Polynomials
Table of Contents
Polynomials.................................................................................................................................................................................1
Graphs of polynomials................................................................................................................................................................2
Roots of polynomials..................................................................................................................................................................2
Factors of a polynomial..............................................................................................................................................................3
Quadratics....................................................................................................................................................................................3
Integral coefficients...............................................................................................................................................................4
Completing the square and the formula...............................................................................................................................5
Graphs of quadratics.............................................................................................................................................................6
Solution for non-integral coefficients...................................................................................................................................7
Repeated roots - discriminant zero......................................................................................................................................8
No real roots...........................................................................................................................................................................8
Difference of two squares.....................................................................................................................................................9
Dividing polynomials...................................................................................................................................................................9
Long division of polynomials................................................................................................................................................9
Remainder Theorem............................................................................................................................................................12
Factor theorem....................................................................................................................................................................12
Roots of a cubic...................................................................................................................................................................13
Factorising a cubic : method 1...........................................................................................................................................13
Factorising a cubic : method 2...........................................................................................................................................14
Polynomials
understanding
Polynomials are a type of function. Examples are:
x2+2x+3
2x4+x3+5x-7
x3-5
These are functions of x. It makes no difference what the independent variable is.
f(t) = t2-4
is a polynomial in t.
In general a polynomial is a function of the form
anxn + an-1xn-1+ an-2xn-2.....a1x+a0
where the ans are constants. Some of them might be zero. For example
x3+x2+x+1
x3+2x-5
x3
are all polynomials. The degree of a polynomial is the largest n for which there is a non-zero
an. So
Degree 2 is a quadratic like x2+x+1
Degree 1 is a linear function like 2x-3
Degree 0 is a constant, like 4
Degree 3 is a cubic, like x3-2x2+3x-7
EXERCISE
Which of these functions are polynomials?
1.
f(x) = x2+x+1
2.
f(x) = 1/(x-1)
3.
f(x) = x3+1
4.
f(p) = √p
5.
f(x) =1+x2
6.
f(t) = t-1
Graphs of polynomials
2
understanding
1
A linear function is a straight line, and a quadratic is a
-3
-2
-1
0
1
2
parabola.
-1
A cubic in general has a maximum and a minimum as
-2
shown.
-3
but these can coincide, producing a 'point of
inflection', as for y=x
2
3
1
-3
-2
-1
0
-1
-2
-3
Roots of polynomials
understanding
The roots of a polynomial (or any other function) p(x) are the x values where
p(x) = 0
1
2
On a graph, these are the x values where the graph crosses the x axis.
Factors of a polynomial
understanding
Factorising an integer means finding other integers which when multiplied give the original
integer, such as
12 = 4 X 3
and we can continue this to prime factors
12 = 223
We can do a corresponding process with factorising a polynomial.
For example
x3-6x2+11x-6
= (x-1)(x2-5x+6)
= (x-1)(x-2)(x-3)
If a polynomial p(x) has a factor of, say x-1:
p(x) = (x-1) ( .. another polynomial..)
then clearly at x=1, p(x)=0. We come back to this
later.
9
8
If a polynomial has a factor x-a
7
p(x)=(x-a) ( .. another polynomial.. )
6
5
then at x=a, p is zero. In other words, the graph
4
3
of p crosses the axis at x=a, and a is a root of the
2
polynomial.
1
-3
For example this polynomial:
-2
-1
0
-1
-2
crosses the x axis at -1, +2 and +3.
-3
So it is
-5
(x+1)(x-2)(x-3)
Quadratics
For example
2x2+3x+4
We usually write this as
ax2+bx+c
with a,b and c constants, and a ≠ 0
Solving a quadratic means solving
ax2+bx+c = 0
-4
1
2
3
4
Thinking about the shapes of quadratics,
the graph might cross the x axis twice, or
not at all, or just touch the axis.
Quadratics have 0 or 2
roots
4
So that means a quadratic might have
3
1. Two different factors
2
2. One repeated factor (such as (x-1)2 )
3. No (real) factors
1
Putting x=0 in
-3
-2
-1
0
1
2
3
2
y = ax +bx+c we can see the graph intercepts the y
axis at y=c
-1
-2
Integral coefficients
technique
Exam questions often mean we need to factorise a quadratic where the coefficients are
integers. For example
15x2 -2x -8
We need to find a b c and d in (ax+b)(cx+d)
To do this systematically we have to go through all possibilities. When we multiply the ax
and bx we get 15x2, so and b could be
15 and 1
5 and 3
b times d is -8, so b and d can be
8 and 1 or
2 and 4
But to get -8, we must consider
8 and -1 or -8 and 1
2 and -4 or -2 and 4
We decide which pairs will give us -2x as the cross term.
From 15,1 and -8, 1 we get -119. From 15,1 and 8,-1 we get +119.
No combination of 15, 1 and 8,1 give us 2
Similarly no combination of 15,1and 4,2 give us 2
So a,b must be 5,3
The options are
5,3 and 8,-1 or -8, 1
5,3 and 4,-2 or -4, 2
4X3 = 12 and 5X2 =10 have a difference of 2, so this fits:
15x2 -2x -8 = (5x-4)(3x+2)
If the quadratic has a constant factor common to all terms, factorise it out first - for
example
2x2+6x+4 = 2(x2+3x+2) = 2(x+1)(x+2)
EXERCISE
Factorise these quadratics
1.
x2+3x+2
2.
x2+5x+6
3.
x2-5x+6
4.
x2+2x-3
5.
x2-2x-3
6.
2x2+5x+2
7.
2x2+8x+6
8.
3x2+5x-2
9.
4x2+6x+2
10. 4x2+2x-6
11. 4x2-4x-3
12. 4x2+9x+2
13. 4x2-8x-12
Completing the square and the formula
A quadratic with b=0 is very easy to solve. Such as
x2-9=0 so x2=9 and x = ± 3.
It is easy because we have a square: x2=..something, and we can take the square root.
Suppose we have
x2+2x-3=0
This is like
(x+1)2.....=0 since when we multiply it out we get x2+2x. But we also get +1, and we have to
remove it again:
(x+1)2 -1 -3 =0 so
(x+1)2=4 so x+1 = ± 2 and x=-3 or +1
This method is called 'completing the square'. We do this:
Completing the square
1.
If a≠ 1, divide through by a
2.
Write it as (x+ something)2 choosing something to get the term in x.
3.
Add or subtract the something2
4.
Re-arrange to (x+1)2 = something
5.
Take the square root
For example
2x2+5x-7=0
x2+ 5/2 x -7/2 = 0
( x + 5/4)2 -25/16 -7/2 = 0
(x+ 5/4 )2 = 25/16 + 7/2 = 25/16 + 56/16 =81/16
x+5/4 = ± 9/4
x= 4/4 = 1 or -14/4 = -7/2
We can generalise this:
ax 2 +bx+c=0
a is not zero, so
c
2 b
x + x+ =0
a
a
The formula
( x+
b 2 b2 c
) − 2 + =0
2a
4a a
( x+
b 2 b2 c
) = 2−
2a
4a a
√
√
x+
b
b2 c
=
−
2a
4 a2 a
x+
b
b 2 4 ac √ b 2−4 ac
=
−
=
2a
2a
4 a2 4 a2
so
x=
−b±√ b2 −4 ac
2a
which is just the standard formula
for the roots of a quadratic. So the
formula is just the same as
4
completing the square.
Positive a
3
EXERCISE
2
Solve these by completing the
1
square:
-5
1.
x2-3x+6=0
2.
x2-7x+12=0
3.
x2-x-2=0
4.
x2-3x+3=0
Graphs of
quadratics
-4
-3
-2
-1
0
1
2
3
4
-1
-2
Negative a
-3
-4
-5
For very large x, x2 is greater than bx or c, and the function becomes approximately
ax2
If a is positive, then for very large x, or very large negative x, this is positive. But if a is
negative, these values are negative
In the formula, we get
( x+
b 2
) =a constant
2a
so this is similar to
x2= a constant
with x shifted by b/2a
This means the curve is
symmetrical about x=-b/2a, and
this is a maximum or minimum
5
point.
y=
a
c
4
At x=-b/2a, this value is:
b2
b2
−
+c
4 a2 2 a
c-b2/4a
3
2
2
=
2
b
b
− +c
4a 2a
1
2
=
c−
b
4a
-3
-2
-1
x=-b/2a
Solution for non-integral coefficients
technique
For example
x2+0.1x-0.02
The factors correspond to the roots. So we can find the solutions of
x2+0.1x-0.02 = 0
using the formula:
x=
−0.1±√ .01+ 4 X 0.02
2
=
−0.1± √ .01+0.08
2
=
−0.1± √ .09
2
0
-1
=
−0.1±0.3
2
=
0.2
2
or
−0.4
2
so the roots are x=0.1 and -0.2, and the factors are
(x-0.1)(x+0.2)
We might have factorised this directly - but using the formula works in more difficult cases.
EXERCISE
Solve these
1.
2.5 x2 + 3x - 1.5 =0
2.
0.7 x2 +x -2.3 = 0
3.
1.2 x2 +2x - 4.5 = 0
Repeated roots - discriminant zero
technique
in ax2+bx+c, the expression b2-4ac is called the discriminant (as used in the formula).
If the discriminant is zero, we have repeated roots (a double root).
For example
x2+2x+1=0
so a = 1, b=2 and c=1
so b2-4ac = 0
We can factorise this as
(x+1)2 = 0
so the double root is at x=-1
No real roots
technique
If the discriminant is negative there are no real roots (graph does not cross the x axis).
For example x2+1=0
so a=1, b=0, c=1, then the discriminant = 0 2- 4 X 1 X 1 <0
EXERCISE
How many real roots do these quadratics have?
1.
x2+2x+3=0
2.
x2-2x-3 = 0
3.
2x2-3x+1=0
4.
x2-4x+4=0
Difference of two squares
technique
For example
x2-9 = x2-32 = (x+3)(x-3)
and in general
x2-a2 = (x-a)(x+a)
Another example - factorising x2-16 = (x+4)(x-4)
We can also use this if the constant term is not a square number, by using surds.
For example x2-3 = x2-(√3)2 = (x-√3)(x+√3)
Exercise
1. Simplify
x 2−a2
+a
x +a
2
x −5+ x+ √5
2. Factorise
Dividing polynomials
understanding
We can factorise an integer like 15 = 3 X 5
and we can also factorise polynomials like x2-9 = (x+3)(x-3)
We can also divide polynomials the same way we divide integers.
If we divide, say 47 by 9, we get 5 (the quotient) and remainder 2. In other words
47 = 5 X 9 + 2
When we divide one polynomial by another, we also get a quotient and remainder
Long division of polynomials
technique
For example, what is
6 x 2 +5 x+ 4
2 x+ 1
?
We start off
?
2x+1
6x2
+5x
+4
We need to find ? so that ? times 2x+1 is 6x2. We need to multiply 2x+1 by 3x for that:
3x
2x+1
6x2
+5x
6x2
+3x
+4
We have written 3x times (2x+1) underneath in red. Next we subtract them to see what is
left:
3x
2x+1
6x2
+5x
6x2
+3x
+4
2x
The +4 is also left : bring it down so we do not forget it:
2x+1
3x
+?
6x2
+5x
6x2
+3x
2x
+4
+4
Now we need the ?. When we multiply the 2x+1 by it, we should get 2x. So the ? is just 1:
2x+1
3x
+1
6x2
+5x
6x2
+3x
+4
2x
+4
2x
+1
and we subtract again to see what is left:
2x+1
3x
+1
6x2
+5x
6x2
+3x
+4
2x
+4
2x
+1
3
We cannot multiply anything by 2x+1 to get this - we would have to get an x. So the
remainder is 3:
6x2+5x+4 = (2x+1)(3x+1) + 3
or
6 x 2 +5 x+ 4
3
=(3 x+ 1)+
2 x+ 1
2 x +1
Another example : what is 4x3+5x-7 divided by 2x-1? We write this as 4x3+0x2+5x-7 and use
the same method:
2x-1
2x2
+x
+3
4x3
+0
+5x
4x3
-2x2
2x2
+5x
2x2
-x
-7
6x
-7
6x
-3
-4
So
4x3+5x-7 = (2x-1)(2x2+x+3) -4
This technique can also be used if we know one factor of a polynomial, and we want to find
the rest. For example, if x3-6x2+11x-6 has a factor (x-1), what are the other factors? We
divide the polynomial by x-1:
x-1
x2
-5x
+6
x3
-6x2
+11x
x3
-x2
-5x2
+11x
-5x2
+5x
-6
6x
-6
6x
-6
0
so the remainder is zero, and
x3-6x2+11x-6 = (x-1)(x2-5x+6)
and we can factorise the quadratic by inspection to get
x3-6x2+11x-6 = (x-1)(x-2)(x-3)
Remainder Theorem
technique
Suppose when we divide a polynomial (or another type of function) by (x-a), we get a
remainder R. so
p(x) = (x-a)q(x)+R
where q(x) is some other polynomial (of degree 1 less).
Now let x=a. Then
p(a) = R
For example
p(x) = 4x3+5x-7 = (2x-1)(2x2+x+3) -4
so if x=1/2, p(1/2) = -4
Check
p(1/2) = 4 /8 +5/2 - 7 = 3-7 = -4
We might often use this to find the remainder, without doing the division.
For example, what is the remainder when p(x)=x3-x2-3 is divided by (x-3)?
Simply p(3) = 27-9-3 = 15
Factor theorem
technique
Suppose there is no remainder, so (x-c) is a factor of the polynomial:
p(x) = (x-c) q(x)
Then when x=c,
p(c) = 0
So, c is a root of the polynomial.
Roots of a cubic
understanding
2
If the cubic is
ax3+bx2+cx+d
and a is positive, for very large x
this is positive, and for very
negative x, negative - so it
crosses the x axis
1
-3
-2
-1
0
If a is negative, for large x this is
negative, and positive for very
negative x. So it crosses the x
axis.
-1
So a cubic must have at least
one root - that is, one factor.
-2
1
2
3
If that root is r, then the cubic is
(x-r)( .. some quadratic ... )
Now the quadratic might have 0, 2 or a double root.
So a cubic can have 1 factor, 3 factors, or 1 factor and a repeated factor.
Graphically, this is shown here. The green cubic has one root, the blue has one root plus a
double root, and the red cubic has three distinct roots.
Factorising a cubic : method 1
technique
•
Factorise the x3 and x2 terms
•
Factorise the x and constant term
•
Look for a common factor.
•
Factorise the quadratic which is left if possible
For example
x3+2x2 -9x-18
=x2(x+2) -9(x+2)
=(x+2)(x2-9)
=(x+2)(x-3)(x+3)
Another example:
x3-x2+3x-3
=x2(x-1) +3(x-1)
=(x-1)(x2+3)
This cubic has only 1 root.
Factorising a cubic : method 2
technique
•
Find all factors of the constant term
•
Try each and see if any make the cubic 0
•
If c makes f(c)=0, then x-c is a factor
Suppose the cubic is actually
(x-a)(x-b)(x-c)
then the constant term is abc. So the factors of the constant term includes the roots.
For example
p(x) = x3+4x2+x-6
The factors of the constant term are 1,2 and 3
p(1)=0
so 1 is a root and x-1 is a factor.
We could find the co-factor by polynomial division of x3+4x2+x-6 by x-1. Or:
x3+4x2+x-6 = (x-1)(ax2+bx+c)
To get x3, a must be 1. And to get -6, c must be 6. So
x3+4x2+x-6 = (x-1)(x2+bx+6)
the term in x2 is 4x2 = (-1+b)x2
so b=5, so
x3+4x2+x-6 = (x-1)(x2+5x+6)
and the complete factorisation is then
x3+4x2+x-6 = (x-1)(x+2)(x+3)