Algebraic Expressions: Algebraic Expressions: Expanding and factorizingExpanding and factorizing

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To distribute or not to distribute

A key property you will use over and over again for therest of your mathematics education is something youlearned long ago:

4 ×

(sometimes in a galaxy far away)

(5 + 7) = ?·

You were taught to always do what is in brackets first...

(12) = 48

But let’s try it the “other” way...

4 (5 + 7)

·4

·5

+

4 7· = 20 + 28 = 48

This is the law.DISTRIBUTIVE=

(That is... multiplication distributes over addition.)

To distribute or not to distribute

Now let us apply this to the more general case of algebra:

4 (x + y) = ?·

Now we are unable to simplify what is in brackets further...

4 (x + y)

·4

·x

+

4 y·=

But the distributive law allows us to expand this expression:

= 4x + 4y That’s it!

Let’s keep going...

Distribute !

4x2y3 (2x + 3y2) = ?·4x2y3 (2x + 3y2)

·4x2y3

·2x+

4x2y3 3y2

·=

Again we expand this expression with the distributive law:

= (4x2y3)2x + (4x2y3)3y2

Let’s try a more complicated expression:

Q: Now what? A: Simplify each term using exponent laws

=

(4x2y3)2x

8x3y3

+ (4x2y3)3y2

12x2y5

A Return to Number

(5 + 7)How could we use the distributive law to expand:

(4 + 3)· (without simplifying first)?Well, let’s call (5 + 7) say... ☺

(4 + 3)(5 + 7)

☺ ·But we already know how to expand this:

=

☺☺

· ·4 3+(5 + 7) (5 + 7) But this is the same as...(5 + 7) 4 3(5 + 7)= +· · Now distribute again!

=

4·54·7 3·53·7

+ + +

= 20 + 28 + 15 + 21 = 84

= (12)·(7)

(commutative law: a·b = b·a)

Can you guess what’s next?

(x + y)Distributive law applied to products of binomials:

(z + w)· ?This time we’ll call (x + y) say... ☼

(z + w)(x + y)

☼ ·=

☼ ☼ z w(omitting ·’s henceforth...)+

=

(x + y) (x + y)

z w+

=

xzyz xw

Notice how all four combinations of variables arise...

yw

+ + +

(distributive law once more...)

(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:

(x + y)·(z + w) = xz + yz + xw + yw

(x + y)·(x + y) = xx + yx + xy + yy

(simplify!)

= x2 + 2xy + y2

2) Suppose we replace z by x, and w by – y:

(x + y)·(z + w) = xz + yz + xw + yw

(x + y)·(x – y) = xx + yx + x(–y) + y(–y)

= x2 + xy – xy – y2

= x2 – y2

(x + y)2 =

Generalize to a few new ones...

Our result: (x + y)·(z + w) = xz + yz + xw + yw

Maybeeee... x = 3ab , y = 2cd, z = 2ab, w = 4cd

Then we get:

(3ab + 2cd)·(2ab + 4cd) = = 3ab·2ab + 2cd·2ab + 3ab·4cd + 2cd·4cd

= 6a2b2 + 4abcd + 12abcd + 8c2d2

= 6a2b2 + 16abcd + 8c2d2

For future thought: how canwe go backwards?

OR what if w = a + b?

(x + y)·(z + a + b) = xz + yz + x(a+b) + y(a+b)

Apply the distributive law yet again to the rhs!binomial × trinomial

As you can see, starting with the very simple rule thatmultiplication distributes over addition – which you know fromarithmetic, you can build arbitrarily complex expressions by multiplying polynomials together.

And so on...

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 thatwe will encounter in greater detail later.

a) Factoring out a common monomial

-when every term in an algebraic expression has a commonnumerical factor, variable, or any product thereof, we can“pull out” those common elements:

4x + 4y 4 (x + y)

2xy + 4x2

44

2x 2x 2x (y + 2x)

3xyz2 (xy + 2yz + 4xz ) 3x2y2z2 + 6xy2z3 + 12x2yz33xyz2 3xyz2 3xyz2

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 amonomial, then reading the equation above from right toleft allows us to factor it immediately as follows...

a2x2 – b4y4

two termsdifference of

= (ax + b2y2)(ax – b2y2)

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 expressionsof the form:

ax2 + bx + c

The idea will be to first to solve the simpler problem offactoring:

x2 + bx + c

which in turn will require us to find integers d and e such that

d + e = b and d · e = c

since then: (x + d) (x + e) = (x + d) x + (x + d) e

= x2 + (d+e)x + d·e

x2 + bx + c

=by the distributive law!