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A L G E B R A

19

THE DIFFERENCEOF TWO SQUARES

Geometrical algebra

2nd level:

The form (a + b)(ab)

WHEN THE SUM of two numbers multiplies their difference --

(a + b)(ab)

-- then the product is the difference of their squares:

(a + b)(ab) = a2b2

For, the like terms will cancel. (Lesson 16.)

Symmetrically, the difference of two squares can be factored:

x2 − 25 = (x + 5)(x − 5)

x2 is the square of x.  25 is the square of 5.

Example 1.   Multiply  (x3 + 2)(x3 − 2).

Solution.   Recognize the form:  (a + b)(ab).  The product will be the difference of two squares:

(x3 + 2)(x3 − 2) = x6 − 4.

x6 is the square of x3.  4 is the square of 2.

Upon seeing the form (a + b)(ab), the student should not do the FOIL method.  The student should recognize immediately that the product will be a2b2.

That is skill in algebra.

And the order of factors never matters:

(a + b)(ab) = (ab)(a + b) = a2b2.

Problem 1.   Write only final product..

 a) (x + 9)(x − 9) = x2 − 81 b) (y + z)(y − z) = y2 − z2 c) (6x − 1)(6x + 1) = 36x2 − 1 d) (3y + 7)(3y − 7) = 9y2 − 49 e) (x3 − 8)(x3 + 8) = x6 − 64 f) (xy + 10)(xy − 10) = x2y2 − 100 g) (xy2 − z3)(xy2 + z3) = x2y4 − z6 h) (xn + ym)(xn − ym) = x2n − y2m

Problem 2.   Factor.

 a) x2 − 100  = (x + 10)(x − 10) b) y2 − 1 = (y + 1)(y − 1) c) 1 − 4z2 = (1 + 2z)(1 − 2z) d) 25m2 − 9n2 = (5m + 3n)(5m − 3n) e) x6 − 36 = (x3 + 6)(x3 − 6) f) y4 − 144  = (y2 + 12)(y2 − 12) g) x8 − y10 = (x4 + y5)(x4 − y5) h) x2n − 1 = (xn + 1)(xn − 1)

Problem 3.   Factor completely.

 a)  x4 − y4 = (x2 + y2)(x2 − y2) = (x2 + y2)(x + y)(x − y)
 b)  1 − z8 = (1 + z4)(1 − z4) = (1 + z4)(1 + z2)(1 − z2) = (1 + z4)(1 + z2)(1 + z)(1 − z)

Problem 4.    Completely factor each of the following.  First remove a common factor.  Then factor the difference of two squares.

a)  xy2xz2  = x(y2z2) = x(y + z)(yz)

b)  8x2 − 72  = 8(x2 − 9) = 8(x + 3)(x − 3)

c)  64zz3  = z(64 − z2) = z(8 + z)(8 − z)

d)  rs3r3s  = rs(s2r2) = rs(s + r)(sr)

e)  32m2n − 50n3  = 2n(16m2 − 25n2) = 2n(4m + 5n)(4m − 5n)

f)  5x4y5 − 5y5  = 5y5(x4 − 1) = 5y5(x2 + 1)(x + 1)(x − 1)

Geometrical algebra

The entire figure on the left is a square on side a.  The square b2 has been inserted in the upper left corner, so that the shaded area is the difference of the two squares, a2b2.

Now, in the figure on the right, we have moved the rectangle (ab)b to the side.  The shaded area is now equal to the rectangle

(a + b)(ab).

That is,

a2b2 = (a + b)(ab).

*

The Difference of Two Squares completes our study of products of binomials.  Those products come up so often that the student should be able to recognize and apply each form.

Summary of Multiplying/Factoring

In summary, here are the four forms of Multiplying/Factoring that characterize algebra.

 1.  Common Factor 2(a + b) = 2a + 2b 2.  Quadratic Trinomial (x + 2)(x + 3) = x2 + 5x + 6 3.  Perfect Square Trinomial (x − 5)2 = x2 − 10x + 25 4.  The Difference of Two Squares (x + 5)(x − 5) = x2 − 25

Problem 5.   Distinguish each form, and write only the final product.

a)  (x − 3)2  = x2 − 6x + 9.   Perfect square trinomial.

b)  (x + 3)(x − 3)  = x2 − 9.   The difference of two squares.

c)  (x − 3)(x + 5)  = x2 + 2x − 15.   Quadratic trinomial.

d)  (2x − 5)(2x + 5)  = 4x2 − 25.   The difference of two squares.

e)  (2x − 5)2  = 4x2 − 20x + 25.   Perfect square trinomial.

f)  (2x − 5)(2x + 1)  = 4x2 − 8x − 5.    Quadratic trinomial.

Problem 6.   Factor.  (What form is it?  Is there a common factor?  Is it the difference of two squares? .  .  . )

a)  6x − 18  = 6(x − 3).   Common factor.

b)  x6 + x5 + x4 + x3  = x3(x3 + x2 + x + 1).   Common factor.

c)  x2 − 36  = (x + 6)(x − 6).   The difference of two squares.

d)  x2 − 12x + 36  = (x − 6)2.   Perfect square trinomial.

e)  x2 − 6x + 5  = (x − 5)(x − 1).   Quadratic trinomial.

f)  x2x − 12  = (x − 4)(x + 3)

g)  64x2 − 1  = (8x + 1)(8x − 1)

h)  5x2 − 7x − 6  = (5x + 3)(x − 2)

i)  4x5 + 20x4 + 24x3  = 4x3(x2 + 5x + 6) = 4x3(x + 3)(x + 2)

Next Lesson:  Algebraic fractions

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