WEEK 3
TOPIC: Polynomials
Consider the expressions formed from the sum of integral non negative powers of a variable x taken together with some numerical constants. Such expressions are called Polynomials.
The general polynomial takes the form anXn + an1Xn1 + … a2X2 + a1X + a0 where a11 an-1…a0(a0 ≠0) are numerical constants.
The numerical constants an’ an1…. a2, a1 are called coefficients of Xn, Xn1, … X2, X respectively while a0 is called the constant term of the polynomial.
The highest power of the variable is n and is called the degree of the polynomial. Let us designate the general polynomial by p(x). Thus:
P(x) =anXn + an-1 Xn-1 + … + a2X2 + a1X + a0.
The following are examples of polynomials:
(a) P1(x) = 3x2 – 2x + 4
(b) P2(x) = 3x4 – 2x2 + x – 1
(c) P3(x) = x + 1
(d) P1(x) = 2x3 + x – 3
The following are not polynomials:
(a) f1(x) = (x2 + 2x – 3)
(b) f1(x) = 3x2 – 4x2 + 2x – 1
2x + 3
(c) f1(x) = (2x – 3)1
Equality of Polynomials
The polynomial p(x) = a11X11 + an-1 Xn-1 + … + a2X2 + a1X + a0 is said to be equal to the polynomial.
Q(x) =bnXn + bn-1X-n1 + … b2X2 + b1 X + b0 provided
an = bn’ an 1 = bn 1 … a2 = b2′ a1 = b1, a0 = b0
Addition and Subtraction of Polynomials
Let P(x) = anXn + an-1 Xn-1 + … + a2X2 + a1X + a0
Q(x) = bnXn + bn-1X-n1 + … b2X2 + b1 X + b0 then,
P(x) + Q(x) = (an + bn)Xn + (an-1 + bn-1) Xn1 + … + (a2 + b2) X2 + (a1 + b1) X + a0 + b0.
Also,
P(x) – Q(x) = (an-bn) Xn– (an-1– bn-1) Xn1– … + (a2– b2) X2+ (a1– b1) X + a0– b0.
Given that P1(x) = 7x3 – 4x2 + 3x + 4;
P2(x) = 5x2 + 6x + 1 and P3(x) = 4x3 + 2x – 3.
Find:
(a) P1(x) + P2(x)
(b) P1(x) + P3(x)
(c) P1(x) – P2(x)
(d) P3 (x) – P2(x)
(e) P1(x) + P2(x) + P3(x)
Solution
(a) P1(x) + P2(x)
= (7x3 – 4x2 + 3x + 4) + (5x2 + 6x + 1)
= 7x3 + (-4x2 + 5x2) + (3x + 6x) + (4 + 1)
= 7x3 + x2 + 9x + 5
(b) P1 (x) + P3(x)
(7x3 – 4x2 + 3x + 4) + (4x3 + 2x – 3)
= (7x3 + 4x3) + (-4x2( + (3x + 2x) + (4-3)
= 11x3 – 4x2 + 5x + 1
(c) P1 (x) – P2 (x)
= (7x3 – 4x2 + 3x + 4) + (5x2 + 6x + 1)
= 7x3 + (-4x3 – 5x2) + (3x – 6x) + (4 – 1)
= 7x3 – 9x2 – 3x + 3
(d) P3 (x) – P2 (x)
= (4x3 + 2x – 3) – (5x2 + 6x + 1)
= (4x3 – 5x2 + (2x – 6x) + (-3 – 1)
= 4x3 – 5x2 – 4x – 4
(e)P1 (x) + P2 (x) + P3(x)
= (7x3 – 4x2 + 3x + 4) + (5x2 + 6x + 1) + (4x3 + 2x – 3)
= (7x3 + 4x3) + (-4x2 + 5x2) + (3x + 6x + 2x) + (4 + 1 – 3)
= 11x3 + x2 + 11x + 2
Given that P1(x) = 2x3 + 4x2 – x + 1
P2(x) = 3x4 + x3 – 2x2 + x – 3 and
P3(x) = 4x3 + 2x – 4
Find:
(a) 2P1(x) + P2(x)
(b) 3P2(x) + 2P3(x)
(c) 3P1(x) – 3P3(x)
(d) P3(x) + 2P1(x) – 3P2(x)
Solution
(a) 2P1(x) + P2(x)
= 2(2x3 + 4x2 – x + 1) + (3x4 + x3 – 2x2 + x – 3)
= (4x3 + 8x2 – 2x + 2) + (3x4 + x3 – 2x2 + x – 3)
= 3x-1 + 5x3 + 6x2 – x – 1
(b)3P2(x) + 2P3(x)
= 3(3x4 + x3 – 2x2 + x – 3) + 2 (4x3 + 2x – 4)
= 9x4 + 3x3 – 6x2 + 3x – 9 + 8x3 + 4x – 8
= 9x4 + 11x3 – 6x2 + 7x – 17
(c) 3P1(x) – 3P3(x)
= 3(2x3 + 4x2 – x + 1) -3 (4x3 + 2x – 4)
= 6x3 + 12x2 – 3x + 3 – 12x3 – 6x + 12
= -6x3 + 12x2 – 9x + 15
(d) P3(x) + 2P1(x) – 3P2(x)
= (4x3 + 2x – 4) + 2 (2x3 + 4x2 – x + 1) – 3(3x4 + x3 – 2x2 + x – 3)
= 4x3 + 2x – 4 + 4x3 + 8x2 – 2x + 2 – 9x4 – 3x3 + 6x2 – 3x + 9
: P3(x) + 2P1(x) – 3P2(x) = -9x4 + 5x3 + 14x2 – 3x + 7
The value of p(x) at x = a is denoted by p(a) and is obtained by substituting a for x in the polynomial.
Example
Giventhat p(x)= 2x3 + 5x2 – 9x – 18, find;
(a) P(1)
(b) P(-1)
(c) P(2)
(d) P(0)
Solution
(a) P(1) = 2(1)3 + 5(1)2 – 9(1) – 18
= 2 + 5 – 9 – 18
= -20
(b) P(-1) = 2(-1)3 + 5 (-1)2 -9(-1) – 18
= -2 + 5 + 9 – 18
= -6
(c) P(2) = 2(2)3 + 5(2)2 – 9(2) – 18
= 16 + 20 – 18 – 18
= 0
(d) P(0) = 2(0)3 + 5(0)2 – 9(0) – 18
= -18
Multiplication of polynomials
The product of two polynomials of degrees m and n is another polynomial of degree m + n.
Example
Given that P1(x) = 2x2 + 5x + 6 and
P2(x) = 3x2 – 2x + 1, find P1P2
Solution
Method 1
P1 x P2
= (2x2 + 5x + 6)x (3x2 – 2x + 1)
= 2x2(3x2 – 2x + 1) + 5x(3x2 – 2x + 1) + 6(3x2 – 2x + 1)
= 6x4 – 4x3 + 2x2 + 15x3 – 10x2 + 5x + 18x2 – 12x + 6
= 6x4 +(4x3 + 15x3) +(2x2 – 10x2 + 18x2)+ 5x – 12x + 6
= 6x4 + 11x3 + 10x2 – 7x + 6
Method 2
2x2 + 5x + 6
3x2 – 2x + 1
2x2 + 5x + 6
-4x3 – 10x2 – 12x
6x4 + 15x3 + 18x2
6x4 + 11x3 + 10x2 – 7x + 6
Method 2 is usually called Long Multiplication method.
Given that P1(x) = 4x3 – 2x2 + 3x – 1 and P2(x) = 3x3 – 4 find P1(x) x P2(x)
Solution
Method 1
P1P2 = (3x2 – 4) (4x3 – 2x2 – 3x – 1)
= 3x2 (4x3 – 2x3 + 3x – 1)
-4(4x3 – 2x2 + 3x – 1)
= 12x5 – 6x4 + 9x3 – 3x2 – 16x3 + 8x2 – 12x + 4
= 12x5 – 6x4 – 7x3 + 5x2 – 12x + 4
Method 2
4x3 – 2x2 – 3x – 1
3x2 – 4
-16×3+8×2 – 12x + 4
12x5 – 6x4 – 9x3 + 3x2
12x5 – 6x4 – 7x3 + 5x2 – 12x + 4
Division of Polynomials
A polynomial of degree n can be divided by another polynomial of degree m if n ≥ m.
Divide the polynomial P(x) = 3x2 -2x + 4 by the polynomial P(x) = x + 2
Solution
Since P1 is being divided by P2 it is called the dividend while P2 is called the divisor. The result of division of P1 by P2 is called the quotient and whatever is left after division is called the remainder.
The procedure of division of P1 by P2 can best be demonstrated step by step as follows:
Step 1
Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
3x
x + 2 3x2 – 2x + 4
Step 2
Multiply each of the terms of the divisor by the quotient.
3x
x + 2 3x2 – 2x + 4
3x2 + 6x
Step 3
Subtract the product obtained in step 2 from the first two terms of the dividend and add the next term of the dividend.
3x
x + 2 3x2 – 2x + 4
3x2 + 6x
-8x + 4
Step 4
Using -8x + 4 as a new dividend repeat step 1, 2 and 3.
3x – 8(c)
x + 2 3x2 – 2x + 4 (b)
(a) 3x2 + 6x
-8x + 4
-8x – 16
20 (d)
Note that:
(a) x + 2 is the divisor;
(b) 3x2 – 2x + 4 is the dividend;
(c) 3x – 8is the quotient
(d) 20 is the remainder
The expression 3x2 – 2x + 4 can be written as:
3x2 – 2x + 4 = (x + 2) (3x – 8) + 20
Dividend Divisor Quotient Remainder
In general P(x) = D(x) x Q(x) + R
P(x) = Dividend
D(x) = Divisor
Q(x) = Quotient
R = Remainder
Divide 4x3 + 6x2 – 2x + y by 2x – 3 and hence find the quotient and the remainder
Solution
2x2 + 6x + 8
2x – 3 4x3 + 6x2 – 2x + 7
4x3 + 6x2
12x2 – 2x
12x2 – 18x
16x + 7
16x – 24
31
Quotient = 2x2 + 6x + 8
Remainder = 31
Find the quotient and remainder when 2x4 – 3x3 + x2 – 4x + 5 is divided by x2 + 3x + 1
Solution
2x2 – 9x + 26
x2 + 3x + 1 2x4 – 3x3 + x2 – 4x + 5
2x4 – 6x3 + 2x2
-9x3 – x2 – 4x
-9x3 – 27x2 – 9x
26x2 + 5x + 5
26x2 + 78x + 26
-73x – 21
Hence, Quotient = 2x2 – 9x + 26
Remainder = -73x – 21
Find the quotient and remainder when x3 + 8 is divided by x2 – 2x + 4
Solution
x + 2
x2 – 2x + 4 x3 +8
x3 – 2x2 + 4x
2x2 + 4x + 8
2x2 + 4x + 8
0
Quotient = x + 2
Remainder = 0
Hence, x3 – 2x2 + 4x is a factor of x3 + 8.
The Remainder Theorem
The Long Division Method.Although a little cumbersome enables us to find, not only the quotient, but the remainder as well. Suppose we are given the polynomial f(x) = 2x3 – 3x2 + 4x – 1 and we wish to find the remainders when f(x) is divided by:
(a) x – 1
(b) x – 2
(c) x – 3
By using the long division method we have:
2x2 – x + 3
x– 1 2x2 – 3x2+ 4x – 1
2x2 – 2x2
-x2 + 4x
-x2 + x
3x – 1
3x – 3
2
So when f(x) is divided by x – 1, the remainder is 2.
2x2 – x + 6
x – 2 2x3 – 3x2 + 4x – 1
2x3 – 4x2
-x2 + 4x
-x2 + 2x
6x – 1
6x – 12
11
So when f(x) is divided by x – 2, the remainder is 11.
2x2 – 3x + 13
x – 3 2x2 – 3x2 + 4x – 1
2x2 – 6x2
3x2 + 4x
-x2 + 9x
13x – 1
13x – 39
38
So when f(x) is divided by x – 3, the remainder is 38.
Now, find:
a. f(1)
b. f(2)
c. f(3)
d. f(1) = 2 – 3 + 4 – 1 = 2
e. f(2) = 16 – 12 + 8 – 1 = 11
f. f(3) = 54 – 27 + 12 – 1 = 38
Compare the answers in each of the following pairs
i. (a) and (d)
ii. (b) and (e)
iii. (c) and (f)
What do you notice?
This is a remarkable result and it is not a mere coincidence.
Let us try another one.
Given that H(x) = x3 + 2x3 – 4x + 3, find the remainder when H(x) is divided by
(a) x + 1
(b) x + 2
(c) x + 3
x2 + x – 5
(a) x + 1 x3 + 2x3 – 4x + 3
x3 + x2
x2 – 4x
x2 + x
-5x + 3
-5x – 5
8
When H(x) is divided by x + 1, the remainder is 8.
x2 -4
(b) x + 2 x3 + 2x3 – 4x + 3
x3 + 2x2
-4x + 3
-4 – 8
11
So, when H(x) is divided by x + 2, the remainder is 11.
x2 + x – 1
(c) x + 3 x3 + 2x3 – 4x + 3
x3 + 3x2
x2 – 4x
– x2 + 3x
-x + 3
-x – 3
6
When H(x) is divided by x + 3, the remainder is 6. Now, find:
(d) H(-1)
(e) H(-2)
(f) H(-3)
(d) H(-1) = -1 + 2 + 4 + 3 = 8
(e) H(-2) = -8 + 8 + 8 + 3 = 11
(f) H(-3) = -27 + 18 + 12 + 3 = 6
Compare again the answers in each of the following pairs:
(i) (a) and (d)
(ii) (b) and (e)
(iii) (c) and (f)
What again do you notice?
(a) When H(x) is divided by x + 1 the remainder is H(-1)
(b) When H(x) is divided by x + 2 the remainder is H(-2)
(c) When H(x) is divided by x + 3 the remainder is H(-3)
So we can safely conclude that if H(x) is divided by x – a the remainder is H(a) and this forms the basis of what is called the remainder theorem.
Theorem
If the polynomial f(x) is divided by x – a, the remainder is f(a).
Proof
The polynomial function f(x) can be written as:
f(x) = (x – a) Q(x) + R …(1)
where x – a is the divisor, Q(x) the quotient and R the remainder.
Put x = a in (1)
f(a) = (a – a) Q(a) + R
f(a) = R
The theorem whose proof we have just established, is called the RemainderTheorem. In general, if f(x) is divided by ax + b then the remainder is
f
A special case of the remainder theorem is when f(x) leaves no remainder when it is divided by x – a. We therefore say that x – a is a factor of f(x). The modified theorem is called, Factor Theorem and it states:
If f(a) = 0 then x – a is a factor of f(x).
Example
Given that f(x) = 3x3 – 4x2 + 2x + 3, find the remainder when f(x) is divided by x – 1.
Solution
Let R be the remainder. Then using the remainder theorem, R = f(1)
f(1) = 3(1)3 – 4(1)2 + 2(1) +3
= 3 – 4 + 2 + 3
= 4
Find the remainder when
f(x) = (x + 3)(x – 2)(x + 2) is divided by x + 1.
Solution
Let R be the remainder when f(x) is divided by (x + 1) then R = f(-1)
f(-1) = (-1 + 3)(-1 -2)(-1 + 2)
= (2)(-3)(+1)
= -6
Hence the remainder is -6
Find the remainder when f(x) = 2x3 + 3x2 – 4x + 1 is divided by 2x – 1. What conclusion can you draw?
Solution
Let R be the remainder when f(x) is divided by 2x – 1 then R = f
f = 2 f3 + 3 f3– 4 f + 1
= ¼ + ¾ – 2 + 1
= 0
Hence2x – 1 is a factor of f(x).
Evaluation
- If g(x) is the quotient when f(x) = 2X3 + 3x2 -2x +1 is divided by x+2 find the remainder when g(x) is divided by x-1
General Evaluation
If f(x) = 6x3 + 13x2 + 2x – 5,
(a) show that f(-1) = 0 (b) find the factor of f(x)
(1) When the polynomial f(x) = x4 + px3 + x2 + qx + 1 is divided by x2 + 3x + 2 quotient is x2 – 1 and the remainder is 5x + 3. Find the value of constants p and q.
(2) If (x + 1) is a factor of the polynomial f(x) = x3 + kx2 + 3x + 10. Find the value of the constant k, and factorise the polynomial completely.
Reading assignment
New Further Maths Project 1 pages71 – 75 Exercise 6b Q1, 8, 22 and 26
Weekend Assignment
1. Given that f(x) = x5 + 4x4 – 6x2 + 2x + 2, find (-1)
(a) 2 (b) -3 (c) -4 (d) -2
(2) Determine the values of p and q if(x – 1) and (x – 2) are factors of 2x3 + px – 4 + q
(a) p = 12, q = 13 (b) p = -13, p = -12 (c) p = -13, q = 12
(d) p = 10, q = 8
(3) Find zero of the polynomial p(x) = x3 + 4x2 + x – 6
(a) x = -1, 2, 3 (b) x = -1, -3 (c) x = 1, -2, -3 (d) x = 1, 2, -3
(4) If f(x) = 6x3 + 13x2 + 2x – 5, find the factors of f(x)
(a) (x – 1) (2x – 1) (3x – 5) (b) (x + 1) (2x – 1) (3x + 5) (c) (x + 1) (2x + 1) (3x – 5)
(d) (x + 1) (1 – 2x) (5 – 3x)
(5) If (2x + 1) is a factor of the polynomial f(x) = 2x3 – 8x + x2 + k, find the value of the constant k. (a) -2 (b) -3 (c) -4 (d) 3
Theory
(1) Find the values of the constant p, q and r such that, when the polynomial
f(x) = x3 + px2 + qx + r is divided by (x + 2), (x – 1) and (x – 3), the remainder are respectively -48, 0 and 2. Hence factorise f(x) completely.
(2) If x – 2 is a factor of f(x) = x3 – x2 + px + q and f(x) leaves a remainder of 12 when it is divided by (x – 3), find
(a) the values of the constant p and q
(b) the three values of c for which x3 – x2 + px + q = 0