Division with polynomials

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Division with polynomials

• from the remainder theorem to the Factor Theorem

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The Remainder Theorem
When p(x) is divided by (x-a) . the remainder
is p(a)
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The Factor Theorem example
For bigger values of x the x3 term will
dominate and make p(x) larger
What value of p(...)0, hence will give no
remainder?
If we calculate p(0) 2(0)3 5(0)2 0 - 12
-12
p(1) 2(1)3 5(1)2 1 - 12 2-51-12 -14
p(2) 2(2)3 5(2)2 2 - 12 16-202-12 -16
p(3) 2(3)3 5(3)2 3 - 12 54-453-12 0
By the Remainder Theorem - the factor (x-3)
gives no remainder
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The Factor Theorem example
p(3) 2(3)3 5(3)2 3 - 12 54-453-12 0
By the Remainder Theorem - the factor (x-3)
gives no remainder
So (x-3) divides exactly into p(x) (x-3)
is a factor
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The Factor Theorem
For a given polynomial p(x) If p(a) 0 then
(x-a) is a factor of p(x)
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Page 95 Ex.G Q3
When is p(x) divided by x2 the remainder is 5
Which theorem?
The Remainder Theorem
If we calculate p(-2) ..
p(-2) (-2)3 b(-2)2 b(-2) 5 -8 4b
- 2b 5 2b - 3
By the Remainder theorem 2b - 3 5
2b 8 b 4
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Page 97 Q2 AQA 2002
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
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Page 97 Q2 AQA 2002
c) Express f(x) as a product of 3 linear factors
.. means (x-a)(x-b)(x-c)x3 3x2 - 6x - 8
We know (x-2)(x-b)(x-c)x3 3x2 - 6x - 8
. consider (x-2)(ax2bxc)x3 3x2 - 6x - 8
a?
a1 so x x ax2 x3
(x-2)(x2bxc)x3 3x2 - 6x - 8
c4 so -2 x 4 -8
c?
(x-2)(x2bx4)x3 3x2 - 6x - 8
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Page 97 Q2 AQA 2002
c) Express f(x) as a product of 3 linear factors
(x-2)(x2bx4)x3 3x2 - 6x - 8
Expand need only check the x2 or x terms
bx2 -2x2 3x2 .
EASIER
b - 2 3 b5
Or - 2bx 4x - 6x .
-2b 4 -6 b5
HARD
(x-2)(x25x4)x3 3x2 - 6x - 8
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Page 97 Q2 AQA 2002
c) Express f(x) as a product of 3 linear factors
(x-2)(x25x4)x3 3x2 - 6x - 8
(x25x4) (x4)(x1)
So,
(x-2)(x4)(x1) x3 3x2 - 6x - 8
. a product of 3 linear factors
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Page 97 Q2 AQA 2002 - extra bit
(x-2)(x4)(x1) x3 3x2 - 6x - 8
. Sketch x3 3x2 - 6x - 8
y x3 3x2 - 6x - 8
y (x-2)(x4)(x1)
Where does it cross the x-axis (y0) ?
(x-2)(x4)(x1) 0
Either (x-2) 0 x2
Or (x4) 0 x-4
Or (x1) 0 x-1
Where does it cross the y-axis (x0) ?
y (0)3 3(0)2 - 6(0) - 8 -8
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Page 97 Q2 AQA 2002 - extra bit
Either (x-2) 0 x2
Where does it cross the x-axis (y0) ?
Or (x4) 0 x-4
Or (x1) 0 x-1
y -8
Where does it cross the y-axis (x0) ?
Goes through these -sketch a nice curve
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Have a go

• Page 93
• Exercise F
• Q1, 2, 5, 8

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Good News Everybody
You have a test on Monday (6th November), covering
the following topics .

• Simultaneous equations
• linear and quadratic
• mainly using the substitution method
• Inequalities
• linear and quadratic
• Polynomials and functions
• division
• remainder theorem
• factor theorem

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Some targeted revision questions(do these as an
absolute minimum)

• Simultaneous equations
• Page 59-60 Exercise C Q3, Q6
• Inequalities
• Page 70 Exercise C Q2, Q9
• Polynomials
• Page 93 Exercise F Q4
• Page 95 Exercise G Q1 Q2