9 Polynomials 2 TERMINOLOGY
Continuous function: A function: A function is continuous over an interval if it has no break in its graph. For every x value value on the graph the limit exists and equals the function value
Newton’s method of approximation: An approximation: An approximation method to find a solution to an equation. If x a is close to the root of a polynomial, then the x -intercept -intercept of the tangent at that point will generally be closer
Halving the interval: Taking interval: Taking an x -value -value halfway between 2 other values and determining the sign of the y -value -value in order to find an approximate solution to an equation
Polynomial: A Polynomial: A function of the form + a x where P (x) = a0 + a 1 x + a x + a , a , a 0 1 are real numbers and n is a positive integer or zero
Interval: A section of a line or curve over a restricted Interval: A domain including the endpoints
Root of a polynomial: polynomial: The The solution of a polynomial equation. Graphically, Graphically, it is where the polynomial crosses the x -axis -axis
Iteration: Starting with an initial value, often a guess, and Iteration: Starting using a process over and over again to produce a closer approximation
=
2
2
n
…
…
n
n
Chapter 9 Polynomials 2
INTRODUCTION POLYNOMIALS ARE POLYNOMIALS ARE AN IMPORTANT part of algebra and
are used in many branches of mathematics. You studied polynomials in the Preliminary Course. In this chapter you will study the estimation of roots of polynomials by using the methods of halving the interval and Newton’s method.
DID YOU KNOW? The word ‘polynomial’ means an expression with many terms. (A binomial has 2 terms and a trinomial has 3 terms). ‘Poly’ means ‘many’, and is used in many words, for example, polyanthus, polygamy, polyglot, polygon, polyhedron, polymer, polyphonic, polypod and polytechnic. Do you know what all these words mean? Do you know any others with ‘poly-’?
Estimation of Roots The roots of a polynomial equation a0 + a1 x + a 2 x2 + … + a x-intercepts of the graph P(x) = a0 + a1 x + a2 x2 + . . . + an xn
n
x
n
=
0 are
the
In some cases, we can find the roots of the equation by using algebra.
447 44 7
448
Maths In Focus Mathematics Extension 1 HSC Course
EXAMPLES 1. Find the exact roots of the quadratic equation x2 −
3x
+
1
=
0.
Solution x=
=
=
=
2
−b ±
b −
4ac
(− 3 ) 2
−
2a − (− 3 ) ±
3± 3± 2
4 (1) (1)
2 ( 1) 4
9− 2 5
Note: The roots are the x-intercepts of the graph of the parabola y
=
x
2
−
3x
+
1.
2. For the polynomial P(x) = x3 − 2x2 − 5x + 6 (a) show that x − 1 is a factor of the polynomial. (b) write the polynomial as a product of its factors. (c) find the roots of the polynomial equation x3 − 2x2 − Remember from the Preliminary Course that x − a is a factor of P(x) if P(a) = 0.
Solution (a) If x − 1 is a factor of the polynomial, then P (1) P (x) P (1)
=
3
x
3
1 = 0 =
− −
2x2
−
2
2(1)
5x −
+
6
5(1) + 6
So x − 1 is a factor.
=
0.
5x
+
6
=
0.
Chapter 9 Polynomials 2
(b) Divide the polynomial by x
1
x −
g
x x
2
−
3
−
3
x −
2x
− x −x
−
You learned how to do this in the Preliminary Course.
− 1 to find other factors.
6 5x
+
6
2
2
−x
2
x
5x
−
2
+ x − 6x +
6
− 6x +
6 0
So P (x) (c)
x
3
−
(x = (x =
2x2
(x − 1) (x x −
−
−
1) (x 1) (x
−
−
x
−
6)
3) (x + 2)
5x + 6 = 0 3) (x + 2) = 0 −
1 = 0, x x =
−
2
−
1,
3
=
x =
0, x
+
3,
2
=
0
x = −
2
Note: The roots are the x-intercepts of the graph of the polynomial P (x)
=
x
3
−
2x2 − 5x
+
6
It is not always possible to find the roots of a polynomial equation by algebraic methods. We can estimate the roots by looking at the graph.
EXAMPLE Use stationary points to sketch the curve y = 2x3 + 3x2 − 12x + 5 and use the graph to give a rough estimate of the roots of the equation 2x
3
+
3x
2
−
12x
+
5
=
0
Solution y = 2x3 + 3x2 − 12x + dy 2 = 6x + 6x − 12 dx
449
5
CONTINUED
450
Maths In Focus Mathematics Extension 1 HSC Course
For stationary points 6x2 + 6x
12
=
+ x −
2 2) (x − 1)
=
0,
1
x
(x + x +
2
2
=
−
x = −
When x
x −
2,
= −
2(− 2)3 = 25
y =
When x y = 2(1) = −
dx
=
0.
0 = 0 =
0
x =
1
2
+
=
3
0
dy
2
3(− 2)
−
12(− 2) + 5
1
3(1)2
+
−
12(1) + 5
2
So (− 2, 25) and (1, − 2) are stationary points. d2 y dx2
12x
=
+
6
At (− 2, 25) d2 y dx2
12(− 2) + 6
=
= −
18
(concave downwards)
0
<
So (− 2, 25) is a maximum turning point. At (1, − 2) d2 y dx2
=
12(1) + 6
=
18
(concave upwards) So (1, − 2) is a minimum turning point. >
0
For x-intercept, y 0
=
2x
3
+
3x
2
=
12x
−
0
5
+
This will not factorise so we cannot find the x-intercepts. For y -intercept, x 3
y = 2(0) =
+
2
3(0)
−
=
0
12(0) + 5
5
There are 3 x-intercepts, so 3 roots of the equation 2x
3
+
3x
2
−
12x
+
5 = 0.
One root is where x < − 2, say around x 3. Another root is between x 0 and x 1, say x 0.2. The third root is where x > 1, say x 3. = −
=
=
=
=
Chapter 9 Polynomials 2
There are two iterative methods of estimating roots of polynomial equations that you will study in this course. These are called halving the interval and Newton’s method of approximation. In these methods, we guess a solution and then use a process over and over to produce closer approximations. These methods can also be used to find roots of functions that are not polynomials, as long as the function is continuous over the interval.
DID YOU KNOW? Niels Abel (1802–29), a
Norwegian mathematician, proved that it is impossible to find a general method to solve quintic equations (equations of degree 5). He was 22 years old when he made this discovery. Abel established many theorems in his short life, especially in the area of group theory, which is one of the fundamentals of abstract algebra. He died at the age of 26.
Halving the interval
If f (x) is continuous for a ≤ x ≤ b and f (a) and f (b) have opposite signs, then there is at least one root of f (x) 0 in that interval. =
451
452
Maths In Focus Mathematics Extension 1 HSC Course
This method is also called bisection.
We can find an approximation to the root between a and b by halving this interval.
If f (a) < 0 and f (b) > 0, then there are three possibilities: 1. f
ca bm
=
0 means
2. f
ca bm
<
0
>
0 means
x
3. f x
+
2
+
2
=
+
2
b
is a root of the equation
means the root lies between
x
a =
+
b
2
b
ca bm +
2
a =
a
+
2
b
the root lies between x
=
and
a
and
Chapter 9 Polynomials 2
If we halve the interval several times, the approximation to the root will usually, but not always, become more accurate.
EXAMPLES 1. (a) Show that a root of x3 − 3x2 − 9x + 1 = 0 lies between x 4 and x 5. (b) By halving the interval, show that the root lies between 4.75 and 4.875. =
Solution (a) f (4) = 43 − 3(4)2
−
9(4) + 1
−
9(5) + 1
19
= −
f (5) = 53 = 6
−
3(5)2
Since f (4) < 0 and f (5) > 0, the root lies between 4 and 5.
(b) f 4 + 5
c
2
m
=
f (4.5)
4.53 − 3(4.5)2 = − 9.125 =
−
9(4.5) + 1
Since f (4.5) < 0, the root lies between 4.5 and 5.
CONTINUED
=
453
454
Maths In Focus Mathematics Extension 1 HSC Course
f
c 4.52 5 m +
=
f (4.75)
4.753 − 3(4.75)2 = − 2.265625 =
−
9(4.75) + 1
Since f (4.75) < 0, the root lies between 4.75 and 5.
f
c 4.752 5 m +
=
f (4.875)
4.8753 = 1.686 =
−
3(4.875)2
−
9(4.875) + 1
Since f (4.875) > 0, the root lies between 4.75 and 4.875.
Note: Since f (4.875) is closer to 0 than f (4.75), x approximation than 4.75.
=
4.875 is
a better
2. (a) Show that 3x4 + 4x3 − 12x2 − 1 = 0 has a root between x 3 2. and x (b) Use the method of halving the interval to show that the root lies between − 3 and − 2.75. =
=
−
Solution (a) f (− 3) = 3(− 3)4 =
+
4(− 3)3
−
12(− 3)2
−
1
+
4(− 2)3
−
12(− 2)2
−
1
26
f (− 2) = 3(− 2)4 = − 33
−
Chapter 9 Polynomials 2
Since f (− 3) > 0 and f (− 2) < 0, the root lies between − 3 and − 2.
(b) f
c
−
3 + −2 2
m
=
f (− 2.5)
3(− 2 .5) 4 + 4(− 2 .5)3 = − 21.3125 =
−
12(− 2.5)2
−
1
Since f (− 2. 5) < 0, the root lies between − 3 and − 2.5.
f
c
−
3 + −2.5 2
m
=
f (−2.75)
3(−2.75)4 = − 3.363 =
+
4(− 2.75)3
−
12 (−2.75)2
−
1
Since f (− 2.75) < 0, the root lies between − 3 and − 2.75.
Note: f (− 2.75) is closer to zero than f (− 3) . So − 2.75 is a closer approximation to the root than − 3. This does not always happen. Sometimes the first or second value of f (x) found can be closer to the root than subsequent values.
455
456
Maths In Focus Mathematics Extension 1 HSC Course
Computer Application The method of approximation by halving is tedious and can easily be done on a computer. Some computer programs already have this method built in. Otherwise, you could put the formula into a spreadsheet.
9.1 Exercises 1.
Show that there is at least one root between the x-values given for each function. (a) f (x) x2 2x 5 between x 3 and x 4 (b) f (x) = 7 + x − x2 between x 3 and x 4 (c) p(x) = 7 + x − x 2 between x 2 3 and x 3 (d) P (x) = 2x − 3x2 − 72x + 1 between x 0 and x 1 (e) f (x) = x4 − 5x2 + 2 between x 2 and x 3. =
−
=
6.
The function f (x) = x2 + 5x + 1 has 5 and x 4. a root between x By halving the interval, show 5 that the root lies between x and x 4.5. Which is the better approximation? =
−
−
=
=
=
=
=
−
−
−
=
=
−
=
2.
=
Show that there is no root between x 0 and x 1 on the curve f (x) = 2x3 − 3x2 + x + 1. =
(a) Show P (x)
=
4x3
+
15x2 −
21x − 5 has
a
root between x 1 and x 2. (b) By halving the interval twice, show that P (x) has a zero of 1.25. =
5.
=
(a) Show f (x) = x2 + x − 3 has a root between x 1 and x 2. (b) By halving the interval once, show that the root lies between 1.5. x 1 and x =
=
=
9.
−
=
The curve y x2 x 7 has a root between x 3 and x 4. By halving the interval twice, find an approximation to the root. =
−
−
=
=
=
=
4.
8.
=
=
3.
=
=
(a) Show f (x) = 2x2 + 5x − 2 has a root between x 0 and x 1. (b) By halving the interval once, show that the root lies between 0 and x 0.5. x
Find an approximation to the root of the function f (x) x7 2 that lies between x 1 and x 2, by halving the interval once. =
−
=
=
7.
=
The function f (x) = x3 + 4x2 + 1 has a root between x 5 and 4. Halve the interval twice x and find an approximation to the root. =
=
−
−
10. (a) Sketch the function f (x) = 2x3 + 9x2 + 12x + 1. (b) How many zeros does f (x) have? Between which x-values do they lie? (c) By halving the interval twice, find approximations to these zeros.
=
11. (a) Show that e = sin x + 3 has a root between x 1 and x 2. (b) Use the method of halving the interval three times to find a closer approximation to the root. x
=
=
Chapter 9 Polynomials 2
12. For the polynomial P(x) (a) show that when P(x)
=
x
3 =
x
=
3
−
14. (a) Show that a root of ln (x + 3) − e = 0 lies between 0.1 and x 0.2. x (b) By halving the interval, show that the root lies between 0.125 and x 0.15. x
12
x
0,
12
=
(b) show that a root of P(x) 0 lies between x 2 and x 3 (c) use the method of halving the interval twice to find an approximation to 3 12 . =
=
=
x −
5
=
0 lies
between
x
=
=
=
=
15. A root of P(x) 2x x 3 lies between x 1 and x 2. Show that the root lies between 1.25 and x x 1.375 by using the method of halving the interval. =
=
13. (a) Show that a root of sin x
457
=
2
and x 3. (b) Use the method of halving the interval twice to find an estimate of the root. =
−
=
=
Newton’s method of approximation Newton’s method is a different way of approximating the root of a polynomial equation. It generally gives a more accurate approximation than the method of halving the interval, as well as taking fewer steps to get this approximation. Sketching y f (x), a continuous function, shows how Newton’s method works. =
The function only needs to be continuous in the interval near the root.
If x a is close to the root of the equation f (x) (a1) of the tangent at a is usually closer to the root. =
a1
=
a
f (a) −
l
f (a)
, where x
=
=
0, then the x-intercept
is close to the root
a
458
Maths In Focus Mathematics Extension 1 HSC Course
We often halve the interval to get the first approximation of x a.
Proof
=
Let P [a, f (a)] be a point on the curve y f (x) and let point R be the x-intercept of the tangent at P. The gradient of the tangent to the curve is given by f (x) \ the gradient of the tangent at P is given by f (a) The equation of the tangent at P is given by y y1 m(x x1) =
=
l
l
y At R, i.e.
−
−
=
f (a)
=
y
0
−
−
l
a f (a) l
a f (a)
−
−
=
f ( a)
=
f (a) x
f (a)
=
f ( a) x
=
x
=
x
=
x
l
−
a
−
l
f (a)(x
f (a)
l
f (a) f ( a) l
f (a)
−
a)
0
=
l
l
l
f (a) (x
f ( a)
f (a) a f (a) f ( a) f (a)
−
l
−
−
a) l
af (a)
l
the x-intercept a1 of the tangent at P is given by
\
a1
You could use a computer for these calculations.
=
a
f ( a) −
l
f (a) Several applications of Newton’s method will often give a closer approximation.
EXAMPLES 1. Find an approximation to the root of x3 + x − 1 = 0 by using Newton’s method once and starting with an approximation of x 0.5. =
459
Chapter 9 Polynomials 2
Solution
f (0.5) = 0.53 + 0.5 − 1 = − 0.375 f (x) = 3x2 + 1 l
f (0.5) = 3(0.5) 2 + 1 = 1.75 l
a1 = a − =
f ( a) l
f (a) f (0.5)
0. 5 −
=
0. 5 −
=
0.714
l
f (0.5) 0.375 1.75
f (0.714)
−
x
=
Z
0.079, so
0.714 is
a good approximation.
2. (a) Sketch f (x) x2 4x 1 and show on your sketch that one root of 2 5. 4x 1 0 lies between x x 4 and x (b) By choosing an approximation of x 4.5, use Newton’s method once to find an approximation to this root. =
−
−
=
−
−
=
=
=
Solution (a) Sketch y f (x) by finding its axis of symmetry, or using a table of values. =
CONTINUED
460
Maths In Focus Mathematics Extension 1 HSC Course
(b) f (4.5)
=
=
l
f (x)
2x
=
2(4.5)
=
5
=
a
=
4.5
=
4.5
=
4.25
l
f (4.25)
=
0.0625, so x
=
4.25 is
−
=
f (4.5)
a1
4.52 4(4.5) 1.25
−
1
4
−
4
−
f (a) −
l
f (a) f (4.5) −
−
l
f (4.5) 1.25 5
So an approximation to the root is x
a good approximation.
3. Find an approximation to the root of The approximate root of cos x
x =
3
was found
graphically to be x Chapter 5.
=
4.25.
cos
x x
Solution
1.1 in
cos
i.e. cos x
x x
=
3
x −
f (x)
Then
f (x)
l
0
=
3
Let
Use
=
cos x
=
−
=
1.1 as
f (1.1)
=
cos 1.1
=
0.0869
=
−
sin 1.1
=
−
1.2245
=
a
=
1. 1
l
a1
−
sin x
x
f (1.1)
1.17 is a much closer approximation to the root than 1.1.
=
x 3 −
1 3
the first approximation. −
1.1 3
−
1 3
f ( a) −
l
f (a) f (1.1) −
l
f (1 .1)
Z 1.17
Newton’s method doesn’t work in some cases.
=
3
near x
=
1.1.
461
Chapter 9 Polynomials 2
EXAMPLE Find an approximation to the root of y = x3 + x − 1 by using Newton’s method once and starting with an approximation of x 0.2. =
Solution
=
f (0.2) = 0.23 + 0.2 − 1 = − 0.792 f (x) = 3x2 l
+
=
=
1
f (0.2) = 3 (0.2) 2 = 1.12 l
a1
The root lies between x 0 and x 1. The approximation of x 0.2 is not a good choice as it is too far from the root.
+
1
f ( a)
=
a
=
0. 2
−
l
f (a) f (0.2) −
l
f (0.2) 0.792 0. 2 1.12 0.907 −
=
=
−
This is not a close approximation to the root.
Class Investigation Here are some examples where the approximation of a root either is not very close to the root, or cannot be found. Discuss these examples. 1.
In this case, the approximation of a1 will be further away from the root than a. Can you see why?
Z 0 .65 , which is not very close to 0.
f (0 .9 07 )
462
Maths In Focus Mathematics Extension 1 HSC Course
2.
In this case, a1 cannot be found. 3.
The x-intercept a1 is too far away from the root in this case.
Can you find any more examples where Newton’s method doesn’t work?
9.2 Exercises 1.
The polynomial equation 2 2x 9 0 has a root near x 2.5. Find an approximation x to the root, correct to 2 decimal places, by using 1 application of Newton’s method. −
−
4.
=
=
2.
The equation f (x) 0 where f (x) = x3 − 3x2 + 9 has a zero near 1.5. By using 1 application x of Newton’s method, find an approximation to the root, correct to 2 significant figures.
3.
=
=
=
=
=
=
−
The polynomial equation 3 2 4x − x + x − 2 = 0 has a root near x 0.6. Use 2 applications of Newton’s method to find an approximation to the root, correct to 3 decimal places.
(a) Show that the polynomial equation x3 − 2x2 − 6x + 3 = 0 has a root between x 0 and x 1. (b) Use Newton’s method with 1 application to find an approximation to the root, correct to 3 decimal places, using 0.5 as an approximation to x the root. 1
5.
(a) Show that x 2 + x − 1 = 0 has a root between x 0 and x 1. (b) Use 1 application of Newton’s method to find an approximation of the root to 3 decimal places. =
=
Chapter 9 Polynomials 2
6.
(a) Show that the equation 3 2 x − 2x + x + 5 = 0 has a root 1. 2 and x between x (b) Use the method of halving the interval twice to find an approximation to this root of the equation. (c) Use Newton’s method once to find an approximation to this root of the equation, correct to 2 decimal places, starting with 1.5. x =
=
7.
=
−
−
(a) Show that the equation 2 2x + x − 2 = 0 has a root between 0 and x 1. x (b) Use the method of halving the interval twice to find an approximation to this root of the equation. (c) Use Newton’s method once to find an approximation to this root of the equation, correct to 1 decimal place. (d) Use the quadratic formula to find an approximation to this root of the equation, correct to 2 decimal places. =
8.
−
=
(a) For the polynomial 3 P(x) x 9, show that the equation P (x) 0 can be written as x 3 9 . (b) Between which 2 integers does 3 9 fall? (c) By using Newton’s method once, find an approximation to 3 9 , correct to 2 decimal places. =
−
=
10. The curve y (2x 3)5 x has a root between x 2 and x 3. Use Newton’s method with 1 application to find an approximation to the root, correct to 2 decimal places. Is this a good approximation? =
−
−
=
=
11. The equation log x x3 2 has a root between x 1 and x 2. Use f (x) = loge x − x3 + 2 and Newton’s method once to find an approximation to the root, correct to 2 decimal places. e
=
−
=
=
12. The root of the equation 3 e x 1 and 0 lies between x 2. Use Newton’s method once x to find an approximation to the root, correct to 2 decimal places. x
−
=
=
=
13. Use a first approximation of 0.6 with 1 application of x Newton’s method to solve tan x x, correct to 2 decimal places. =
=
14. (a) Find two integers between which the root of ln (x + 3) = x − 1 lies. (b) Use Newton’s method once to find an approximation, correct to 2 decimal places.
=
9.
Use Newton’s method with 1 application to find, correct to 2 decimal places, (a) 3 5 (b) 5 400 (c) 4 55 (d) 7 50
15. Use x 0.5 to find an approximation to the root of cos x x, correct to 2 decimal places, using 2 applications of Newton’s method. =
=
463
464
Maths In Focus Mathematics Extension 1 HSC Course
Test Yourself 9 1.
The polynomial equation 3x2 2x 3 0 has a root near x 1. Use two applications of Newton’s method to find a closer approximation to the root, correct to 2 decimal places. −
−
=
=
2.
7. Solve log x x3 2 by using two applications of Newton’s method, starting with a first approximation of x 1. 8.
=
=
(a) Show that there is a root of f (x) x4 3x2 5 between x 2 and =
x
=
−
−
=
3.
(b) Using x 2 as a first approximation, use Newton’s method to find a closer approximation. =
4.
(a) Show that the function f (x) = x3 − 5x + 7 has a root between 2. x 3 and x (b) By halving the interval twice, find the closest approximation to the root. =
5.
6.
−
=
−
(a) Sketch the curve y = 2x3 − 9x2 + 7 showing any stationary points. (b) Find the x-intercepts on the curve, using Newton’s method where necessary, to find an approximation to 1 decimal place. Find an approximation to the root of 3 x + 3x − 7 = 0 by halving the interval 3 times.
(a) Use x 2 as a first approximation to find the solution to x5 10, using two applications of Newton’s method (answer correct to 2 decimal places). (b) Find the solution of x5 10 directly. =
=
x
3.
−
=
Use one application of Newton’s method to find an approximation to the root of e x3 0, starting with a first approximation at x 2. −
=
e
=
9.
(a) Show that a root of x3 − 3x + 1 = 0 lies between x 0 and x 1. (b) Explain why Newton’s method will not work with a first approximation of x 1. (c) Use Newton’s method with x 0 as a first approximation to find the root of the equation. (d) Halve the interval twice to find an approximation to the root. =
=
=
=
10. A root of sin x x
=
x =
3
lies between
x
=
2 and
3.
(a) Halve the interval twice to find an approximation to the root. (b) Use one application of Newton’s method to find the root, starting with x 2 as a first approximation. =
Chapter 9 Polynomials 2
Challenge Exercise 9 1.
(a) Sketch the curve y = x4 − 6x 2 + 8x + 1. (b) Estimate the roots of equation 4 2 x − 6x + 8x + 1 = 0 , by choosing appropriate approximations for the roots and using 1 application of Newton’s method. (c) One of the roots is a poor estimation. Which one is it? (d) By halving the interval twice, find the best estimate of this root.
2. Given P (x) x3 a, the equation P(x) 0 has a root near x b where b ≠ 0. Show that an approximation to the root is =
−
=
=
given by
x1
2b =
3
+
3b
3
a
.
3.
(a) Show that a root of
sin
x
x
3
−
=
0 lies
between x 6 and x 6.5. (b) Use the method of halving the interval to show that this root lies between x 6.25 and x 6.375. (c) Use Newton’s method to find an approximation of this root correct to 3 decimal places using a first approximation of 6.25. =
=
=
=
465
