polynomials expressions like 3x 4 + 2x 3 – 6x 2 + 11 and m 6 – 4m 2 +3 are called polynomials....
TRANSCRIPT
![Page 1: Polynomials Expressions like 3x 4 + 2x 3 – 6x 2 + 11 and m 6 – 4m 2 +3 are called polynomials. (5x – 2)(2x+3) is also a polynomial as it can be written](https://reader030.vdocument.in/reader030/viewer/2022032805/56649ee45503460f94bf3c69/html5/thumbnails/1.jpg)
Polynomials
Expressions like 3x4 + 2x3 – 6x2 + 11 and m6 – 4m2 +3 are called polynomials.
(5x – 2)(2x+3) is also a polynomial as it can be written 10x2 + 11x - 6.
The degree of the polynomial is the value of the highest power.
3x4 + 2x3 – 6x2 + 11 is a polynomial of degree 4.
m6 – 4m2 + 3 is a polynomial of degree 6.
In the polynomial 3x4 + 2x3 – 6x2 + 11, the coefficient of x4 is 3 and the coefficient of x2 is -6.
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A root of a polynomial function, f(x), is a value of x for which f(x) = 0
3 2 2
1. Express each polynomial as a poduct of factors.
(a) 5 10 ( ) 9 16 ( ) 6 7 5x x b x c x x
2( ) 5 ( 2)a x x
( ) (3 4)(3 4)b x x
( ) (2 1)(3 5)c x x
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22. Find the roots of the function ( ) 3 12f x x
23 12 0x 23( 4) 0x
3( 2)( 2) 0x x
2 0 2 0x x 2 2x x
Hence the roots are 2 and –2.
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Division by (x – a)
Dividing a polynomial by (x – a) allows us to factorise the polynomial.
You already know how to factorise a quadratic, but how do we factorise a polynomial of degree 3 or above?
We can divide polynomials using the same method as simple division.
8 3504
323 0
3
246 350 8 43 remainder 6
Conversely, 350 8 43 6
Divisor Quotient Remainder
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3 2( 2) 5 2 8 2x x x x
25x
3 25 10x x212x 8x
12x
212 24x x32x 2
32
32 64x 62
3 2 2(5 2 8 2) ( 2) ( 2)(5 12 32) 62x x x x x x x
Divisor Quotient Remainder
Note: If the remainder was zero, (x – 2) would be a factor.
This is a long winded but effective method. There is another.
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3 2 2(5 2 8 2) ( 2) ( 2)(5 12 32) 62x x x x x x x
Let us look at the coefficients.
2 5 2 8 -2
5 10
12 24
32 64
62
quotientdivisor remainder
This method is called synthetic division.
If the remainder is zero, then the divisor is a factor.
Let us now look at the theory.
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3 2( ) 5 2 8 2f x x x x
We divided this polynomial by (x – 2).
(2)f 62
If we let the coefficient be Q(x), then
( ) ( ) ( ) ( )f x x h Q x f h
Remainder Theorem
If a polynomial f (x) is divided by (x – h) the remainder is f (h).
We can use the remainder theorem to factorise polynomials.
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Remainder Theorem
If a polynomial f (x) is divided by (x – h) the remainder is f (h).
If ( ) 0 then ( ) is a factor of ( )f h x h f x
Conversely, if ( ) is a factor of ( ) then ( ) 0.x h f x f h
4 3 21. Show that ( 4) is a factor of 2 9 5 3 4x x x x x
4 2 -9 5 -3 -4
28
-1-41
41
40
Since the remainder is zero, (x – 4) is a factor.
3 2( ) ( 4)(2 1)f x x x x x
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3 22. Factorise fully 2 5 28 15x x x
To find the roots we need to consider the factors of -15.
1, 3, 5, 15
3 2 5 -28 -15
2611
335
150
2( ) ( 3)(2 11 5)f x x x x ( 3)(2 1)( 5)x x x
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4 23. Factorise fully 7 18.x x
To find the roots we need to consider the factors of -18.
1, 2, 3, 6, 9, 18
3 1 0 -7 0 -18
133
92
66
3 2( ) ( 3)( 3 2 6)f x x x x x
180
-3 1 3 2 6
1-30
02
-60
2( ) ( 3)( 3)( 2)f x x x x
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Finding a polynomial’s coefficientsWe can use the factor theorem to find unknown coefficients in a polynomial.
4 3 21. If ( 3) is a factor of 2 6 4 15, find . x x x px x p
Since we know (x + 3) is a factor, the remainder must be zero.
-3 2 6 p 4 -15
2-60
0p
-3p
4 - 3p
9p - 12
9p - 27 9 27 0p 9 27p
3p
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4 3 22. If ( 2) and ( 4) are factors of 8, find and . x x x ax x bx a b
2 1 a -1 b -8
12
2 + a4 + 2a3 + 2a
6 + 4a
6 + 4a + b
12 + 8a + 2b
4 + 8a + 2b = 0
-4 1 a -1 b -8
1-4
a - 416 - 4a15 - 4a
16a - 60
16a + b - 60
240 - 64a - 4b
232 - 64a - 4b = 0
8 2 4
64 4 232
a b
a b
( 2) 16 4 8
64 4 232
a b
a b
48 240a
5a 40 2 4b 22b
5 and 22a b
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Solving polynomial equations3 21. Find the roots of ( ) 4 6 0f x x x x
3 1 -4 1 6
13-1
-3-2
-60
3 2 24 6 ( 3)( 2)x x x x x x ( 3)( 2)( 1)x x x
Roots are 1, 2 and 3.
If we sketch the curve of f (x), we see that the roots are where f (x) crosses the X axis.
y
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Functions from Graphs
f (x)
x
d
a b c
The equation of a polynomial may be established from its graph.
( ) ( )( )( )f x k x a x b x c
is found by substituting the point (0, )k d
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1. From the graph, find an expression for f (x).
f (x)
x
-12
-3 2
( ) ( 3)( 2)f x k x x
Substituting (0, -12)
12 (0 3)(0 2)k 12 6k
2k ( ) 2( 3)( 2)f x x x
22( 6)x x 22 2 12x x
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2. From the graph, find an expression for f (x).
f (x)
x
30
-2 1 5
( ) ( 2)( 1)( 5)f x k x x x
Substituting (0, 30)
30 (0 2)(0 1)(0 5)k 30 10k
3k
( ) 3( 2)( 1)( 5)f x x x x 23( 2)( 6 5)x x x
3 23 12 21 30x x x
3 2 23( 6 5 2 12 10)x x x x x 3 23( 4 7 10)x x x
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Curve sketchingThe factor theorem can be used when sketching the graphs of polynomials.
3 21. Sketch the graph of 8 12y x x x
The Y axis intercept is (0, 12)
Y axis Intercept
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We will use synthetic division to find the roots of the function. This will tell us where the graph crosses the X axis.
2 1 -1 -8 12
121
2-6
-120
2( 2)( 6)y x x x ( 2)( 2)( 3)x x x
This give us the points (0,2) and (0, -3)
X axis Intercept
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Stationary Points
3 2 8 12y x x x
23 2 8dy
x xdx
0 for s.p.
(3 4)( 2) 0x x
3 4 0x 3 4x
4
3x
2 0x 2x
3 24 4 4 4
When , 8 123 3 3 3
x y
1418
27
3 2When 2, 2 2 8 2 12x y 0
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Nature of Stationary Points
4 4 42 2
3 3 3x
(3 4)x ( 2)x
dy
dx
Slope
-
-
+ 0
+
-
- 0
+
+
+
4 14Maximum T.P. at ,18
3 27
Minimum T.P. at 2,0
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y
x
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1
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3 27
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Approximate Roots
When the roots of f (x) = 0 are not rational, we can find approximate values by an iterative process.
We know a root exists if f (x) changes sign between two values.
f (x)
xa b
f (x)
xa b
( ) 0f x at x a ( ) 0f x at x b
A root exists between a and b.
( ) 0f x at x a ( ) 0f x at x b
A root exists between a and b.
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3 21. For the function ( ) 4 2 7 show that there is a real
root between 1 and 2. Find this root to two decimal places.
f x x x x
(1) 2f (2) 5f
Hence the graph crosses the x - axis between 1 and 2.
( ) Root lies betweenx f x
1 22 -5 1 and 2
1.5 -1.625 1 and 1.51.4 -0.896 1 and 1.41.3 -0.163 1 and 1.31.2 0.568 1.2 and 1.31.25 0.203 1.25 and 1.31.26 0.130 1.26 and 1.31.27 0.057 1.27 and 1.31.28 -0.016 1.27 and 1.281.275 0.020 1.275 and 1.28 Hence the root is 1.28 to 2 d.p.