FUNCTIONS: POLYNOMIALS: REMAINDER THEOREM
FUNCTIONS
FUNCTIONS
Example I: F: x→2x – 1. This reads as:
the function F such that (:) any
input x is mapped into (→) the value
2x – 1. Hence, if x = 3, the function, f: (x
= 3) → 2(3) – 1 = 6 – 1 = 5
Example 2: if g(x) = 2x3 – 3x + 1, find the values of g(0),
g(-1), g(t) and g(2x)
Solutions:
g(0) = 2(0)3
– 3(0) + 1 = 1
g(-1) = 2(-1)3
– 3(-1) + 1 = -2 + 3 + 1 = 2
g(t) = 2(t)3 – 3(t) + 1
= 2t3 – 3t + 1
g(2x) = 2(2x)3 – 3(2x) +
1 = 16x3 – 6x + 1
Example
3: if f(x) = x +
1
x2 + x - 2
Find
(a) the image of -1 and (b) f(x – 1)
and (c) what values of x will have no
images for this function
Solutions:
(a)
f(-1) = -1 + 1 = 0 = 0
(-1)2 + (-1) – 2 -2
(b)
f(x
– 1) = x
– 1 + 1 = x = x .
(x – 1)2 + (x –
1) – 2 x2
– 2x + 1 + x – 1 – 2 x
2 – x – 2
(c)
To find the values of x that have no
images for this function, equate the denominator to zero, hence x2 + x – 2 = 0
or (x + 2) (x - 1) = 0
Hence,
the values of x = -2 and +1 have no
images for this function.
DISTINCTION
BETWEEN A FUNCTION AND AN EQUATION
FUNCTION: Example y = f(x) = x2
– 3x – 10 = (x – 5)(x + 2). A function cannot be solved but is used
to find the images of x. The values
of x which makes f(x) = 0 are called zeros of f. The zeros of this
function are x = 5 and x = -2.
EQUATION: Example x2 – 3x – 10 = 0. An equation is solved to find its roots, which are the values of x
which satisfy the equation. The roots of this equation are x = 5 or x = -2
TYPES OF
FUNCTIONS
(a)
Linear function eg y = 2x – 3
(b)
Quadratic function eg y = x2
– x – 10
(c)
Polynomial function eg y = 2x3
– 3x2 + x – 4
(d)
Trigonometric function eg y = sin x0
(e)
Logarithmic function eg y = log x
(f)
Exponential function eg y = 2x
QUESTIONS
1.
Find the values stated for these functions:
(a)
f(x)
= 3x – 1: f(0), f(-1), f(1)
(b)
f(x) = x2 + x + 1: f(-1), f(0), f(x + 1)
(c)
g(x) = x2 – x – 2: g(-1), g(½), g(0), g(-½)
(d)
f(x)
= (x + 1)2: f(-1), f(0), f(3)
(e)
f(x)
= x
: f(0), f(1), f(x
– 1)
x + 1
2.
Find the zeros of the functions:
(a)
f(x)
= x – 1 (b)
f(x)
= (x + 1)2 (c)
f(x)
= x2 – x – 6
(d)
g(x)
= x – 1 (e) f(x) = x2
+ 2x – 15
x + 2 x2 – 1
3.
Given the function f(x) = x + 1 , find the images of x = -1, 1, ½
x – 2
What
value of x has no image?
4.
If f(x) = 3x + 2, what is the value of x
whose image is 5?
5.
Functions f and g are given by f(x) = x2
– x and g(x) = 2x – 3. Evaluate f(0), f(-1), g(0), and g(-1). If f(x) + g(x) = 3, find x.
6.
If h(x) = ax + b, find the values of a and b given that h(2) = 4 and
h(4) = 10.
7.
If f(x) = 2x
– 1 , find the zero of f(x) and the values of x which have no images
x2 – 3x – 4
8.
Given the function f(x) = x2
– 3x – 2, express f(2a)
– f(a) in terms of a.
9.
If f(x) = x2 – x + 1, f(a), f(b)
and f(a + b). Is f(a
+ b) = f(a) + f(b)?
10.
f(x)
= ax2 + bx + c
where a, b and c are constant
numbers. If f(0) = 5, what is the
value of c? Given also that f(1) = 6 and f(-1) = 8, find the values of a
and b.
11.
If f(x) = 2x2 – x + 1,
express f(x + h) – f(x) in its simplest form.
h
POLYNOMIALS
This
is the sum of separate terms of positive powers of a variable with a constant
term (which may be zero) eg 4x3
– 2x2 + 7x – 5. The general form of polynomials
is anxn + an-1xn-1 + ….. + a1x + a0 where an, an-1, …. a1
are the coefficients and a0
is the constant term. Any of these could be zero, except an
ADDITION OF
POLYNOMIALS
Example
I: Add P1 = 2x3
– 3x2 + 5x – 7 and P2 = x3 + x2 – x + 1
Solution
2x3 – 3x2 + 5x – 7
+ x3 + x2 – x + 1
3x3
– 2x2 + 4x
– 6
Or (2x3
+ x3) + (– 3x2 + x2) + {5x + (–x)} + (–7 + 1)
= 3x3 – 2x2 + 4x – 6
DIVISION
OF POLYNOMIALS
Example 2: Divide (x3 + 3x2
– x + 1) by (x – 2)
Solution
x2 + 5x + 9 .
x – 2 x3
+ 3x2 – x + 1
x3
– 2x2
5x2
– x
5x2
– 10x
9x
+ 1
9x
– 18
19
So the quotient is x2 + 5x + 9
and the remainder is 19. We can write:
x3 + 3x2 – x + 1 = (x2 +
5x + 9)( x – 2) + 19.
Polynomial = Divisor x Quotient +
Remainder
P = D
x Q +
R
REMAINDER
THEOREM
This Theorem states that when a
polynomial f(x) is divided by a Divisor (x
– a), the remainder is the value of f(a);
and if it is divided by (x + a), the remainder is the value of f(-a).
Example I: f(x) = x3 – 3x2 + x – 5, what are the remainders when f(x) is divided by (a) x – 1 (b) x
+ 2 (c) 2x – 1 (d) 3x + 2
Solution:
(a) Divisor = x – 1
Remainder = f(1) = 13 – 3(1)2 + 1 – 5
1 – 3 + 1 – 5 = –6
(b) Divisor = x + 2
Remainder = f(–2) = (–2)3 –
3(–2)2 – 2 – 5
–8 – 12 – 2 – 5 = –27
(c) Divisor = 2x – 1
Remainder
= f(½) = (½)3 – 3(½)2 – ½ – 5
=
1 – 3 + 1 – 5 = - 41
8
4 2 8
(d) Divisor = 3x + 2
Remainder
= f(- 2/3) = (- 2/3)3 – 3(- 2/3)2 + (- 2/3) – 5
=
8 - 4 - 2 - 5 = -197
27
3 3 27
USE
OF REMAINDER THEOREM: THE FACTOR THEOREM
If f(x) is divided by (x – a) and f(a)
= 0, then there is no remainder. This means that (x – a) is a factor of f(x).
Hence, the factor theorem states that if f(a) = 0, x – a is a factor of f(x)
Example I: Factorize x3 – 2x2 – 5x + 6
Solution:
Since f(x) is of degree 3, then
f(x)
= (x – a)(x – b)(x
– c) where a,b,c are positive integers. Also multiplying the last term of each
factors, a x b x c = 6. Hence the
possible factors of f(x) will be:
(x
+ 1)( x + 2)( x
+ 3
.
.
First find the first one by trial method
eg (x + 1), hence f(-1) = (-1)3 – 2(-1)2
– 5(-1) + 6
= -1 – 2 + 5 + 6 = 8 ≠ 0, hence it’s not
a factor
Try (x
– 1), hence f(1) =(1)3 – 2(1)2 – 5(1) + 6
= 1 – 2 – 5 + 6 = 0, hence this a factor
Hint: Once you get the first factor,
then divide the polynomial by it and factorize the quotient.
Hence, (x3 – 2x2
– 5x + 6) ÷ (x – 1) = (x2 –
x – 6 )
Factorize x2 – x – 6 = (x + 2)(x – 3)
Hence, x3 – 2x2
– 5x + 6 = (x – 1)(x + 2)(x – 3)
Example 2: It is known that (x – 1) and (x – 2) are factors of x3
+ ax2 – 7x + b,
where a and b are constants. Find a
and b and then the third factor.
Solution:
(x
– 1) is a factor. Therefore, putting x = 1 in the polynomial
= (1)3 + a(1)2 – 7(1) + b
= 1 + a – 7 + b = 0
= a
+ b = 6……………………………(1)
(x
– 2) is also a factor, therefore putting x
= 2 in the polynomial
= (2)3 + a(2)2 – 7(2) + b
= 8 + 4a – 14 + b = 0
= 4a
+ b = 6…………………………(2)
(2) – (1), 3a = 0, Hence a = 0
Substituting for a in (1)
= 0 + b = 6, hence b = 0
Hence a = 0, b = 6, and f(x)
= x3 – 7x + 6
To find the third factor, divide x3 – 7x + 6 by (x – 1) (x – 2)
= (x3
– 7x + 6)/(x2 – 3x + 2) =
x + 3
SOLVING
EQUATIONS
We can use the factor theorem to solve
polynomial equations if we can factorize the polynomial.
Example I: Solve the equation 2x3 – 3x2 – 5x + 6 =
0
Solution:
Factorize f(x) = 2x3 – 3x2 – 5x + 6 =
0
Try (x
– 1), hence f(1) = 2(1)3 – 3(1)2 – 5(1) + 6 = 0
Hence (x – 1) is a factor.
Divide 2x3 – 3x2
– 5x + 6 by (x – 1)
= (2x3
– 3x2 – 5x + 6) ÷ (x – 1) = 2x2 –
x – 6
Factorize 2x2 – x – 6 =
(2x + 3) (x – 2)
Hence 2x3 – 3x2
– 5x + 6 = (x – 1)(2x + 3)(x – 2)
Example 2: x + 1 is a factor of a + 3x +
3x2 + bx3 and the remainder when this expression is
divided by x + 2, is 20.
(a) Find the value of a and b
(b) With these values factorize the
expression completely
(c) Hence solve the equation a + 3x + 3x2 + bx3 =
0
Solution:
(a) If (x + 1) is a factor of f(x) = a
+ 3x + 3x2 + bx3, f(-1)
= 0
Then f(-1)
= a -
3 + 3 - b = 0,
which gives a = b or
a
– b
= 0…………………………………………(1)
When f(x) is divided by (x + 2), the remainder is 20
Hence, f(-2) = a + 3(-2) + 3(-2)2 + b(-2)3 = 20
a
- 6 + 12 - 8b = 20
a
- 8b = 14 (2)
(2) – (1), -7b = 14, Hence b = -2
But a
= b = -2
Hence f(x) = -2x3 + 3x2 + 3x – 2
(b) Divide -2x3
+ 3x2 + 3x – 2 by x + 1
. -2x2
+ 5x – 2 .
x + 1
-2x3 + 3x2 + 3x – 2
-2x3 – 2x2
5x2
+ 3x
5x2 + 5x
-2x
– 2
-2x
– 2
The quotient is 2x2 + 5x – 2 = (2x – 1)(-x + 2)
= 2x2
+ 5x – 2 = (2x – 1)(2 – x)
Hence f(x) = (x + 1)(2x – 1)(2 – x)
(c) To solve a + 3x + 3x2 + bx3 = 0
= -2x3
+ 3x2 + 3x – 2 = 0
= (x
+ 1)(2x – 1)(2 – x) = 0
The roots of the equation are x = -1, -½ or 2
IDENTITIES
Two polynomials of the same degree are
identical if corresponding coefficients are equal. For example if the
polynomial 3x2 – 2x2 + x – 5 and ax3
+ bx2 + cx +d, then a = 3, b = –2, c = 1 and d = –5
Example: If the polynomial 2x2 – 4x + 3 can be expressed in the form a(x – 1)(x – 2) + b(x – 1) + c(x
– 2), find the values of a, b and c.
Solution:
The polynomials 2x2 – 4x + 3
and a(x – 1)(x – 2) + b(x – 1) + c(x – 2) are identical.
But 2x2
– 4x + 3 = ax2 + x(–3a + b
+ c) + 2a – b – 2c.
By comparison 2 = a, –4 = –3a + b + c, 3
= 2a – b – 2c
Since a = 2, then – 4 = –6 + b
+ c and 3 = 4 – b – 2c. This will give:
b + c = 2………………….(1)
b + 2c = 1…………….….(2)
(2) – (1), c = –1
From (1), b – 1 = 2. Hence b = 3
Therefore: a = 2, b = 3, c = –1
QUESTIONS
1. For the function f(x) = 2x2 – x – 2, find the values of f(-2),
f(-1), f(0) and f(2)
2. If f(x) = x – 2, find f(0), f(1), and f(2). What
value of x has no image for this
function?
x
+ 3
3. If f(x) = x2 – 2x, find the values of x
for which f(x) = 15
4. Given that f(x) = ax2 + bx – 1 and f(- 1) = 4, f(2) = 7, find the values of a and b.
5. If f(x) = x2 – x – 1, find in its simplest form f(x + 1) – f(x
-1)
6. If f(x) = x +
1 , find the value of k other than k = 1, such
that f(k) = f(1)
x2 – x + 1
7. If P = x3 – x2
+ 3x – 4 and Q = x – 3, find (a) 3P – Q2, (b) PQ, (c) P
Q
8. State the coefficients of x2 and x3 in the product
(x3
– 2x2 + x – 1) (x2 + x + 1).
9. (a) Multiply x2 + ax – 3 by
x2 + x + b arranging your
answer in descending powers of x
(b) If the coefficients of x3
and x in the product are each equal
to – 1, find the values of a and b
(c) With these values of a and
b, what is the coefficient of x2?
10. Divide (a) x2 – x + 3 by x – 1
(b) x3 + 2x2 – x – 5 by x + 4
(c) 2x3 – 3x2 + x – 4 by 2x – 3
11. Factorize (a) x3 – 3x –
2 (b) x3 + 6x2
+ 11x + 6 (c) x3 + x2 – 6x + 4
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