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A comprehensive guide on multiplying polynomials, focusing on monomials, binomials, and special products such as the difference of two squares. It includes various examples and formulas to help understand the concepts.
Typology: Slides
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From ยง1.6 ๏ Exponent Properties
1 as an exponent a^1 = a
0 as an exponent a^0 = 1
Negative exponents
The Product Rule
The Quotient Rule
The Power Rule ( a m ) n^ = a mn Raising a product to a power
( ab ) n^ = a n^ b n
Raising a quotient to a power.
n (^) n n
a a b (^) b
๏ฃซ๏ฃฌ ๏ฃถ๏ฃท (^) = ๏ฃญ ๏ฃธ
.
m (^) m n n a (^) a a =^ โ
a m^ โ a n^ = a m^ + n.
n^1 , n^ m ,^ n^ n n m n a a^ b^ a^ b a b a b^ a
โ โ^ โ = (^) โ = ๏ฃซ๏ฃฌ๏ฃญ^ ๏ฃถ๏ฃท๏ฃธ^ =๏ฃซ๏ฃฌ๏ฃญ^ ๏ฃถ๏ฃท๏ฃธ
denominators are 0 and that 0 This summary assumes that no
is not^0
considered. For any integers
(^) m (^) and (^) n
c) (โ 8 x^6 )(3 x^4 )
๏ง Solution a) (6 x )(7 x ) = (6 โ 7) ( x โ x ) = 42 x^2
๏ง Solution b) (5 a )(โ a ) = (5 a )(โ 1 a )
= (5)(โ1)( a โ a ) = โ 5 a^2
๏ง Solution c) (โ 8 x^6 )(3 x^4 ) = (โ 8 โ 3) ( x^6 โ x^4 )
= โ 24 x 6 + 4^ = โ 24 x^10
a) x ( x + 7) = x โ x + x โ 7 = x^2 + 7 x
b) 6 x ( x^2 โ 4 x + 5) = (6 x )( x^2 ) โ (6 x )(4 x ) + (6 x )(5)
= 6 x^3 โ 24 x^2 + 30 x
5 x^2 ( x^3 โ 4 x^2 + 3 x โ 5) =
= 5 x^5 โ 20 x^4 + 15 x^3 โ 25 x^2
๏ง Multiply: (โ 3 x^2 โ 4)(2 x^2 โ 3 x + 1)
๏ง Solution
2 x^2 โ 3 x + 1 โ 3 x^2 โ 4 โ 8 x^2 + 12 x โ 4 โ 6 x^4 + 9 x^3 โ 3 x^2 โ 6 x^4 + 9 x^3 โ 11 x^2 + 12 x โ 4
F O I L ( x + 4)( x^2 + 3) = x^3 + 3 x + 4 x^2 + 12
O
I
F L
= x^3 + 4 x^2 + 3 x + 12
๏ง The terms are rearranged in descending order for the final answer
FOIL applies to ANY set of TWO BiNomials, Regardless of the BiNomial Degree
(x + 4)(x โ 4) = x^2 โ 4x + 4x โ 16 = x^2 โ 16
(x + 3)(x โ 3) = x^2 โ 3x + 3x โ 9 = x^2 โ 9
(5 โ y)(5 + y) = 25 +5y โ 5y โ y^2 = 25 โ y^2
What do all of these have in common?
๏ง ALL the Results are Difference of 2-Sqs: