Polynomials SS2 Further Mathematics Lesson Note

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) = 3x² – 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-ⁿ–¹ + … 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-1Xⁿ–¹ + … b2X² + b1 X + b0 then,

P(x) + Q(x) = (an + bn)Xn + (an-1 + bn-1) Xⁿ–¹ + … + (a2 + b2) X² + (a1 + b1) X + a0 + b0.

Also,

P(x) – Q(x) = (an-bn) Xn- (an-1- bn-1) Xⁿ–¹ … + (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+ 5×2) + (3x + 6x) + (4 + 1)

= 7x³ + x2 + 9x + 5

(b) P1 (x) + P3(x)

(7x³ – 4x² + 3x + 4) + (4x² + 2x – 3)

= (7x³ + 4×3) + (-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) + (5×2 + 6x + 1) + (4x³ + 2x – 3)

= (7x³ + 4x³) + (-4x² + 5x²) + (3x + 6x + 2x) + (4 + 1 – 3)

= 11x³ + x² + 11x + 2

ASSIGNMENT 

(1) Find the quotient and remainder when 2x⁴ – 3x³+ x² – 4x + 5 is divided by x² + 3x + 1

(2) If p1 = 3x³ + 2x² – x + 2, 

p2 = 2x² + x – 6 and 

p3 = x³ + 3x² + 2x – 4, find

(i) p3(p1 + p2)

(ii) p2 + p3 – 3p1

(iii) p2 x (p3 + p1)