remainder and factor theorem polynomials polynomials polynomials combining polynomials combining...
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Remainder and Factor Remainder and Factor TheoremTheorem
•PolynomialsPolynomials•Combining polynomialsCombining polynomials
•Function notationFunction notation•Division of PolynomialDivision of Polynomial•Remainder TheoremRemainder Theorem
•Factor TheoremFactor Theorem
Polynomials
• An expression that can be written in the form
• a + bx + cx2 + dx3 +ex4 + ….
• Things with Surds (e.g. x + 4x +1 ) and reciprocals (e.g. 1/x + x) are not polynomials
• The degree is the highest index• e.g 4x5 + 13x3 + 27x is of degree 5
Polynomials can be combined to give new
polynomials
Set-up a multiplication table
3x2 -5x -32x4
+4x2
6x6 -10x5 -6x4
+12x4-20x3 -12x2
Gather like terms= 6x6 - 10x5 + 6x4 - 20x3 -12x2
They can be added:2x2 - 5x - 3 + 2x4 + 4x2 + 2x
= 2x4 + 6x2 -3x -3
(2x4 + 4x2)(3x2 - 5x - 3)Or multiplied:
6x6 -10x5 -6x4+12x4-20x3 -12x2
Function Notation
• An polynomials function can be written as
• f(x) = a + bx + cx2 + dx3 +ex4 + ….• f(x) means ‘function of x’• instead of y = ….
• e.g f(x) = 4x5 + 13x3 + 27x
• f(3) means ….– “the value of the function when x=3”– e.g. for f(x) = 4x5 + 13x3 + 27x – f(3) = 4 x 35 + 13 x 33 + 27 x 3– f(3) = 972 + 351 + 81 = 1404
x
f(x)
Combining Functions (1)
• Suppose– f(x) = x3 + 2x +1– g(x) = 3x2 - x - 2
• g(x) or p(x) or q(x) or ….. Can all be used to define different functions
We can define a new function by any linear or multiplicative combination of these…
• e.g. 2f(x) + 3g(x) = 2(x3 + 2x +1) + 3(3x2 - x - 2)• e.g. 3 f(x) g(x) = 3(x3 + 2x +1)(3x2 - x - 2)
Combining Functions (2)We can define a new function by any linear or multiplicative combination of these…
• e.g. 2f(x) + 3g(x) = 2(x3 + 2x +1) + 3(3x2 - x - 2)= 2x3 + 4x + 2 + 9x2 -3x -6GATHER LIKE TERMS
x3 x2 x
= 2x3 + 9x2 + x - 4
2 + 9 + - 4
Combining Functions (3)e.g. 3 f(x) g(x)
= 3(x3 + 2x +1)(3x2 - x - 2)
x3 +2x +13x2 -x -2
Do multiplication table; gather like terms and then multiply through by 3
Finding one bracket given the other
Fill in the empty bracket:
x2 - x - 20 = (x + 4)( )
To get x2 the x must be multiplied by another x
x2 - x - 20 = (x + 4)(x )
To get -20 the +4 must be multiplied by -5
x2 - x - 20 = (x + 4)(x - 5)
Expand it to check: (x + 4)(x - 5) = x2 + 4x - 5x -20 = x2 - x - 20
We can do division now
f(x) = (x2 - x - 20) (x + 4)
f(x) = (x2 - x - 20) (x + 4)
x (x+4) x (x+4)
(x + 4) x f(x) = (x2 - x - 20)
It is exactly the same question as:=Fill in the empty bracket:
x2 - x - 20 = (x + 4)( )
Finding one bracket given the other - cubics Fill in the empty bracket:
x3 + 3x2 - 12x + 4 = (x - 2)( )
To get x3 the x must be multiplied by x2
To get +4 the -2 must be multiplied by -2
x3 + 3x2 - 12x + 4 = (x - 2)(x2 )
x3 + 3x2 - 12x + 4 = (x - 2)(x2 -2)
These 2 give us -2x2,
but we need 3x2This x must be multiplied by 5x to give us another 5x2x3 + 3x2 - 12x + 4 = (x - 2)(x2 +5x -2)
Finding one bracket given the other - cubics (2.1)
Fill in the empty bracket:x3 + 3x2 - 12x + 4 = (x - 2)( )
Using a multiplication table:
+4-12x+3x2x3
x
-2
ax2 +bx +c
x3
+4
Can put x3 and +4 in. They can only come from 1 place.So a = 1, c = -2
Finding one bracket given the other - cubics (2.2)
Fill in the empty bracket:x3 + 3x2 - 12x + 4 = (x - 2)(x2 +bx -2 )
Using a multiplication table:
-12x+3x2
x
-2
x2 +bx -2
x3
+4
-2x
-2x2
Complete more of the table by multiplying known values
Complete the rest algebraically
+bx2
-2bx
Finding one bracket given the other - cubics (2.3)
Fill in the empty bracket:x3 + 3x2 - 12x + 4 = (x - 2)(x2 -2 )
Using a multiplication table:
-12x+3x2
x
-2
x2 +bx -2
x3
+4
-2x
-2x2
+bx2
-2bx
+3x2 = bx2 - 2x2
-12x = -2bx - 2x
+3 = b - 2
-12 = -2b - 2
Either way,b = 5
+5x
Gather like terms
Dealing with remainders
Fill in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) + R
Using a multiplication table:
x
+2
ax2 +bx +c
x3
remainder
+x-x2
+15
x3
Can put x3
This can only come from 1 place.So a = 1
Dealing with remainders
Fill in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) + R
Using a multiplication table:
+x-x2
x
+2
x2 +bx +c
x3
remainder
+15
x3
x2
2x2
We can fill this bit in now
The rest of the x2 term must come from here
bx2
bx2 + 2x2 = -x2
b + 2 = -1b = -3
Dealing with remainders
Fill in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) + R
Using a multiplication table:
+x-x2
x
+2
x2 -3x +c
x3
remainder
+15
x3
x2 -3x
2x2
-3x2
-6x
We can fill this bit in nowThe rest of the x term must come from here
cx
cx - 6x = +xc - 6 = 1
c = 7
Dealing with remainders
Fill in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) +R
Using a multiplication table:
+x-x2
x
+2
x2 -3x +7
x3
remainder
+15
x3
x2 – 3x +7
2x2
-3x2
-6x
7x
+14
We can fill this bit in now- and we’ve got our 2nd function
Dealing with remainders
Fill in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) +R
Using a multiplication table:
+x-x2
x
+2
x2 -3x +7
x3
remainder
+15
x3
x2 – 3x +7
2x2
-3x2
-6x
7x
+14Remainder
The numerical term (+15) comes from the +14 and the remainder R
+15 = +14 + R So, R = 1
Dealing with remainders
Filled in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) +1
Using a multiplication table:
+x-x2
x
+2
x2 -3x +7
x3+15
x3
x2 – 3x +7
2x2
-3x2
-6x
7x
+14Remainder
The numerical term (+15) comes from the +14 and the remainder R
+15 = +14 + R So, R = 1
Fill in the empty bracket:
f(x) = x3 - x2 + x + 15 = (x + 2)( ) + R
If:f(x) = x3 - x2 + x + 15
Division with Remainders
What is f(x) divided by x+2 …. and what is the remainder
This is exactly the same as ……………
We found:x3 - x2 + x + 15 = (x + 2)( ) +1x2 – 3x +7
If:f(x) = x3 - x2 + x + 15
What is f(x) divided by x+2? …. and what is the remainder?
What is f(x) divided by x+2? (x2 – 3x +7)
…. and what is the remainder? 1
If:f(x) = x3 - x2 + x + 15 = (x + 2)( ) +1 What is f(x) divided by x+2? (x2 – 3x +7)
…. and what is the remainder? 1
The Remainder Theorem – example 1
x2 – 3x +7
If we calculate f(-2) …..
f(-2) = (-2)3 – (-2)2 + -2 + 15 = -8 -4 - 2 +15 = 1
-2 from (x+2)=0Our remainder
If:p(x) = x3 + 2x2 - 9x + 10= (x - 2)( ) +8 What is p(x) divided by x-2? (x2 + 4x - 1)
…. and what is the remainder? 8
The Remainder Theorem – example 2
x2 + 4x - 1
If we calculate p(2) …..
p(2) = (2)3 + 2(2)2 - 9(2) + 10 = 8 + 8 - 18 + 10 = 8
2 from (x-2)=0Our remainder
When p(x) is divided by (x-a) …. the remainder is p(a)
The Remainder Theorem
Given:p(x) = 2x3 - 5x2 + x - 12
What value of p(...)=0,hence will give no remainder?
The Factor Theorem – example
If we calculate p(0) = 2(0)3 – 5(0)2 + 0 - 12 = -12
p(1) = 2(1)3 – 5(1)2 + 1 - 12 = 2-5+1-12 = -14
p(2) = 2(2)3 – 5(2)2 + 2 - 12 = 16-20+2-12 = -16
p(3) = 2(3)3 – 5(3)2 + 3 - 12 = 54-45+3-12 = 0
By the Remainder Theorem :- the factor (x-3) gives no
remainder
For bigger values of ‘x’ the x3 term will dominate and make p(x) larger
Given:p(x) = 2x3 - 5x2 + x - 12
The Factor Theorem – example
p(3) = 2(3)3 – 5(3)2 + 3 - 12 = 54-45+3-12 = 0
By the Remainder Theorem :- the factor (x-3) gives no
remainderSo (x-3) divides exactly into p(x)
……… (x-3) is a factor
For a given polynomial p(x)
If p(a) = 0
… then (x-a) is a factor of p(x)
The Factor Theorem
If:p(x) = x3 + bx2 + bx + 5 When is p(x) divided by x+2 the remainder is 5
If we calculate p(-2) …..
p(-2) = (-2)3 + b(-2)2 + b(-2) + 5 = -8 + 4b - 2b + 5 = 2b - 3-2 from (x+2)
Which theorem?The Remainder Theorem
By the Remainder theorem: 2b - 3 = 5 Our remainder
2b = 8 b = 4
If:f(x) = x3 + 3x2 - 6x - 8 a) Find f(2) f(2) = (2)3 + 3(2)2 - 6(2) - 8
= 8 + 12 - 12 - 8 = 0
b) Use the Factor Theorem to write a factor of f(x)
For a given polynomial p(x)If p(a) = 0… then (x-a) is a factor of p(x)
f(2) = 0…. so (x-2) is a factor of x3 + 3x2 - 6x - 8
If:f(x) = x3 + 3x2 - 6x - 8 b) (x-2) is a factor of x3 + 3x2 - 6x - 8c) Express f(x) as a product of 3 linear factors
.. means (x-a)(x-b)(x-c)=x3 + 3x2 - 6x - 8We know (x-2)(x-b)(x-c)=x3 + 3x2 - 6x - 8
…. consider (x-2)(ax2+bx+c)=x3 + 3x2 - 6x - 8
a=? a=1 : so x x ax2 = x3
(x-2)(x2+bx+c)=x3 + 3x2 - 6x - 8
c=? c=4 : so -2 x 4 = -8 (x-2)(x2+bx+4)=x3 + 3x2 - 6x - 8
If:f(x) = x3 + 3x2 - 6x - 8 b) (x-2) is a factor of x3 + 3x2 - 6x - 8c) Express f(x) as a product of 3 linear factors
(x-2)(x2+bx+4)=x3 + 3x2 - 6x - 8Expand : need only check the x2 or x terms
… + bx2 -2x2 + … = … + 3x2 + ….
Or … - 2bx + 4x + … = … - 6x + ….
b - 2 = 3 b=5
-2b + 4 = -6 b=5
EASIER
HARD
(x-2)(x2+5x+4)=x3 + 3x2 - 6x - 8
If:f(x) = x3 + 3x2 - 6x - 8 b) (x-2) is a factor of x3 + 3x2 - 6x - 8c) Express f(x) as a product of 3 linear factors
(x-2)(x2+5x+4)=x3 + 3x2 - 6x - 8
(x2+5x+4) = (x+4)(x+1)
So, (x-2)(x+4)(x+1) = x3 + 3x2 - 6x - 8
……. a product of 3 linear factors
(x-2)(x+4)(x+1) = x3 + 3x2 - 6x - 8
……. Sketch x3 + 3x2 - 6x - 8
y = x3 + 3x2 - 6x - 8y = (x-2)(x+4)(x+1)
Where does it cross the x-axis (y=0) ?
(x-2)(x+4)(x+1) = 0 Either (x-2) = 0 x=2 Or (x+4)= 0 x=-4Or (x+1) = 0 x=-1
Where does it cross the y-axis (x=0) ?
y = (0)3 + 3(0)2 - 6(0) - 8 = -8
Factor and Remainder Theorem
Where does it cross the x-axis (y=0) ? Either (x-2) = 0 x=2 Or (x+4)= 0 x=-4Or (x+1) = 0 x=-1
Where does it cross the y-axis (x=0) ? y = -8
x
y
Goes through these-sketch a nice curve
x-1
x-4
x2
x-8