Transcript 4 - Brodish

Multiplying polynomials
In this PowerPoint presentation you
will see two alternative approaches to
polynomial multiplication.
To see multiplication using a table
click here.
To see multiplication “in your head”
click here.
Using a table/box
Multiply x² + 3x – 2 by 2x² – x + 4
x²
2x²
-x
4
3x
-2
Using a table/box
Multiply
x² by 2x²
x²
2x²
-x
4
3x
-2
Using a table/box
Multiply
x² by 2x²
x²
2x²
-x
4
2x4
3x
-2
Using a table/box
Multiply
3x by 2x²
x²
2x²
-x
4
2x4
3x
-2
Using a table/box
Multiply
3x by 2x²
2x²
-x
4
x²
3x
2x4
6x³
-2
Using a table/box
Fill in the rest of the table
in the same way
2x²
-x
4
x²
3x
2x4
6x³
-2
Using a table/box
Fill in the rest of the table
in the same way
2x²
-x
4
x²
3x
-2
2x4
6x³
-4x²
Using a table/box
Fill in the rest of the table
in the same way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
4
Using a table/box
Fill in the rest of the table
in the same way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
4
Using a table/box
Fill in the rest of the table
in the same way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
Using a table/box
Fill in the rest of the table
in the same way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
Using a table/box
Fill in the rest of the table
in the same way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
Using a table/box
Fill in the rest of the table
in the same way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
Using a table/box
First the term
in x4
Now add up all the
terms in the table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
(x² + 3x – 2)(2x² - x + 4) = 2x4
Using a table/box
Now add up all the
terms in the table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
then the terms
in x³
(x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³
Using a table/box
Now add up all the
terms in the table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
then the terms
in x²
(x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³ - 3x²
Using a table/box
Now add up all the
terms in the table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
then the terms
in x
(x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³ - 3x²+ 14x
Using a table/box
Now add up all the
terms in the table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
and finally the
constant term
(x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³ - 3x²+ 14x - 8
Multiplying in your head
Multiply x² + 3x – 2 by 2x² – x + 4
(x² + 3x – 2)(2x² – x + 4)
Multiplying in your head
Start by multiplying the
first terms in each bracket
to give a term in x4
(x² + 3x – 2)(2x² – x + 4)
= 2x4
Multiplying in your head
Now look for pairs of terms
which multiply together to give a
term in x³. There are two pairs.
(x² + 3x – 2)(2x² – x + 4)
= 2x4 – x³
Multiplying in your head
Now look for pairs of terms
which multiply together to give a
term in x³. There are two pairs.
(x² + 3x – 2)(2x² – x + 4)
= 2x4 – x³ + 6x³
Multiplying in your head
Now look for pairs of terms which
multiply together to give a term in
x². There are three pairs.
(x² + 3x – 2)(2x² – x + 4)
= 2x4 – x³ + 6x³ + 4x²
Multiplying in your head
Now look for pairs of terms which
multiply together to give a term in
x². There are three pairs.
(x² + 3x – 2)(2x² – x + 4)
= 2x4 – x³ + 6x³ + 4x² - 3x²
Multiplying in your head
Now look for pairs of terms which
multiply together to give a term in
x². There are three pairs.
(x² + 3x – 2)(2x² – x + 4)
= 2x4 – x³ + 6x³ + 4x² - 3x²- 4x²
Multiplying in your head
Now look for pairs of terms
which multiply together to give
a term in x. There are two pairs.
(x² + 3x – 2)(2x² – x + 4)
= 2x4 – x³ + 6x³ + 4x² - 3x²- 4x²+ 12x
Multiplying in your head
Now look for pairs of terms
which multiply together to give
a term in x. There are two pairs.
(x² + 3x – 2)(2x² – x + 4)
= 2x4 – x³ + 6x³ + 4x² - 3x²- 4x²+ 12x + 2x
Multiplying in your head
Finally multiply the last two
terms in each bracket to give
the constant term.
(x² + 3x – 2)(2x² – x + 4)
= 2x4 – x³ + 6x³ + 4x² - 3x²- 4x²+ 12x + 2x - 8
Multiplying in your head
Now simplify by collecting like
terms.
(x² + 3x – 2)(2x² – x + 4)
= 2x4 – x³ + 6x³ + 4x² - 3x²- 4x²+ 12x + 2x - 8
= 2x4 + 5x³ - 3x² + 14x - 8
A metal work wants to make an open box from a 12 in x 16 in
sheet of metal by cutting equal squares from each corner.
Draw a
picture.
16
12
A metal work wants to make an open box from a 12 in x 16
in sheet of metal by cutting equal squares from each corner.
Draw a
picture.
x
16 - 2x
x
x
x
12 -2x
x
x
x
x
A metal work wants to make an open box from a 12 in x 16
in sheet of metal by cutting equal squares from each corner.
Find the volume.
x
16 - 2x
x
x
x
12 -2x
x
x
x
V = lwh = (16-2x)(12-2x)(x)
x
V = lwh = (16-2x)(12-2x)(x) Find the zeros!
x= 8; x = 6; x= 0
Find the maximum
volume of the box and
the side length of the
cut out squares that
generates that
volume.
Use 2nd CALC -> MAX
Practical Domain:
(0,6)
Practical Range:
(0, 194.07]
(2.26, 194.07)
Cut Out = 2.26
Max volume = 194.07
A rectangular picture is 12in by 16 in. When each dimension
is increased by the same amount, the area is increased by
60 in2. If x represent the number of inches by which each
dimension is increased, which equation could be used to find
the value for x?
16
Draw a
picture.
12
A = 16(12)
A= 192 in2
(x + 16)(x + 12) = 252
x2 + 12x + 16x + 192 – 252 = 0
x2 + 28x – 60 = 0
(x + 30)(x -2) = 0 Find the zeros: x = -30 and x = 2
Solution: x = 2
12 +x
16 + X
A= 192 + 60
A = 252 in2
Introducing: Pascal’s Triangle
Row 5
Row 6
Take a
moment to
copy the
first 6
rows. What
patterns do
you see?
The Binomial Theorem
Use Pascal’s Triangle to expand (a + b)5.
Use the row that has 5 as its second number.
The exponents for a begin with 5 and decrease.
1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5
The exponents for b begin with 0 and increase.
In its simplest form, the expansion is
a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5.
Row 5
The Binomial Theorem
(not in your notes)
Use Pascal’s Triangle to expand (x – 3)4.
First write the pattern for raising a binomial to the fourth power.
1
4
6
4
1
Coefficients from
Pascal’s Triangle.
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
Since (x – 3)4 = (x + (–3))4, substitute x for a and –3 for b.
(x + (–3))4 = x4 + 4x3(–3) + 6x2(–3)2 + 4x(–3)3 + (–3)4
= x4 – 12x3 + 54x2 – 108x + 81
The expansion of (x – 3)4 is x4 – 12x3 + 54x2 – 108x + 81.
The Binomial Theorem
Use Pascal’s Triangle to expand (2x + 4)3.
First write the pattern for raising a binomial to the fourth power.
1
3
3
1
Coefficients from
Pascal’s Triangle.
(2x + 4)3 = 1(2x)3 + 3(2x)24 + 3(2x)142 + 1(4)3
(2x + 4)3 = 8x3 + 3(4x2)(4) + 3(2x1)(16) + (64)
= 8x3 + 48x2 + 96x + 64
The expansion of (2x + 4)3 is 8x3 + 48x2 + 96x + 64.
The Binomial Theorem
Use Pascal’s Triangle to expand (-3x + 4)4.
First write the pattern for raising a binomial to the fourth power.
1
4
6
4
1
Coefficients from
Pascal’s Triangle.
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
Since (-3x + 4)4 = (x + (–3))4, substitute x for -3 and 4 for b.
= (-3x)4 + 4(-3x)3(4) + 6(-3x)2(4)2 + 4(-3x)(4)3 + (4)4
= 81x4 – 216x3 + 216x2 – 96x + 16
The expansion of (-3x + 4)4 is 81x4 – 216x3 + 216x2 – 96x + 16.