Quadratic Algebra Lesson #1
Products And Expansions
• A product of two (or more) factors is the result obtained when
multiplying them together.
• Consider the factors -3x and 2x2. Their product -3x × 2x2 can be
simplified by following the steps below:
o Step 1: Find the product of the signs.
o Step 2: Find the product of the numerals (numbers).
o Step 3: Find the product of the pronumerals (letters).
• So,
• Example 1:
Simplify the following products:
(a.) −3 × 4x (b.) 2x × − x 2 (c.) −4x × −2x 2
(b.) 2x × − x 2
−3 × 4x
(c.) −4x × −2x 2
= −2x 3
= −12x
= 8x 3
• Other simplifications of two factors are possible using the
expansion (or distributive) rules:
o a(b + c) = ab + ac and a(b − c) = ab − ac
• Example 2:
Expand the following:
(a.) 3(4x + 1) (b.) 2x(5 − 2x) (c.) −2x(x − 3)
(a.)
(a.)
(c.)
3(4x + 1)
= 3 × 4x + 3 × 1
= 12x + 3
−2x(x − 3)
= ( −2x) × x − ( −2x) × 3
= −2x 2 + 6x
(b.)
2x(5 − 2x)
= 2x × 5 − 2x × 2x
= 10x − 4x 2
• Example 3:
Expand and simplify:
(a.) 2(3x − 1) + 3(5 − x) (b.) x(2x − 1) − 2x(5 − x)
(a.)
2(3x − 1) + 3(5 − x)
= 6x − 2 + 15 − 3x
= 3x + 13
(b.)
x(2x − 1) − 2x(5 − x)
= 2x 2 − x − 10x + 2x 2
= 4x 2 − 11x
The Product (a + b)(c + d)
• Consider the factors (a + b) and (c + d) .
• The product (a + b)(c + d) can be found using the distributive
law several times.
o
(a + b)(c + d) = a(c + d) + b(c + d)
= ac + ad + bc + bd
o i.e., (a + b)(c + d) = ac + ad + bc + bd
• Notice that the final result contains four terms:
o ac is the product of the First terms of each bracket.
o ad is the product of the Outer terms of each bracket.
o bc is the product of the Inner terms of each bracket.
o bd is the product of the Last terms of each bracket.
• This is sometimes called the FOIL rule.
• Example 1:
Expand and simplify: (x + 3)(x + 2)
(x + 3)(x + 2) = x × x + x × 2 + 3 × x + 3 × 2
= x 2 + 2x + 3x + 6
= x 2 + 5x + 6
• Example 2:
Expand and simplify: (2x + 1)(3x − 2)
(2x + 1)(3x − 2) = 2x × 3x − 2x × 2 + 1× 3x − 1× 2
= 6x 2 − 4x + 3x − 2
= 6x 2 − x − 2
• Example 3:
Expand and simplify:
(a.) (x + 3)(x − 3) (b.) (3 x − 5)(3x + 5)
(x + 3)(x − 3)
= x 2 − 3x + 3x − 9
= x2 − 9
• Example 4:
Expand and simplify:
(a.) (3x + 1)2 (b.) (2x − 3)2
(a.)
(a.)
(3x + 1)2
= (3x + 1)(3x + 1)
= 9x 2 + 3x + 3x + 1
= 9x 2 + 6x + 1
(b.)
(3 x − 5)(3x + 5)
= 9x 2 + 15x − 15x − 25
= 9x 2 − 25
(b.)
(2x − 3)2
= (2x − 3)(2x − 3)
= 4x 2 − 6x − 6x + 9
= 4x 2 − 12x + 9
Difference Of Two Squares
• a2 and b2 are perfect squares and so a2 – b2 is called the
difference of two squares.
o Notice that:
• Example 1:
Expand and simplify:
(a.) (x + 5)(x − 5) (b.) (3 − y)(3 + y)
(a.)
(x + 5)(x − 5)
(b.)
= x 2 − 52
= x 2 − 25
• Example 2:
Expand and simplify:
(a.) (2x − 3)(2x + 3) (b.) (5 − 3y)(5 + 3y)
(3 − y)(3 + y)
= 32 − y 2
= 9 − y2
(a.)
(2x − 3)(2x + 3)
(b.)
(5 − 3y)(5 + 3y)
= (2x) − 3
= 52 − (3y)2
= 4x 2 − 9
= 25 − 9y 2
• Example 3:
Expand and simplify: (3x + 4y)(3x − 4y)
2
2
(3x + 4y)(3x − 4y)
= (3x)2 − (4y)2
= 9x 2 − 16y 2
Perfect Squares Expansion
• (a + b)2 and (a − b)2 are called perfect squares.
o Notice that:
o Also,
• Therefore, we can state the perfect square expansion rules:
o (a + b)2 = a2 + 2ab + b2
o (a − b)2 = a2 − 2ab + b2
• The following is a useful way of remembering the perfect
square expansion rules:
o Step 1: Square the first term.
o Step 2: Add or subtract twice the product of the first and
last terms depending on the sign between the terms.
o Step 3: Add on the square of the last term.
• Example 1:
Expand and simplify:
(a.) (x + 3)2 (b.) (x − 5)2
(b.) (x − 5)2
(x + 3)2
= x 2 + 2 × x × 3 + 32
= x 2 − 2 × x × 5 + 52
= x 2 + 6x + 9
= x 2 − 10x + 25
• Example 2:
Expand and simplify using perfect square expansion rules:
(a.) (5x + 1)2 (b.) (4 − 3x)2
(a.)
(5x + 1)2
= (5x)2 + 2 × 5x × 1 + 12
= 25x 2 + 10x + 1
• Example 3:
Expand and simplify:
(a.) (2x 2 + 3)2 (b.) 5 − (x + 2)2
(a.)
(a.)
(2x 2 + 3)2
= (2x 2 )2 + 2 × 2x 2 × 3 + 32
= 4x 4 + 12x 2 + 9
(b.)
(4 − 3x)2
= 42 − 2 × 4 × 3x + (3x)2
= 16 − 24x + 9x 2
(b.)
5 − (x + 2)2
= 5 − [x 2 + 4x + 4]
= 5 − x 2 − 4x − 4
= 1 − x 2 − 4x
Further Expansion
• Consider the expansion of (a + b)(c + d + e) .
• Now: (a + b)(c + d + e)
Compare: p(c + d + e)
= (a + b)c + (a + b)d + (a + b)e
= pc + pd + pe
= ac + bc + ad + bd + ae + be
• Notice that there are 6 terms in this expansion and that each
term within the first bracket is multiplied by each term in the
second, i.e., 2 terms in first bracket multiplied by 3 terms in
second bracket gives 6 terms in expansion.
• Example 1:
Expand and simplify: (2x + 3)(x 2 + 4x + 5)
(2x + 3)(x 2 + 4x + 5)
= 2x 3 + 8x 2 + 10x
{all terms of 2nd bracket × 2x}
+ 3x 2 + 12x + 15
{ all terms of 2nd bracket × 3}
= 2x 3 + 11x 2 + 22x + 15
{collecting like terms}
• Example 2:
Expand and simplify: (x + 2)3
(x + 2)3
= (x + 2)2 × (x + 2)
= (x 2 + 4x + 4)(x + 2)
= x 3 + 4x 2 + 4x
{all terms in 1st bracket × x}
+ 2x 2 + 8x + 8
{ all terms in 1st bracket × 2}
= x3 + 6x 2 + 12x + 8
{collecting like terms}
• Example 3:
Expand and simplify:
(a.) x(x + 1)(x + 2) (b.) (x + 1)(x − 2)(x + 2)
(a.)
(b.)
x(x + 1)(x + 2)
= (x 2 + x)(x + 2)
{all terms in 1st bracket × x}
{expanding remaining factors}
= x 3 + 2x 2 + x 2 + 2x
3
2
= x + 3x + 2x
{collecting like terms}
(x + 1)(x − 2)(x + 2)
{expanding first two factors}
= (x 2 − 2x + x − 2)(x + 2)
2
{collecting like terms}
= (x − x − 2)(x + 2)
= x 3 + 2x 2 − x 2 − 2x − 2x − 4 {expanding remaining factors}
= x 3 + x 2 − 4x − 4
{collecting like terms}