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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
Completing the square
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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
Subtract this value squared
Completing the square
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2x 2 +10x −1
Take out the 2
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2(x 2 + 5x) −1
Completing the square
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2x 2 +10x −1Half the coefficient of x
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2 x+52
⎛
⎝ ⎜
⎞
⎠ ⎟2
−2×52
⎛
⎝ ⎜
⎞
⎠ ⎟2
−1
Subtract this value squared
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2(x 2 + 5x) −1
Completing the square
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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
Subtract this value squared
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2(x 2 + 5x) −1
Completing the square
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−3x 2 −12x + 2Take out the -3
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−3(x 2 + 4 x) + 2
Completing the square
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−3x 2 −12x + 2Half the coefficient of x
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−3 x+ 2( )2−(−3)×22 + 2
Subtract this value squared
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−3(x 2 + 4 x) + 2
Completing the square
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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
Subtract this value squared
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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
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
f (1) = 4 1( )3
−13 1( ) + 6 ≠ 0
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
f (1) = 4 1( )3
−13 1( ) + 6 ≠ 0
f (−2) = 4(−2)3 −13(−2) + 6 = 0
⇒ x + 2 is a factor
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
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