Calculus 3.4 Manipulate real and complex numbers and solve equations AS 90638.

Calculus 3.4Manipulate real and complex numbers and solve equations

AS 90638

Recognising terms

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x 3 + px 2 + 2x + q = Ax 3 + 2x 2 + Bx −1

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1x 3 = Ax 3

⇒ A =1

Recognising terms

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x 3 + px 2 + 2x + q = Ax 3 + 2x 2 + Bx −1

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px 2 = 2x 2

⇒ p = 2

Recognising terms

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x 3 + px 2 + 2x + q = Ax 3 + 2x 2 + Bx −1

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2x = Bx

⇒ B = 2

Recognising terms

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x 3 + px 2 + 2x + q = Ax 3 + 2x 2 + Bx −1

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q = −1

Worksheet 1

Quadratics

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y = ax 2 + bx + c

General formula:

Example 1.

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y = x 2 − 2x − 3Solving for the roots (i.e. y = 0)Method 1: Factorising

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x2 −2x−3=0

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⇒ x−3( ) x+1( ) =0

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x=3, -1

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x 2 − 2x − 3 = 0Method 2: Quadratic Formula

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x =−b ± b2 − 4ac

2a

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x 2 − 2x − 3 = 0

Method 3: Graphics Calculator

•Equation Mode

•F2: Polynomial

•Degree: 2 (F1)

•Enter 1, -2, -3

•Solve (F1)

Equation has 2 real solutionsx = 3, -1

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b2 − 4ac > 0

Example 2

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y = x 2 − 6x + 9Method 1: Factorize

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x2 −6x+ 9 =0

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⇒ x−3( ) x−3( ) =0

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⇒ x=3

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x 2 − 6x + 9 = 0Method 2: Quadratic Formula

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b2 − 4ac = 0

Equation has two equal real solutions:x = 3, 3

Example 3

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y = x 2 + 3x + 3

Equation cannot be factorised.

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x 2 + 3x + 3 = 0Using quadratic formula

Equations has no real solutions

Equation has no real roots.

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b2 − 4ac < 0

Forming quadratic equations from 2

solutions.

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ax 2 + bx + c = 0

⇒ x 2 +b

ax +

c

a= 0

If solutions are

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x = α , β

Then x - α( ) x − β( ) = 0

Multiplying out x 2 − α + β( )x + αβ = 0

Example 1

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x = 3, 5

⇒ x 2 − 3+ 5( )x + 3× 5 = 0

Quadratic equation is

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y = x 2 − 8x +15

Example 2

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x = 6, −9

⇒ x 2 − 6 − 9( )x + 6×−9 = 0

Quadratic equation is

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y = x 2 + 3x − 54

Example 3

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x =3

2,2

5

⇒ x 2 −3

2+

2

5

⎛

⎝ ⎜

⎞

⎠ ⎟x +

3

2×

2

5= 0

⇒ x 2 −19

10x +

6

10= 0

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y=10x2 −19x+ 6

Factorise

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32x − 3x +2 −10

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3x( )

2− 32 × 3x

( ) −10

Form the quadratic

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3x −10( ) 3x +1( )

Solve

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x − 3( ) x +1( ) ≤ 0Sketch the graph

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−1 ≤ x ≤ 3

Solve

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x − 3( ) x +1( ) > 0Sketch the graph

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x < −1

x > 3

Completing the square

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x 2 − 6x + 2Half the coefficient of x

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x − 3( )2

− 32 + 2

= x − 3( )2

− 7

Subtract this value squared

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x − 3( )2

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x − 3( )2

− 32


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x 2 + 8x − 3Half the coefficient of x

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x + 4( )2

− 42 − 3

= x + 4( )2

−19



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2x 2 +10x −1

Take out the 2

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2(x 2 + 5x) −1


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2x 2 +10x −1Half the coefficient of x

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2 x+52

⎛

⎝ ⎜

⎞

⎠ ⎟2

−2×52

⎛

⎝ ⎜

⎞

⎠ ⎟2

−1


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2(x 2 + 5x) −1


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2x 2 +10x −1Half the coefficient of x

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2 x +5

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

− 2 ×5

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

−1

= 2 x +5

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

−11.5


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2(x 2 + 5x) −1


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−3x 2 −12x + 2Take out the -3

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−3(x 2 + 4 x) + 2


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−3x 2 −12x + 2Half the coefficient of x

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−3 x+ 2( )2−(−3)×22 + 2


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−3(x 2 + 4 x) + 2


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−3x 2 −12x + 2Half the coefficient of x

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−3 x+ 2( )2−(−3)×22 + 2

=−3 x+ 2( )2+14


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−3(x 2 + 4 x) + 2

Worksheet 2

Factor Remainder Theorem

Division of polynomials

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2x 3 + 3x 2 − 29x + 20 by x − 2

Write down the coefficients2 3 -29

20 Solve

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x − 2 = 0

⇒ x = 2

2 2

x

4

7

14

-15

-30

-10

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2x 2 + 7x −15 Remainder -10

Divide

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2x 3 + 3x 2 − 29x + 20 = (x − 2)(2x 2 + 7x −15) −10

Substituting x = 2

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2 2( )3

+ 3 2( )2

− 29 2( ) + 20 = (2 − 2)(2 2( )2

+ 7 2( ) −15) −10

= 0 −10

i.e. substituting into the original gives us the

remainder

Example 2

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3x 3 − 2x 2 − 7x − 2 by x +1

Write down the coefficients

3 -2 -7 -2

Solve

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x +1= 0

⇒ x = −1

-1 3

x

-3-5

5

-2

2

0

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3x 2 − 5x − 2 Remainder 0

A remainder of 0 means is a factor and -1 is a solution

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x +1

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⇒ 3x 3 − 2x 2 − 7x − 2 = x +1( )(3x 2 − 5x − 2)

Substituting x = -1

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⇒ 3 −1( )3

− 2 −1( )2

− 7 −1( ) − 2 = −1+1( )(3 −1( )2

− 5 −1( ) − 2) = 0

When -1 is substituted, remainder is 0

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⇒ 3x 3 − 2x 2 − 7x − 2 = x +1( )(3x 2 − 5x − 2)

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=(x +1)(3x +1)(x − 2)

Factorising the quadratic gives

To factorise a cubic first find a value that will give a

remainder of 0.

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f (x) = 4x 3 −13x + 6


remainder of 0.

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f (x) = 4x 3 −13x + 6

f (1) = 4 1( )3

−13 1( ) + 6 ≠ 0


remainder of 0.

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f (x) = 4x 3 −13x + 6

f (1) = 4 1( )3

−13 1( ) + 6 ≠ 0

f (−2) = 4(−2)3 −13(−2) + 6 = 0

⇒ x + 2 is a factor


remainder of 0.

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f (x) = 4x 3 −13x + 6

f (1) = 4 1( )3

−13 1( ) + 6 ≠ 0

f (−2) = 4(−2)3 −13(−2) + 6 = 0

⇒ x + 2 is a factorCreate the cubic by equating coefficients

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x + 2( ) 4x 2 − 8x + 3( )

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f (x) = 4 x 3 + 0x 2 −13x + 6

Calculus 3.4 Manipulate real and complex numbers and solve equations AS 90638.

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Transcript of Calculus 3.4 Manipulate real and complex numbers and solve equations AS 90638.