POLYNOMIAL
A polynomial in x is an expression of the form anxn + an–1xn–1+ an–2 x n–2 + – – – + a1x + a0 where a0, a1 a2 – – – an are real number and n is positive integer The polynomial is of degree n (i.e. the highest power of in the polynomial), an is the leading coefficient and a0 is the constant term . In standard form, a polynomial is written with descending power of x
Depending on the value of the integer n (which specifies the highest power of x), we have several subclasses of polynomial function
Case of n = 0: y = a0 [Constant function]
Case of n = 1: y = a1x+a0[Linear function]
Case of n = 2: y = a2x2+a1x+a0[Quadratic function]
Case of n = 3: y = a3x3+a2x2+a1x+a0 [Cubic function]
Standard form Degree
5x4 – 4x3 + 4 4
x5 – 2x2+x+2 5
3x3+3x2+x+1 1
Polynomials with one, two or three terms are called monomials, binominals or trinomials respectively.
OPERATION WITH POLYNOMIAL
ADDITION AND SUBTRACTION: The same way you add and subtract real numbers, simply add or subtract the like terms (terms having the same variable to the same powers) by adding or subtracting their coefficient.
Example
f1(x) = 3x3 – 3x2+5x–5
f2(x) = 4x3 + 6x2+3x–7
1. f1(x) + f2(x)= (3x3–3x2+5x–5) +(4x3+6x2+3x–7) = (3+4)x3+(–3+6)x2+(5+3)x+(–5–7) = 7x3+3x2+8x–12
2. f1(x) – f2(x) = (3x3–3x2+5x–5) – (4x3+6x2+3x–7) = (3 – 4)x3+(–3–6)x2+(5–3)x+(–5–(–7))
= –x3–9x2+2x+2
MULTIPLICATION OF POLYNOMIAL
Example, Multiply (2x2+6x+4)(2x+5)
Solution 2x2+6x+4
× 2x+5
Multiply throughout by 2x: 4x3+12x2+8x
Multiply by 5 10x2+30x+20
Add the two lines 4x3+22x2+38x+20
(2x2+6x+4) (2x+5) = 4x3+22x2+38x+20
Example
Evaluate (4x3+x–6)(3x2+2x–1)
Solution
You will notice that there is no x2 in 4x3+x–6 so we insert 0x2 to aid our working
4x3+0x2+x–6
3x2+2x–1
Multiply by 3x2 12x5+ 0x4+3x3–18x2
Multiply by 2x +8x4+0x3+2x2–12x
Multiply by –1 –4x3– 0x2–x+6
Add the three lines 12x5+8x4–x3–16x2–13x+6
(4x3+x–6)(3x2+2x–1)=12x5+8x4–x3–16x2–13x+6
3. DIVISION OF POLYNOMIALS
Rules to follow when dividing one polynomial by another
1. Arrange each polynomial in their descending powers
- Divide the first term of the dividend (the polynomial to be divided by the first term of the divisor. This gives the first term of the quotient.
- Multiply the divisor by the first term of the quotient and subtract the product from the dividend
- Using the remainder, thus obtained as new dividend, repeat the above mentioned process to find the second term of the quotient.
5. Repeat the process until a remainder is obtained which is either zero or is lower degree than the divisor.
Example
Divide the polynomial f(x) = 4x3+3x2–2x+1 by x+2
Solution
Step 1
Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
Step 2
Multiply each of the term of the divisor by the first term of the quotient and write below the first two terms of the dividend and subtract the product from the dividend.
Step 3
Using –5x2– 2x as a new dividend repeat step 1, 2
Note that
a. x + 2 is divisor
b. 4x3+3x2–2x+1 is the dividend
c. 4x2–5x+8 is the quotient
d. –15 is the remainder
Example
Divide x3 – 7x + 6 by x – 2
Solution
There is no x2–term in the dividend, the coefficient of that term is zero, and we have
Polynomial of degree m can only be divided by polynomial of n if m > n
ZEROS OF POLYNOMIAL FUNCTION
A zero of a function f is a number for which f(x) = 0 for example 4 is a zero for the function f(x) = x–4 because
f(4) = 4 – 4= 0
Example
Find all the zero of f(x)
a. x2+2x–3
b. x2–4
Solution
(a) f(x) = x2+2x–3
Let f(x) = 0
x2+2x–3=0
(x+3)(x–1) = 0
x = –3 or x = 1
(b) f(x) = x2 – 4
Let f(x) = 0
x2 – 4 = 0
x2 = 4
x = + 4
Note: The following hold true for a polynomial function f(x) of degree n
1. The graph of f(x) will have at least n – 1 turning points 2. The polynomial function of n degrees will have n zeros
POLYNOMIAL FUNCTION WITH GIVEN ZERO
Example
Find the polynomial function with the following zero (a) –3, –2 (b) 2, –4, –3
Solution
(a) For each given zero, we form the corresponding factor e.g. the zero given by x= –3, corresponds to the factor (x+3). Thus we can write the function as
f(x) = (x+3) (x+2) = x2+5x+6
(b) f(x) = (x–2) (x+4) (x+3) = (x2+2x–8) (x+8) = (x3+5x2–2x–24)
THE REMAINDER THEOREM
The remainder theorem states that if a polynomial f(x) is divided by (x–a), the quotient will be a polynomial q(x) of a degree one less than that of f(x) together with a remainder R still to be divided by (x–a).
Example
Apply the remainder theorem to determine the remainder in each of the following case
(1) (5x3– 4x2–3x+ 6) (x–2)
R = f(2) = 5x3 – 4x2–3x+6
f(2) = 5(2)3 – 4(2)2 – 3(2) +6 = 40 – 16 – 6 + 6= 24
(2) (4x3–3x2+5x–3) (x–4)
R = f(4) = 4x3 – 3x3 + 5x – 3
f(4) = 4(4)3 – 3(4)2 + 5(4) – 3
= 256 – 48 + 20 – 3
= 228
(3) (2x3 + 3x2–x+4) (x +2)
f(x) = 2x3 + 3x2 – x + 4
f(–2) = 2(–2)3 + 3(–2)2 – (–2) +4
= –16 + 12 + 2 + 4
= 2
FACTOR THEOREM
If f(x) is a polynomial and is divided by (x–a) gives a remainder of zero i.e. f(a) = 0 then (x–a) is a factor of f(x) or A linear function ax + b is a factor of polynomial f(x) if and only if f(–b/a) = 0, i.e.
Example
Test whether (x – 2) is a factor of f(x) = 2x3 + 2x2 – 17x +10 .If so factorise f(x) as far as possible
Solution
f(x) = 2x3 + 2x2 – 17x + 10
f(2) = 2(2)3 + 2(2)2 – 17(2) + 10
= 16 + 8 – 34 + 10
= 0
There is no remainder, therefore x – 2 is a factor f(x)
Using long division to find the quadratic factor
2x3 + 2x2 – 17x + 10 = (x–2) (2x2 + 6x –5)
Example
Show that (x–4) is a factor of f(x) = x3–6x2–7x+60 and as far as possible factorise f(x) into linear factor
Solution
f(x) = x3 – 6x2 – 7x + 60
f(4) = (4)3 – 6(4)2 – 7(4) + 60
= 64 – 96 – 28 + 60
= 0
f(4) = 0
Hence (x – 4) is a factor of f(x)
Using long division to find the quadratic factor
MORE WORKED EXAMPLEON REMAINDER AND FACTOR THEOREM
Example
When the polynomial f(x) = x4 + ax3 + x2 + bx + 1 divided by x2 + 3x + 2, the quotient is x2 – 1 and the remainder is 5x+3. Find the values of the constants a and b.
Solution
f(x) = x4 + ax3 + x2 + bx + 1
Using Remainder theorem
or f(x) = (x–a) q(x) + R
f(x) = (x2+3x+2) (x2–1) + 5x+3
f(x) = x4+3x3+2x2+3x–2+5x+3
= x4+3x3+x2+2x+1
x4 + ax3 + x2+ bx + 1= x4 + 3x3 + x2 + 2x + 1
Comparing coefficient of like terms
x3: a = 3
x: b = 2
Example
Given that x2 – 3x – 10 is a factor of
f(x) = 2x3 – 7x2 + ax + b where a and b are constants. Find the values of the constant and hence factorise f(x) completely.
Solution
f(x) = 2x3–7x2+ax+b
The divisor = x2–3x–10= (x–5)(x+2)
Hence (x–5) and (x+2) are factors of
2x3 – 7x2+ax+b
Using factor theorem
f(5) = 2(5)3 – 7(5)2 + a(5) + b = 0
= 250 – 170 + 5a + b = 0
= 75 + 5a +b = 0
5a + b = –75 ––––––––––––––(1)
f(–2) = 2(–2)3 – 7(–2)2 +a(–2) + b = 0
= –16 – 28 – 2a+b = 0
–44 – 2a + b = 0
–2a+b = 44
Solving (1) and (2) Simultaneously gives
a = –17, b = 10
The polynomial function will be f(x) = 2x3–7x2–17x+10
Using long division
Hence 2x2 – 7x2 – 17x+10 = (x2–3x–10)(2x–1)
f(x) = 2x3–7x2–17x+10=(x–5)(x+2)(2x–1)
Example
When the polynomial f(x) = ax3+bx2–7x–6 is divided by x–1, the remainder is –10 and when it is divided by x –3 the remained is 36. find the values of the constant a and b
Solution
f(x) = ax3+bx2–7x–6
Using Remainder theorem
f(x) = a (1)3 + b(1)2 – 7(1) – 6 = –10
= a + b – 7– 6 = –10
a + b = 3 ––––––––––––––––––––––(1)
f(3) = a(3)3+b(3)2–7(3)–6=36
27a+9b – 21 – 6 = 36
27a +9b = 63 ––––––––––––––––(2)
Solving (1) and (2) simultaneously gives
a = 2, b = 1
Example
When g(x) = x3 – 2x2 – 5x + 6 divided by ax + b, the quotient is x2 – x – 6 and the function leaves no remainder. Find the values of the constant a and b. Factorise f(x) completely and state its zero.
Solution
g(x) = x3–2x2–5x+6
Using factor theorem
g(x) = (x2–x–6)(ax+ b)
=ax3–ax2–6ax + bx2 – bx – 6b
= ax3 + (b – a)x2 –(6a+b)x–6b
x3–2x2–5x+6=ax3+(b – a)x2 – (6ax+b)x– 6b
Comparing coefficient of like terms
x3 : a = 1
constant : + 6 = –6b
b = –1
The divisor ax + b = x – 1
Factorising completely
g(x) = x3–2x2 – 5x +5
=(x2–x–6)(x–1)
=(x–3)(x+2)(x–1)
The zeros of g(x) are 3, –2 and 1
Example
If f(x) = 4x4 + bx3 – 23x2 + cx + 11 and when f(x) is divided by 2x2+7x+3 the remainder is 3x +2 determine the values of b and c
Solution
f(x) = 4x4+bx3–23x2 + cx+11
The divisor = 2x2+7x+3
= (2x+1)(x+3)
The remainder = mx + n = 3x+2
Consider the division has be perform, then
4x4+bx3–23x2+cx+11= (2x2+7x+3) × (quotient) + 3x + 2
Putting x = –3
f(x)=4(–3)4+b(–3)3–23(–3)2+c(–3) +11= –9+2
324 – 27b – 207 – 3c – 11 = –7
128 – 27b – 3c = –7
27b + 3c = 135 –––––––––(1)
Putting x = – ½
f(–½) = 4(–½)4 + b(–½)3 – 23 (–½)2 + c(–½) + 11=
Multiplying through by 8
2 – b – 46 – 4c + 88 = 4
44 – b – 4c = 4
b + 4c---------(2)
Bringing out (1) and (2) we have
27b + 3c = 135-----------(1)
b + 4c = 40 ---------------(2)
Solving (1) and (2) Simultaneously we have
b = 4 and c = 9
Hence f(x) = 4x4 = bx3 – 23x2 + 9x + 11
Example
When the expression x3 + ax2 +bx – c is divided byx2 – 4, the reminder is 18 –x, when it is divide by (x +3), the remainder is 21 find the remainder when the expression is divided by (x+1)
Solution
Let f(x) = x3 +ax2 +bx +c
Using Remainder theorem
f(x) = (x–a)q(x) + R
Let R = mx + n
When the divisor is x2 – 4 = (x–2) (x+2)
Suppose the division to have been performed then
x3 + ax2 + bx +c = (x2– 4)q(x) + 18 – x
x3 + ax2 +bx + c = (x–2) (x+2) q(x) + 18 – x
Putting x = 2
(–2)3 + a(–2)2+b(–2)+c=18–(–2)
–8+4a–2b+c=18+2
4a–2b+c=28 ––––––––––(1)
Putting x = 2
(2)3+a(2)2+b(2)+c=18 –(2)
8+4a+2b+c =16
4a+2b+c=8 –––––––(2)
When the divisor is (x+3)
x3+ax2+bx+c = (x+3)q(x)+R
Putting x = –3
(–3)2 + a(–3)2 +b(–3) +c = 21
=27 + 9a – 3b + c = 21
= 9a – 3b + c = 48 –––(3)
Bring out the 3 equations
4a – 2b + c = 28 ––––––––(1)
4a + 2b + c = 8 –––––––––(2)
9a – 3b + c = 48 ––––––(3)
Subtract (1) from (2)
4b = –20
b = –5
Subtract (1) from (3) and substitute b = –5 into it
5a – b = 20
5a + 5 = 20
a = 3
Substitute a = 3, b = –5 into (1)
4(3) –2(–5) +c = 28
12+10+c=28
c = 6
The required equation = x3 + 3x2 – 5x + 6
(b) The remainder when expression is divided by x – 1
f(–1) = (–1)3+3(–1)2–5(–1)+6
= –1 + 3+5+ 6
= 13
The remainder is 13
Example
Find the remainder that results when
1. x4 + 6x3 + x2 – 5x – 2 is divided by (x2–1)
2. x4 +4x3 + 3x2 – 2x – 5 is divided by (x + 1) (x – 2)
Solution
1. (x4+6x3+x2–5x–2) (x2 –1)
Supposing the division to have been performed then
x4+ 6x3 + x2 – 5x –2 = (x2 – 1)q(x) + R
x4 + 6x3 +x2 –5x –2 = (x–1) q(x) + mx + n
Putting x = 1
1 + 6 + 6 – 5 – 2 = m + n
m + n = 1 ---------------(1)
Putting x = –1
1 – 6 + 1 + 5 –2 = –m + n
–m + n= –1-------------------(2)
solving (1) and (2) simultaneously gives
m = 1, n = 0
The remainder is – 4x + 10
2. x4+4x3+3x2–2x–5 [(x+1) (x–2)]
x4+4x3+3x2–2x–5 = [(x+1)(x–2)] q(x)+R
x4+4x3+3x2–2x–5 = (x+1)(x–2) q(x) + mx+n
put x = –1
1 – 4 + 3 + 2 – 5 = m + n
– 3 = –m+n ––––––(1)
put x = 2
16 + 32 + 12 – 4 – 5 = 2m+n
35 = 2m+n ––––––(2)
Solving (1) and (2)
Hence the remainder is
Example
When a polynomial expression f(x) is divided by (x–2), it leaves a remainder of 5, when the same expression is divided by (x+3), the remainder is 15. Find what is left behind when the expression is divided by (x2+x–6)
Solution
f(x) (x2+x–6)
Suppose the division to have been performed then
f(x) = q (x) (x2+x–6) + R
f(x) = (x2+x–6)q (x)+R
f(x) = (x+3) (x–2)q(x) + mx +n
Putting x = 2
f(2) = 2m + n =5 –––––––––––(1)
Putting x = –3
f(–3) = –3m+n = 15 –––––––––(2)
Subtracting (2) from (1)
5m = –10
m = –2
Substituting m = –2 into (1)
2(–2) + n = 5
n = 9
Hence the remainder when the expression is divided by x2 + x – 6 is –2x + 9
Example
When f(x) = x3 + qx2 + rx + 5 is divided by x – 3 the remainder is 30, if x2 – 4 is a factor f(x), find the values q, r and s
Solution
f(x) = x3 + qx2 +rx +5
f(3) = 33 + q(3)2 + r (3) + 5 = 30
= 27 + 9q + 3r + 5 = 30
9q + 3r + 5 = 3 –––––––––(1)
Also
Let x3 + qx2 + rx + 5 = g (x) (x2–4) + R
Since x2 – 4 is a factor the remainder is zero
x3 + qx2 +rx + 5 = (x–2) (x+2)g(x)
f(2) = 23 + q(2)2 + r(2) + 5 = 0
4q + 2r + 5 = –8 –––––––––(2)
f(–2) = (–2)3 + q(–2)2+r(–2) + 5 = 0
4q – 2r + 5 = 8 ––––––––––(3)
Subtract (3) from (2)
4r = – 16
r = –4
Subtract (2) from (1)
5q + r = 11 ––––––––(4)
Substitute r = – 4 into (4)
5q –4 = 11
q =3
substitute q = 3, r = –4 into (1)
9(3)+3(–4)+s=3
s =3–27 +12
s = –12
The values of q, r and s are 3, –4 and –12 respectively
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