Unit 2 – Outcome 1 Polynomials & Quadratics Higher Mathematics

What Is A Polynomial? A polynomial is an expression made up of constants and constants multiplied by powers of a letter.The largest power in the expression gives the DEGREE of the polynomial, and the powers must be positive(can be equal to zero). x 2 + 7x - 1Polynomial of degree 2 x 4 + 3x 2 – 5Polynomial of degree 4 x + 1Polynomial of degree 1 4x -3 NOT a polynomial

Evaluating The Value Of Polynomials For the polynomial f(x) = 5x 4 + 3x 3 – 7x 2 + 8x – 5, what is the value of this polynomial when x = 4 ? To evaluate it as it stands would require some tricky calculations involving powers. f(4) =5(4) 4 + 3(4) 3 - 7(4) 2 + 8(4)-5 = = 1387

An easier way to deal with this is to use what is called the NESTED FORM of the polynomial. f(x) = 5x 4 + 3x 3 – 7x 2 + 8x – 5 f(x) = (5x 3 + 3x 2 – 7x + 8)x – 5 f(x) = [(5x 2 + 3x – 7)x + 8]x – 5 f(x) = {[(5x + 3)x – 7]x + 8}x – 5 The calculations required now to evaluate the polynomial are simple multiplications, additions and subtractions.

f(x) = {[(5x + 3)x – 7]x + 8}x – 5 f(4) = {[(5(4) + 3)(4) – 7](4) + 8}(4) – 5 f(4) = {[(23)(4) - 7](4) + 8}(4) - 5 f(4) = {[85](4) + 8}(4) - 5 f(4) = {348}(4) - 5 f(4) = f(4) = 1387 There is a shorter and easier way to arrange this type of calculation.

EXAMPLE 1 For the polynomial f(x) = 3x 3 + 7x 2 + 6x – 4, evaluate f(4). f(x) = 3x 3 + 7x 2 + 6x – 4 f(x) = [3x 2 + 7x + 6]x – 4 f(x) = [(3x + 7)x + 6]x – Everything on this line gets multiplied by So f(4) = 324

EXAMPLE 2 For the polynomial f(x) = 5x 3 + 6x 2 - 9x – 3, evaluate f(3) Everything on this line gets multiplied by So f(3) = 159

EXAMPLE 3 For the polynomial f(t) = t 4 + 2t 3 – 7t + 9, evaluate f(-2). CAREFUL! There is no t 2 term, which means the COEFFICIENT of t 2 is zero Everything on this line gets multiplied by So f(-2) =

EXAMPLE 4 Show that x = 2 is a root of the equation x 4 + 3x 3 – 5x – 30 = 0. NOTE:If x = 2 is a root, then (x – 2) must be a factor of f(x). i.e f(2) = Everything on this line gets multiplied by Since f(2) = x = 2 is a root of the equation

Consider this calcuation 25  7 = 3, remainder 4 This could also be written as follows: 25 = 7 x 3, remainder 4 divisorquotientremainder Division Of Polynomials

Now consider the polynomial f(x) = ax 2 + bx + c Then f(h) = ah 2 + bh + c – 1 :f(x)-f(h) = a(x 2 – h 2 ) + b(x – h) f(x) - f(h) = a(x + h)(x – h) + b(x – h) f(x) - f(h) = (x – h)[a(x + h) + b ] f(x) = (x – h)[ax + ah + b ] + f(h) divisorquotientremainder

f(x) = ax 2 + bx + c habc ahah 2 + bh ah 2 + bh + c aah + b We can use this method, then, to divide a polynomial f(x) by (x – h), and the remainder will be f(h). = f(h)

EXAMPLE 1 Find the quotient and remainder when x 3 + 8x 2 + 5x – 2 is divided by (x – 3) So (x 3 + 8x 2 + 5x – 2) = (x – 3) ( x x + 38) divisorquotientremainder

EXAMPLE 2 Find the quotient and remainder when x 3 + 5x 2 - x + 9 is divided by (x + 2) So (x 3 + 5x 2 - x + 9) = (x + 2) ( x 2 + 3x - 7) + 23 divisorquotientremainder

= (2x + 1) ( 2x 2 - x - 2) + 9 EXAMPLE 3 Find the quotient and remainder when 4x 3 - 5x + 7 is divided by (2x + 1). -½ = 2(x + ½) (2x 2 - x - 2) + 9 = (x + ½) (4x 2 - 2x - 4) + 9 quotientremainder 4x 3 - 5x + 7 divisor

The Remainder Theorem: The Factor Theorem We have seen that, if a polynomial f(x) is divided by x – h, then f(x) =(x – h) q(x) + R divisorquotientremainder (h – h) q(x) + Rf(h) = R f(x)  (x – h) gives a remainder of f(h) – this is called the REMAINDER THEOREM

What happens if f(h) = 0? This means there is no remainder. So (x – h) must divide EXACTLY into the polynomial. (x – h) is a FACTOR of the polynomial. If f(h) = 0  (x – h) is a factor. This is called the FACTOR THEOREM.

EXAMPLE 1 Factorise fully the polynomial f(x) = 2x 3 – 11x x – x x x - 6 = (x - 2) We want remainder = 0 Remainder is zero so (x – 2) is a factor. (2x 2 – 7x + 3) = (x - 2)(2x – 1) (x – 3)

EXAMPLE 2 Factorise fully the polynomial f(x) = x 3 + 3x 2 - 6x x 3 + 3x 2 - 6x - 8 = (x + 1) We want remainder = 0 (x 2 + 2x - 8) = (x + 1)(x + 4) (x – 2) Remainder is zero so (x + 1) is a factor.

EXAMPLE 3 Find the value of a, given that (x – 3) is a factor of x 3 + x 2 + ax – 12. Hence factorise the polynomial fully. 311a a a a a + 24 = 0 3a = -24 a = -8 We know (x – 3) is a factor So remainder must equal 0.

The polynomial becomes : = (x - 3)(x 2 + 4x + 4) x 3 + x 2 - 8x - 12 = (x - 3)(x + 2) = (x - 3)(x + 2) 2

Solving Polynomial Equations To solve these equations, we use the following fact: (x – h) is a factor of f(x)  x = h will be a root of f(x) = 0

EXAMPLE 1 Solve the equation x 3 + 5x x – 21 = x 3 + 5x x – 21 = 0 (x - 3) (x 2 + 8x + 7) = 0 (x + 1) (x + 7) = 0 x = 3, -1, -7 Remainder is zero so (x - 3) is a factor.

EXAMPLE 2 Solve the equation 2x 3 - 5x x – 10 = x 3 - 5x x – 10 = 0 (x + 2) (2x 2 - 9x - 5) = 0 (2x + 1) (x - 5) = 0 x = -2, -½, 5 Remainder is zero so (x + 2) is a factor.

EXAMPLE 3 Solve the equation x 3 - x 2 - 3x + 3 = x 3 - x 2 - 3x + 3 = 0 (x - 1) x = 1 (x 2 - 3) = 0 or x 2 = 3 x = 1,  3, -  3 Remainder is zero so (x - 1) is a factor.

EXAMPLE 4 Show that x = 4 is the only real root of the equation x 3 – 4x 2 + 7x – 28 = x 3 - 4x 2 + 7x - 28 = 0 (x - 4) x = 4 (x 2 + 7) = 0 or x 2 = -7 x = 4 NOT POSSIBLE Remainder is zero so (x - 4) is a factor.

EXAMPLE 5 Given that x = 2 is a root of the equation x 3 + ax x + 90 = 0, find the value of a, and hence solve the equation. 21a a + 2 2a + 4 2a a -98 4a - 8 4a - 8 = 0 4a = 8 a = 2

The polynomial becomes : = (x - 2)(x 2 + 4x - 45) x 3 + 2x x + 90 = (x - 3)(x + 5)(x - 9) x = 3, -5, 9

Quadratic Theory Higher Mathematics

Solving Quadratic Equations There are three methods we can use to solve a quadratic equation of the form ax 2 + bx + c = 0 1.Factorising x 2 + 6x - 16 = 0 (x + 8)(x - 2) = 0 x = 2 2x x = 9 2x 2 +17x – 9 = 0 (2x - 1)(x + 9) =0 x = -8 & x = ½ &x = -9 Remember to re-arrange the quadratic into its standard form before attempting to solve.

2.Completing The Square x x - 11 = 0 (x + 5) 2 – 11 = 0-25 (x + 5) 2 – 36 = 0 (x + 5) 2 = 36 (x + 5) = ±6 x = ±6 - 5 x = 1 & x = -11 We could have solved this one by factorising

x 2 + 6x + 3 = 0 (x + 3) = 0 (x + 3) 2 – 6 = 0 (x + 3) 2 = 6 -9 (x + 3) = ±  6 x = ±  x = &x = We could NOT have solved this one by factorising

3.The Quadratic Formula x 2 + 8x - 2 = 0 x = -b±  (b 2 – 4ac) 2a x = -8±  [8 2 – 4(1)(-2)] 2(1) x = -8±  (72) 2 x = 2.83 & x = -5.66

The Discriminant Any quadratic equation can be solved by using the quadratic formula x = -b±  (b 2 – 4ac) 2a The expression under the square root sign, b 2 – 4ac, is the key to deciding what type of ROOTS the equation has. This expression is called the DISCRIMINANT, because it allows us to discriminate between different types of roots. Note

x 2 + 4x + 4 = 0 (x + 2)(x + 2)= 0 x = -2 b 2 – 4ac = 4 2 – 4(1)(4) = 0 x 2 - 8x + 16 = 0 (x - 4)(x – 4)= 0 x = 4 b 2 – 4ac =(-8) 2 -4(1)(16) = 0 When the discriminant is equal to zero, there seems to be only one solution to the quadratic equation. Using the quadratic formula, the solution becomes, which clearly gives only one solution to the quadratic equation. -b 2a

x 2 + 5x + 4 = 0 (x + 4)(x + 1)= 0 x = -4, -1 b 2 – 4ac = 5 2 – 4(1)(4) = 9 x 2 - 3x - 10 = 0 (x + 2)(x - 5)= 0 x = -2, 5 b 2 – 4ac =(-3) 2 -4(1)(-10) = 49 When the discriminant is equal to a square number, there seems to be two solutions, and it can be factorised.

x 2 - 7x + 1 = 0 x = 7±  (45) 2 x = 6.85, 0.15 b 2 – 4ac = (-7) 2 – 4(1)(1) = 45 x 2 - 5x + 2 = 0 x = 7±  (17) 2 x = 5.56, 1.44 When the discriminant is a positive number, there seems to be two solutions. If the discriminant is positive but not a square number, then the quadratic cannot be factorised. b 2 – 4ac = (-5) 2 – 4(1)(2) = 17

x 2 + x + 8 = 0 x = -1±  (-31) 2 b 2 – 4ac is negative Cannot take square root of a negative number If the discriminant is negative, then there are NO solutions to the equation.

Using b 2 – 4ac We can demonstrate each of these possibilities using graphs. Remember, if we plot the graph of y = ax 2 + bx + c the solutions to the equation ax 2 + bx + c = 0 are given by the points where the graph crosses the x-axis. y x b 2 – 4ac is zero One solution y x b 2 – 4ac is negative No solutions y x b 2 – 4ac is positive Two solutions

b 2 – 4ac = 0 1 repeated, real root. EQUAL ROOTS b 2 – 4ac > 0 2 REAL ROOTS b 2 – 4ac < 0 NO REAL ROOTS For real roots, b 2 – 4ac  0 If b 2 – 4ac is a square number then it can be factorised. Note

Applications Of The Discriminant EXAMPLE 1 Find the value(s) of m, given that the equation x 2 + 2mx + 9 = 0 has equal roots. For equal roots, b 2 – 4ac = 0 a = 1 b = 2m c = 9 b 2 – 4ac = 0 (2m) 2 – 4(1)(9) = 0 4m 2 – 36 = 0 4(m + 3)(m – 3) = 0 m = ±3 4(m 2 – 9) = 0

EXAMPLE 2 Find the value(s) of a, given that (a - 2)x 2 + (a + 1)x + 2(a - 1 ) = 0 has equal roots. For equal roots, b 2 – 4ac = 0 a = a - 2 b = a + 1 c = 2a - 2

b 2 – 4ac = 0 (a + 1) 2 – 4(a - 2)(2a - 2) = 0 a 2 + 2a a a - 15 = 0 - 4(2a 2 - 6a + 4) = 0 7a a + 15 = 0 (7a – 5)(a - 3) = 0 a = 5 7, a = 3 a 2 + 2a + 1 – 8a a - 16 = 0

Tangents To Curves LOOK AT THIS EXAMPLE Find the coordinates of the point(s) where curve y = x 2 meets the line y = 2x - 1. Curve meets line when... x 2 = 2x - 1 x 2 - 2x + 1= 0 (x - 1)(x - 1)= 0 x = 1(twice) We have equal roots 1 point of contact Line is a tangent

At x = 1 y = 2 ( 1 ) – 1 = 1 pt. of contact is (1,1) In addition... b 2 – 4ac = (-2) 2 – 4(1)(1) = 0 This proves that we have equal roots

SUMMARY If we have equal roots Then there is only one point of contact. Therefore the line will be a tangent to the curve. To prove that the line is a tangent to the curve, we can also use the fact that the discriminant must be equal to zero.

EXAMPLE 2 Find the value of c given that the line y = x + c is a tangent to the curve with equation y = x 2 - 5x, and find the coordinates of the point of contact. Curve meets line when... x 2 - 5x = x + c x 2 - 6x - c = 0 If the line is a tangent Then we have equal roots And b 2 - 4ac = 0

y = x + c x 2 - 6x - c = 0 a = 1 b = -6 c = -c b 2 – 4ac = 0 (-6) 2 – 4(1)(-c) = c = 0 4c = -36 c = -9 x 2 - 6x + 9 = 0 (x - 3)(x - 3)= 0 x = 3 y = -6 pt. of contact is (3, -6) y = x - 9 At x = 3 y = 3 - 9

EXAMPLE 3 Find the values of k given that the line y - x + k = 0 is a tangent to the curve x 2 + y 2 = 18.Find the point of contact for the negative value of k. x 2 + y 2 = 18y -x +k =0 y = x - kx 2 + (x - k) 2 = 18 x 2 + x 2 - 2xk + k 2 = 18 2x 2 - 2kx + k =0

If the line is a tangent Then we have equal roots And b 2 - 4ac = 0 2x 2 - 2kx + k =0 a = 2 b = -2k c = k b 2 – 4ac = 0 (-2k) 2 – 4(2)(k ) = 0 4k 2 – 8k = 0 -4k 2 = -144 k 2 = 36 k = ±6

For k = -6 2x 2 - 2kx + k =0 2x x + 18 = 0 2(x 2 + 6x + 9) = 0 2(x + 3)(x + 3) = 0 x = -3 At x = -3 y = x - ky = x + 6 Point of contact is (-3, 3 ) y = y = 3

Quadratic Inequations As well as the roots of a quadratic, we are sometimes interested in when the parabola is above or below the x-axis. When the parabola is above the x-axis, y > 0 When the parabola is below the x-axis, y < 0 Note

EXAMPLE 1 Solve the quadratic inequation x 2 + 5x + 6 > 0 x 2 + 5x + 6 > 0 (x + 3)(x + 2) > 0 x y O -2-3 When is the graph above the x-axis? x < -3or x > -2

EXAMPLE 2 Solve the quadratic inequation x 2 - 7x + 12 < 0 x 2 - 7x + 12 < 0 (x - 3)(x - 4) < 0 x y O 43 When is the graph below the x-axis? 3 < x < 4

EXAMPLE 3 Solve the quadratic inequation x 2 - 6x + 9 < 0 x 2 - 6x + 9 < 0 (x - 3)(x - 3) < 0 x y O 3 x 2 - 6x + 9 < 0 for NO values of x. When is the graph below the x-axis?

EXAMPLE 4 Solve the quadratic inequation 3 - 5x - 2x 2  x - 2x 2  0 (2x - 1)(x + 3)  0 x y O -3½ When is the graph above or on the x-axis? x  ½or x  -3 -2x 2 – 5x + 3  0 2x 2 + 5x - 3  0

SUMMARY To solve a quadratic inequation, follow the 3 simple steps. 1. FACTORISE 2. SKETCH GRAPH 3. SOLVE Note