1. Algebraic Expressions: Expanding and factorizing with 65% more flying algebra!

2. 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: 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)(5 + 7) · 4 · 5 + 4 7 · = 20 + 28= 48 This is the law. DISTRIBUTIVE = (That is... multiplication distributes over addition.)

3. To distribute or not to distribute Now let us apply this to the more general case of algebra: 4(x + y)(x + y) =?=? · Now we are unable to simplify what is in brackets further... 4(x + y)(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...

4. Distribute ! 4x2y34x2y3 (2x + 3y 2 ) =?=? ·4x2y34x2y3 · 4x2y34x2y3 · 2x + 4x2y34x2y3 3y 2 ·= Again we expand this expression with the distributive law: = (4x 2 y 3 )2x + (4x 2 y 3 )3y 2 Let’s try a more complicated expression: Q: Now what? A: Simplify each term using exponent laws = (4x2y3)2x(4x2y3)2x 8x3y38x3y3 + (4x2y3)3y2(4x2y3)3y2 12x 2 y 5

5. A Return to Number (5 + 7)(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)(5 + 7) ☺ · But we already know how to expand this: = ☺☺ ··43+ (5 + 7) But this is the same as... (5 + 7)4 3 = + · · Now distribute again! = 4·54·54·74·73·53·53·73·7 + + + = 20 + 28 + 15 + 21 = 84 = (12)·(7) (commutative law: a·b = b·a)

6. Can you guess what’s next? (x + y)(x + y) Distributive law applied to products of binomials: (z + w) · ? This time we’ll call (x + y) say... ☼ (z + w) (x + y)(x + y) ☼· = ☼☼ zw (omitting ·’s henceforth...) + = (x + y) zw + = xzxz yzyz xwxw Notice how all four combinations of variables arise... ywyw + + + (distributive law once more...) (How democratic! As it must be...)

7. 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!) = x 2 + 2xy + y 2 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) = x 2 + xy – xy – y 2 = x 2 – y 2 (x + y) 2 =

8. 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 = 6a 2 b 2 + 4abcd + 12abcd + 8c 2 d 2 = 6a 2 b 2 + 16abcd + 8c 2 d 2 For future thought: how can we 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

9. 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. 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 20 th century.

10. 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: 4x + 4y4 (x + y) 2xy + 4x 2 44 2x 2x (y + 2x) 3xyz 2 (xy + 2yz + 4xz ) 3x 2 y 2 z 2 + 6xy 2 z 3 + 12x 2 yz 3 3xyz 2

11. An Introduction to Factorization - II b) Difference of squares From an earlier slide, we showed (x + y)(x – y) = x 2 – y 2 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... a 2 x 2 – b 4 y 4 two terms difference of = (ax + b 2 y 2 )(ax – b 2 y 2 ) each is a square of a monomial: ax or b 2 y 2

12. An Introduction to Factorization - III b) Quadratic expressions All of next class will be devoted to factorizing expressions of the form: ax 2 + bx + c The idea will be to first to solve the simpler problem of factoring: x 2 + bx + c which in turn will require us to find integers d and e such that d + e = bandd · e = c since then:(x + d) (x + e) = (x + d) x + (x + d) e = x 2 + (d+e)x + d·e x 2 + bx + c = by the distributive law!