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
Given that 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
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
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