Transcript pre-made lesson

Algebraic Expressions:
Expanding and factorizing
with 65% more flying algebra!
To distribute or not to distribute
A key property you will use over and over again for the
rest of your mathematics education is something you
learned long ago:
(sometimes in a galaxy far away)
44 ×· (5
(5
(12)
++ 7)
7) = 48?
You
to always
doway...
what is in brackets first...
Butwere
let’s taught
try it the
“other”
=
4 · 5
7
·
·
+
= 20 + 28 = 48
This is the DISTRIBUTIVE law.
(That is... multiplication distributes over addition.)
To distribute or not to distribute
Now let us apply this to the more general case of algebra:
44 · (x
(x ++ y)
y)
=
?
ButNow
the distributive
we are unable
law to
allows
simplify
us to
what
expand
is in this
brackets
expression:
further...
=
4 · x
y
·
·
+
= 4x + 4y
Let’s keep going...
That’s it!
Distribute !
Let’s try a more complicated expression:
4x2y3 · (2x
(2x ++ 3y
3y22)) =
?
Again we expand this expression with the distributive law:
=
2y3·
4x
2
4x y3 2x
·
+
Q: Now what?
=
8x3y3
3y2
·
22y33)2x
2
3
2
2
3
2
(4x
(4x
y
)3y
=
+
A: Simplify each term using exponent laws
12x2y5
A Return to Number
How could we use the distributive law to expand:
?
(5 +
· (4 + 3)
= 7)
(12)·(7)
Well, let’s call (5 + 7) say...
(without simplifying first)
☺
☺
☺
☺·
But we already know how to expand this:
(5 ++ 7)
7) · 4 + (5
= (5
(5 ++ 7)
7) · 3
= 4·5
·
·
4·7
+ 3·5
3·7
=
+
+
But this is the same as...
(commutative law: a·b = b·a)
Now distribute again!
+
= 20 + 28 + 15 + 21
= 84
Can you guess what’s next?
Distributive law applied to products of binomials:
(x + y) · (z + w)
?
This time we’ll call (x + y) say... ☼
☼☼☼ · z
w
= (x + y)
+ (x + y)
xz
xw w
yw
=
yz z +
=
+
+
(omitting ·’s henceforth...)
(distributive law once more...)
+
Notice how all four combinations of variables arise...
(How democratic! As it must be...)
Specialize to a few familiar cases...
Our result:
(x + y)·(z + w) = xz + yz + xw + yw
1) Suppose we replace z by x, and w by y:
(simplify!)
(x + y)2 = (x + y)·(x + y) = xx + yx + xy + yy
= x2 + 2xy + y2
2) Suppose we replace z by x, and w by – y:
(x + y)·(x – y) = xx + yx + x(–y) + y(–y)
= x2 + xy – xy – y2
= x2 – y 2
Generalize to a few new ones...
Our result:
(3ab
= 3ab·2ab
4cd)
2cd·2ab
+ 2cd·4cd
(x++2cd)·(2ab
y)·(z
++w)
=+=xz
+ yz ++3ab·4cd
xw + yw
Maybeeee... x = 3ab , y = 2cd, z = 2ab, w = 4cd
Then we get:
For future thought: how can
we go backwards?
= 6a2b2 + 4abcd + 12abcd + 8c2d2
= 6a2b2 + 16abcd + 8c2d2
OR what if w = a + b?
(x + y)·(z + a + b) = xz + yz + x(a+b) + y(a+b)
binomial × trinomial
Apply the distributive law yet again to the rhs!
And so on...
As you can see, starting with the very simple rule that
multiplication distributes over addition – which you know from
arithmetic, you can build arbitrarily complex expressions by
multiplying polynomials together.
What about the opposite process?
Factorization is to division what expansion is to multiplication
...as a poet might say, it is the memory of expansion
In general, it is MUCH harder... and was the subject of much
of the history of mathematics prior to the 20th century.
An Introduction to Factorization
Here, we will just look at some of the basic patterns that
we will encounter in greater detail later.
a) Factoring out a common monomial
-when every term in an algebraic expression has a common
numerical factor, variable, or any product thereof, we can
“pull out” those common elements:
4 +4
4x
4y
2xy + 2x
2x
4x2
23
2yz
3xyz
3xyz
3x2y22z2 + 3xyz
6xy2z2 3 + 12x
4 (x + y)
2x (y + 2x)
3xyz2 (xy + 2yz + 4xz )
An Introduction to Factorization - II
b) Difference of squares
From an earlier slide, we showed
(x + y)(x – y) = x2 – y2
Thus if an algebraic expression consists of the difference
of two terms, and each of those terms is the square of a
monomial, then reading the equation above from right to
left allows us to factor it immediately as follows...
a2x2 – b4y4
= (ax + b2y2)(ax – b2y2)
two terms
difference of
each is a square of a monomial: ax or b2y2
An Introduction to Factorization - III
b) Quadratic expressions
All of next class will be devoted to factorizing expressions
of the form:
ax2 + bx + c
The idea will be to first to solve the simpler problem of
factoring:
xx22++ bx
bx ++ cc
which in turn will require us to find integers d and e such that
d+e=b
since then:
and
d·e=c
(x + d) (x + e) = (x + d) x + (x + d) e
by the distributive law!
= x2 + (d+e)x + d·e
=