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U1A L1 Examples FACTORING REVIEW EXAMPLES.

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1 U1A L1 Examples FACTORING REVIEW EXAMPLES

2 Factor x2 + 3x – 4 Solve x2 + 3x – 4 = 0 What _____Γ— _____ = – 4 and _____+ _____ = 3 πŸ’ βˆ’πŸ 𝒙+πŸ’ π’™βˆ’πŸ =𝟎 𝒙=βˆ’πŸ’, 𝒙=𝟏 πŸ’ βˆ’πŸ (𝒙+πŸ’)(π’™βˆ’πŸ) Graph Y1 = x2 + 3x – 4 Find x-intercepts (βˆ’πŸ’,0) (𝟏,0)

3 Factor x2 + 3x + 2 What _____Γ— _____ = 2 and _____+ _____ = 3 Solve π‘₯2 + 3π‘₯ + 2=0 𝟐 𝟏 𝒙+𝟐 𝒙+𝟏 =𝟎 𝟐 𝟏 𝒙=βˆ’πŸ, 𝒙=βˆ’πŸ (𝒙+𝟐)(𝒙+𝟏) Factor 2x2 + 6x + 4 by taking out common factor 2 Solve 2π‘₯2 +6π‘₯ +4 = 0 2(x2 + 3x + 2) = 0 2(x2 + 3x + 2) 𝟐 𝒙+𝟐 𝒙+𝟏 =𝟎 𝟐(𝒙+𝟐)(𝒙+𝟏) 𝒙=βˆ’πŸ, 𝒙=βˆ’πŸ Factor –3x2 – 9x – 6 by taking out common factor – 3 Solve βˆ’3π‘₯2 βˆ’9π‘₯ βˆ’6= 0 – 3(x2 + 3x + 2) = 0 – 3(x2 + 3x + 2) βˆ’πŸ‘ 𝒙+𝟐 𝒙+𝟏 =𝟎 βˆ’πŸ‘(𝒙+𝟐)(𝒙+𝟏) 𝒙=βˆ’πŸ, 𝒙=βˆ’πŸ

4 Graph Y1 = x2 + 3x + 2 Y2 = 2x2 + 6x + 4or Y2 = 2(x2 + 3x + 2) Y3 = –3x2 – 9x – 6 orY3 = –3(x2 + 3x + 2) Find x-intercepts (βˆ’πŸ,0) (βˆ’πŸ,0)

5 Factor –x2 – 6x – 8 by taking out common factor –1
βˆ’(𝒙+πŸ’)(𝒙+𝟐) Graph Y1 = –x2 – 6x – 8 Find x-intercepts Solve –x2 – 6x – 8 = 0 – (x2 + 6x + 8) = 0 βˆ’ 𝒙+πŸ’ 𝒙+𝟐 =𝟎 (βˆ’πŸ’,0) (βˆ’πŸ,0) 𝒙=βˆ’πŸ’, 𝒙=βˆ’πŸ

6 Can you factor x2 + 4 ? Can you solve x2 + 4 = 0 NO 𝒙 𝟐 =βˆ’πŸ’ 𝒙= βˆ’πŸ’ Non-Real Answer Graph Y1 = x2 – 4 Find x-intercepts There are NO x - intercepts

7 Can you factor π‘₯2 βˆ’ 4 ? Can you solve π‘₯2 βˆ’ 4 = 0 (π’™βˆ’πŸ)(𝒙+𝟐) π’™βˆ’πŸ 𝒙+𝟐 =𝟎 Difference of Squares 𝒙=𝟐, 𝒙=βˆ’πŸ OR Graph Y1 = x2 – 4 𝒙 𝟐 =πŸ’ Find x-intercepts 𝒙=±𝟐 Using this method it VERY easy to forget BOTH answers!!!! (βˆ’πŸ,0) (𝟐,0)

8 by taking out common factor 2
Factor 8x2 – 18 by taking out common factor 2 ( πŸ‘ 𝟐 ,0) 𝟐 (πŸ’π’™ 𝟐 βˆ’πŸ—) (βˆ’ πŸ‘ 𝟐 ,0) 𝟐(πŸπ’™βˆ’πŸ‘)(πŸπ’™+πŸ‘) Solve 8x2 – 18 = 0 𝟐 πŸπ’™βˆ’πŸ‘ πŸπ’™+πŸ‘ =𝟎 πŸπ’™βˆ’πŸ‘=𝟎 πŸπ’™+πŸ‘=𝟎 πŸπ’™=πŸ‘ πŸπ’™=βˆ’πŸ‘ 𝒙= πŸ‘ 𝟐 𝒙=βˆ’ πŸ‘ 𝟐 Solve 8x2 – 18 = 0 Common factor 2 is positive. Graph opens up. 8 𝒙 𝟐 =πŸπŸ– 𝒙 𝟐 = πŸπŸ– πŸ– = πŸ— πŸ’ 𝒙=Β± πŸ‘ 𝟐

9 by taking out common factor – 1
(πŸβˆ’πŸ‘π’™)(𝟐+πŸ‘π’™) βˆ’πŸ (πŸ—π’™ 𝟐 βˆ’πŸ’) Solve 4βˆ’9π‘₯2 = 0 βˆ’πŸ(πŸ‘π’™βˆ’πŸ)(πŸ‘π’™+𝟐) πŸβˆ’πŸ‘π’™ 𝟐+πŸ‘π’™ =𝟎 πŸβˆ’πŸ‘π’™=𝟎 𝟐+πŸ‘π’™=𝟎 𝟐=πŸ‘π’™ 𝟐=βˆ’πŸ‘π’™ 𝒙= 𝟐 πŸ‘ 𝒙=βˆ’ 𝟐 πŸ‘ (βˆ’ 𝟐 πŸ‘ ,0) ( 𝟐 πŸ‘ ,0) Solve 4βˆ’9π‘₯2 = 0 βˆ’πŸ— 𝒙 𝟐 =βˆ’πŸ’ 𝒙 𝟐 = βˆ’πŸ’ βˆ’πŸ— = πŸ’ πŸ— 𝒙=Β± 𝟐 πŸ‘ Common factor – 1 is negative. Graph opens down.

10 MORE COMMON FACTORING EXAMPLES
When dividing out common factors look for the greatest common numerical factor and the smallest exponent on the variables. 2 π‘₯ 5 βˆ’8 π‘₯ 4 +6 π‘₯ 3 2 π‘₯ 3 (π‘₯ 2 βˆ’4 π‘₯ 1 +3 π‘₯ 𝟎 ) 2 π‘₯ 5 βˆ’8 π‘₯ 4 +6 𝒙 πŸ‘ 2 π‘₯ 3 (π‘₯ 2 βˆ’4π‘₯+3) 2 𝒙 πŸ‘ (π‘₯ 5βˆ’3 βˆ’4 π‘₯ 4βˆ’3 +3 π‘₯ πŸ‘βˆ’πŸ‘ ) 2 π‘₯ 3 (π‘₯βˆ’3)(π‘₯βˆ’1)

11 MORE COMMON FACTORING EXAMPLES
When dividing out common factors look for the greatest common numerical factor and the smallest exponent on the variables. βˆ’3 π‘₯ βˆ’1 βˆ’18 π‘₯ βˆ’2 +6 π‘₯ βˆ’3 βˆ’3 π‘₯ βˆ’1 βˆ’18 π‘₯ βˆ’2 +6 𝒙 βˆ’πŸ‘ βˆ’3 𝒙 βˆ’πŸ‘ (π‘₯ βˆ’1βˆ’ βˆ’3 +6 π‘₯ βˆ’2βˆ’ βˆ’3 βˆ’2 π‘₯ βˆ’πŸ‘βˆ’(βˆ’πŸ‘) ) βˆ’3 𝒙 βˆ’πŸ‘ (π‘₯ 𝟐 +6 π‘₯ 𝟏 βˆ’2 π‘₯ 𝟎 ) βˆ’3 𝒙 βˆ’πŸ‘ (π‘₯ 2 +6π‘₯βˆ’2) This example will NOT factor further.

12 MORE COMMON FACTORING EXAMPLES
When dividing out common factors look for the greatest common numerical factor and the smallest exponent on the variables. 2 𝒙 βˆ’πŸ (π‘₯ 2 βˆ’4 π‘₯ 1 +3 π‘₯ 𝟎 ) 2π‘₯βˆ’8+6 π‘₯ βˆ’1 2 π‘₯ βˆ’1 (π‘₯ 2 βˆ’4π‘₯+3) 2 π‘₯ 1 βˆ’8 𝒙 𝟎 +6 𝒙 βˆ’πŸ 2 π‘₯ βˆ’1 (π‘₯βˆ’3)(π‘₯βˆ’1) 2 𝒙 βˆ’πŸ (π‘₯ 1βˆ’(βˆ’1) βˆ’4 π‘₯ 0βˆ’(βˆ’1) +3 π‘₯ βˆ’πŸβˆ’(βˆ’πŸ) )

13 When subtracting rational exponents use a common denominator.
12 π‘₯ βˆ’24 π‘₯ βˆ’36 π‘₯ 1 3 12 𝒙 𝟏 πŸ‘ ( π‘₯ 𝟐 βˆ’2 π‘₯ 𝟏 βˆ’3 π‘₯ 𝟎 ) 12 π‘₯ βˆ’24 π‘₯ βˆ’36 𝒙 𝟏 πŸ‘ 12 𝒙 𝟏 πŸ‘ ( π‘₯ 𝟐 βˆ’2π‘₯βˆ’3) 12 𝒙 𝟏 πŸ‘ ( π‘₯ πŸ• πŸ‘ βˆ’ 𝟏 πŸ‘ βˆ’2 π‘₯ πŸ’ πŸ‘ βˆ’ 𝟏 πŸ‘ βˆ’3 π‘₯ 𝟏 πŸ‘ βˆ’ 𝟏 πŸ‘ ) 12 π‘₯ (π‘₯βˆ’3)(π‘₯+1) 12 𝒙 𝟏 πŸ‘ ( π‘₯ πŸ” πŸ‘ βˆ’2 π‘₯ πŸ‘ πŸ‘ βˆ’3 π‘₯ 𝟎 )

14 When subtracting rational exponents use a common denominator.
3 𝒙 βˆ’ 𝟏 𝟐 ( π‘₯ πŸ’ 𝟐 βˆ’9 π‘₯ 𝟎 ) 3 π‘₯ βˆ’27 π‘₯ βˆ’ 1 2 3 π‘₯ βˆ’27 𝒙 βˆ’ 𝟏 𝟐 3 π‘₯ βˆ’ 1 2 ( π‘₯ 2 βˆ’9) 3 π‘₯ βˆ’ 1 2 (π‘₯βˆ’3)(π‘₯+3) 3 𝒙 βˆ’ 𝟏 𝟐 ( π‘₯ πŸ‘ 𝟐 βˆ’ βˆ’πŸ 𝟐 βˆ’9 π‘₯ βˆ’ 𝟏 𝟐 βˆ’ βˆ’πŸ 𝟐 )

15 4(x – 5)4 – 6(x – 5)3 2(x – 5)3 [2(x – 5) – 3] 4(x – 5)4 – 6(x – 5)3 2(x – 5)3 [2x – 10 – 3] 2(x – 5)3 [2(x – 5)4-3 – 3(x – 5)3-3] 2(x – 5)3 (2x – 13) 2(x – 5)3 [2(x – 5)1 – 3(x – 5)0]

16 6 π‘₯βˆ’2 βˆ’5 βˆ’24 π‘₯βˆ’2 βˆ’6 6 π’™βˆ’πŸ βˆ’5 βˆ’24 π’™βˆ’πŸ βˆ’6 6 π’™βˆ’πŸ βˆ’6 [ π‘₯βˆ’2 βˆ’5βˆ’ βˆ’6 βˆ’4 π‘₯βˆ’2 βˆ’6βˆ’ βˆ’6 ] 6 π’™βˆ’πŸ βˆ’6 [ π‘₯βˆ’2 1 βˆ’4 π‘₯βˆ’2 0 ] 6 π’™βˆ’πŸ βˆ’6 [π‘₯βˆ’2βˆ’4] 6 π‘₯βˆ’2 βˆ’6 (π‘₯βˆ’6)

17 8 π‘₯+6 βˆ’ 1 2 βˆ’6 π‘₯+6 βˆ’ 3 2 8 𝒙+πŸ” βˆ’ 1 2 βˆ’6 𝒙+πŸ” βˆ’ 3 2 2 𝒙+πŸ” βˆ’ 3 2 [4 𝒙+πŸ” βˆ’ 1 2 βˆ’ βˆ’3 2 βˆ’3 𝒙+πŸ” βˆ’ 3 2 βˆ’ βˆ’3 2 ] 2 𝒙+πŸ” βˆ’ 3 2 [4 𝒙+πŸ” βˆ’3 𝒙+πŸ” 0 ] 2 𝒙+πŸ” βˆ’ 3 2 [4 𝒙+πŸ” 1 βˆ’3 𝒙+πŸ” 0 ] 2 𝒙+πŸ” βˆ’ 3 2 [4π‘₯+24βˆ’3] 2 π‘₯+6 βˆ’ 3 2 [4π‘₯+21]

18 Factor by Decomposition Example
6x2 – 11x + 3 6 π‘₯ 2 βˆ’11π‘₯+3 6 π‘₯ 2 +2π‘₯βˆ’9π‘₯+3 πŸ”Γ—πŸ‘=18 2Γ—βˆ’9=18 2+(βˆ’9)=βˆ’7 2π‘₯(3π‘₯+1)βˆ’3(3π‘₯+1) (2π‘₯βˆ’3)(3π‘₯+1)

19 Quadratic Formula ax2 + bx + c = 0
𝒙= βˆ’π’ƒΒ± 𝒃 𝟐 βˆ’πŸ’π’‚π’„ πŸπ’‚

20 Solve for x 3x2 – 2x – 4 = 0 π‘₯= βˆ’(βˆ’2)Β± (βˆ’2) 2 βˆ’4(3)(βˆ’4) 2(3)
π‘₯= βˆ’(βˆ’2)Β± (βˆ’2) 2 βˆ’4(3)(βˆ’4) 2(3) π‘₯= 2Β± π‘₯= 2Β± 4Γ—13 6 π‘₯= 2Β± π‘₯= 2Β± π‘₯= 1Β± Answers in simplest and exact radical form Approximate decimal answers to nearest hundredth. π‘₯=βˆ’0.95 π‘Žπ‘›π‘‘ π‘₯=1.54

21 π‘₯= βˆ’(βˆ’3)Β± (βˆ’3) 2 βˆ’4(5)(10) 2(5) π‘₯= 3Β± 9βˆ’200 10 π‘₯= 3Β± βˆ’191 6
Β Solve for x x2 – 3x + 10 = 0 π‘₯= βˆ’(βˆ’3)Β± (βˆ’3) 2 βˆ’4(5)(10) 2(5) π‘₯= 3Β± 9βˆ’ π‘₯= 3Β± βˆ’191 6 Non-real answer.

22 SYNTHETIC DIVISION Divide x3 + 4x2 – 5x – 12 by x – 3
Method I: SUBTRACTION – – –12 – 3 –21 –48 Quotient is x2 + 7x + 16 Remainder is 36 NOTE: x3 + 4x2 – 5x 12 (3)3 + 4(3)2 – 5(3) – 12 = 36

23 SYNTHETIC DIVISION Divide x3 + 4x2 – 5x – 12 by x – 3
Method II: ADDITION – –12 Quotient is x2 + 7x + 16 Remainder is 36 NOTE: x3 + 4x2 – 5x 12 (3)3 + 4(3)2 – 5(3) – 12 = 36

24 Divide x3 + 3x2 – 5 by x + 2 SYNTHETIC DIVISION π’™πŸ‘ + πŸ‘π’™πŸ + πŸŽπ’™ βˆ’πŸ“
Method I: SUBTRACTION –5 –4 – –1 Quotient is x2 + x – 2 Remainder is –1 NOTE: x3 + 3x2 – 5 (–2)3 + 3(–2)2 – 5 = –1

25 Divide x3 + 3x2 – 5 by x + 2 SYNTHETIC DIVISION Method II: ADDITION
–5 –2 – – –1 Quotient is x2 + x – 2 Remainder is –1 NOTE: x3 + 3x2 – 5 (–2)3 + 3(–2)2 – 5 = –1

26 Divide x3 – 8 by x – 2 SYNTHETIC DIVISION π’™πŸ‘ +𝟎 𝒙 𝟐 +πŸŽπ’™βˆ’πŸ–
Method I: SUBTRACTION – –8 –2 –4 –8 Quotient is x2 + 2x + 4 Remainder is 0 NOTE: x3– 8 (2)3 – 8 = 0

27 Difference of Cubes Formula
a3 – b3 = (a – b)(a2 + ab + b2) Factor x3 – 8 π‘Ž=π‘₯ 𝑏=2 If we compare this answer to the previous slide we see it is the same. This is a shortcut that will help with more difficult questions. (π‘₯βˆ’2)( π‘₯ 2 +2π‘₯+ 2 2 ) (π‘₯βˆ’2)( π‘₯ 2 +2π‘₯+4) Factor 27x3 – 64 π‘Ž=3π‘₯ 𝑏=4 3π‘₯βˆ’4 [ 3π‘₯ 2 +(3π‘₯)(4)+ 4 2 ] 3π‘₯βˆ’4 (9 π‘₯ 2 +12π‘₯+16)

28 Divide x3 + 27 by x + 3 SYNTHETIC DIVISION Method I: SUBTRACTION
3 – 1 – Quotient is x2 – 3x + 9 Remainder is 0 NOTE: x3+ 27 (–3) = 0

29 Sum of Cubes Formula a3 + b3 = (a + b)(a2 – ab + b2)
Factor x3 + 27 π‘Ž=π‘₯ 𝑏=3 If we compare this answer to the previous slide we see it is the same. This is a shortcut that will help with more difficult questions. (π‘₯+3 )( π‘₯ 2 βˆ’3π‘₯+ 3 2 ) (π‘₯+3)( π‘₯ 2 βˆ’3π‘₯+9) Factor π‘₯3+125 𝑦 3 Factor 27π‘₯3βˆ’64 π‘Ž=π‘₯ 𝑏=5𝑦 π‘Ž=3π‘₯ 𝑏=4 3π‘₯βˆ’4 [ 3π‘₯ 2 +(3π‘₯)(4)+ (4) 2 ] π‘₯+5𝑦 [ π‘₯ 2 βˆ’(π‘₯)(5𝑦)+ (5𝑦) 2 ] 3π‘₯βˆ’4 (9 π‘₯ 2 +12π‘₯+16) π‘₯+5𝑦 ( π‘₯ 2 βˆ’5π‘₯𝑦+25 𝑦 2 )


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