roots.
Example
Q: Show that `x^3+2x+4` has a root between `-2` and `-1` and
find this root to `1` d.p.
A:
| `x` |
`f(x)` |
Root lies between |
| `-1` |
`1` |
|
| `-2` |
`-8` |
`-2 and -1` |
| `-1.1` |
`0.469` |
`-2 and -1.1` |
| `-1.2` |
`-0.218` |
`-1.2 and -1.1` |
| `-1.15` |
`0.719` |
`-1.2 and -1.15` |
Since the root is between `-1.15` and `-1.2` it is `-1.2` to
`1` d.p.
7.6 Graphs of Polynomials
See handout given in class.
