# U1A L1 Examples FACTORING REVIEW EXAMPLES.

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U1A L1 Examples FACTORING REVIEW EXAMPLES

Factor x2 + 3x โ 4 Solve x2 + 3x โ 4 = 0 What _____ร _____ = โ 4 and _____+ _____ = 3 ๐ โ๐ ๐+๐ ๐โ๐ =๐ ๐=โ๐, ๐=๐ ๐ โ๐ (๐+๐)(๐โ๐) Graph Y1 = x2 + 3x โ 4 Find x-intercepts (โ๐,0) (๐,0)

Factor x2 + 3x + 2 What _____ร _____ = 2 and _____+ _____ = 3 Solve ๐ฅ2 + 3๐ฅ + 2=0 ๐ ๐ ๐+๐ ๐+๐ =๐ ๐ ๐ ๐=โ๐, ๐=โ๐ (๐+๐)(๐+๐) Factor 2x2 + 6x + 4 by taking out common factor 2 Solve 2๐ฅ2 +6๐ฅ +4 = 0 2(x2 + 3x + 2) = 0 2(x2 + 3x + 2) ๐ ๐+๐ ๐+๐ =๐ ๐(๐+๐)(๐+๐) ๐=โ๐, ๐=โ๐ Factor โ3x2 โ 9x โ 6 by taking out common factor โ 3 Solve โ3๐ฅ2 โ9๐ฅ โ6= 0 โ 3(x2 + 3x + 2) = 0 โ 3(x2 + 3x + 2) โ๐ ๐+๐ ๐+๐ =๐ โ๐(๐+๐)(๐+๐) ๐=โ๐, ๐=โ๐

Graph Y1 = x2 + 3x + 2 Y2 = 2x2 + 6x + 4or Y2 = 2(x2 + 3x + 2) Y3 = โ3x2 โ 9x โ 6 orY3 = โ3(x2 + 3x + 2) Find x-intercepts (โ๐,0) (โ๐,0)

Factor โx2 โ 6x โ 8 by taking out common factor โ1
โ(๐+๐)(๐+๐) Graph Y1 = โx2 โ 6x โ 8 Find x-intercepts Solve โx2 โ 6x โ 8 = 0 โ (x2 + 6x + 8) = 0 โ ๐+๐ ๐+๐ =๐ (โ๐,0) (โ๐,0) ๐=โ๐, ๐=โ๐

Can you factor x2 + 4 ? Can you solve x2 + 4 = 0 NO ๐ ๐ =โ๐ ๐= โ๐ Non-Real Answer Graph Y1 = x2 โ 4 Find x-intercepts There are NO x - intercepts

Can you factor ๐ฅ2 โ 4 ? Can you solve ๐ฅ2 โ 4 = 0 (๐โ๐)(๐+๐) ๐โ๐ ๐+๐ =๐ Difference of Squares ๐=๐, ๐=โ๐ OR Graph Y1 = x2 โ 4 ๐ ๐ =๐ Find x-intercepts ๐=ยฑ๐ Using this method it VERY easy to forget BOTH answers!!!! (โ๐,0) (๐,0)

by taking out common factor 2
Factor 8x2 โ 18 by taking out common factor 2 ( ๐ ๐ ,0) ๐ (๐๐ ๐ โ๐) (โ ๐ ๐ ,0) ๐(๐๐โ๐)(๐๐+๐) Solve 8x2 โ 18 = 0 ๐ ๐๐โ๐ ๐๐+๐ =๐ ๐๐โ๐=๐ ๐๐+๐=๐ ๐๐=๐ ๐๐=โ๐ ๐= ๐ ๐ ๐=โ ๐ ๐ Solve 8x2 โ 18 = 0 Common factor 2 is positive. Graph opens up. 8 ๐ ๐ =๐๐ ๐ ๐ = ๐๐ ๐ = ๐ ๐ ๐=ยฑ ๐ ๐

by taking out common factor โ 1
(๐โ๐๐)(๐+๐๐) โ๐ (๐๐ ๐ โ๐) Solve 4โ9๐ฅ2 = 0 โ๐(๐๐โ๐)(๐๐+๐) ๐โ๐๐ ๐+๐๐ =๐ ๐โ๐๐=๐ ๐+๐๐=๐ ๐=๐๐ ๐=โ๐๐ ๐= ๐ ๐ ๐=โ ๐ ๐ (โ ๐ ๐ ,0) ( ๐ ๐ ,0) Solve 4โ9๐ฅ2 = 0 โ๐ ๐ ๐ =โ๐ ๐ ๐ = โ๐ โ๐ = ๐ ๐ ๐=ยฑ ๐ ๐ Common factor โ 1 is negative. Graph opens down.

MORE COMMON FACTORING EXAMPLES
When dividing out common factors look for the greatest common numerical factor and the smallest exponent on the variables. 2 ๐ฅ 5 โ8 ๐ฅ 4 +6 ๐ฅ 3 2 ๐ฅ 3 (๐ฅ 2 โ4 ๐ฅ 1 +3 ๐ฅ ๐ ) 2 ๐ฅ 5 โ8 ๐ฅ 4 +6 ๐ ๐ 2 ๐ฅ 3 (๐ฅ 2 โ4๐ฅ+3) 2 ๐ ๐ (๐ฅ 5โ3 โ4 ๐ฅ 4โ3 +3 ๐ฅ ๐โ๐ ) 2 ๐ฅ 3 (๐ฅโ3)(๐ฅโ1)

MORE COMMON FACTORING EXAMPLES
When dividing out common factors look for the greatest common numerical factor and the smallest exponent on the variables. โ3 ๐ฅ โ1 โ18 ๐ฅ โ2 +6 ๐ฅ โ3 โ3 ๐ฅ โ1 โ18 ๐ฅ โ2 +6 ๐ โ๐ โ3 ๐ โ๐ (๐ฅ โ1โ โ3 +6 ๐ฅ โ2โ โ3 โ2 ๐ฅ โ๐โ(โ๐) ) โ3 ๐ โ๐ (๐ฅ ๐ +6 ๐ฅ ๐ โ2 ๐ฅ ๐ ) โ3 ๐ โ๐ (๐ฅ 2 +6๐ฅโ2) This example will NOT factor further.

MORE COMMON FACTORING EXAMPLES
When dividing out common factors look for the greatest common numerical factor and the smallest exponent on the variables. 2 ๐ โ๐ (๐ฅ 2 โ4 ๐ฅ 1 +3 ๐ฅ ๐ ) 2๐ฅโ8+6 ๐ฅ โ1 2 ๐ฅ โ1 (๐ฅ 2 โ4๐ฅ+3) 2 ๐ฅ 1 โ8 ๐ ๐ +6 ๐ โ๐ 2 ๐ฅ โ1 (๐ฅโ3)(๐ฅโ1) 2 ๐ โ๐ (๐ฅ 1โ(โ1) โ4 ๐ฅ 0โ(โ1) +3 ๐ฅ โ๐โ(โ๐) )

When subtracting rational exponents use a common denominator.
12 ๐ฅ โ24 ๐ฅ โ36 ๐ฅ 1 3 12 ๐ ๐ ๐ ( ๐ฅ ๐ โ2 ๐ฅ ๐ โ3 ๐ฅ ๐ ) 12 ๐ฅ โ24 ๐ฅ โ36 ๐ ๐ ๐ 12 ๐ ๐ ๐ ( ๐ฅ ๐ โ2๐ฅโ3) 12 ๐ ๐ ๐ ( ๐ฅ ๐ ๐ โ ๐ ๐ โ2 ๐ฅ ๐ ๐ โ ๐ ๐ โ3 ๐ฅ ๐ ๐ โ ๐ ๐ ) 12 ๐ฅ (๐ฅโ3)(๐ฅ+1) 12 ๐ ๐ ๐ ( ๐ฅ ๐ ๐ โ2 ๐ฅ ๐ ๐ โ3 ๐ฅ ๐ )

When subtracting rational exponents use a common denominator.
3 ๐ โ ๐ ๐ ( ๐ฅ ๐ ๐ โ9 ๐ฅ ๐ ) 3 ๐ฅ โ27 ๐ฅ โ 1 2 3 ๐ฅ โ27 ๐ โ ๐ ๐ 3 ๐ฅ โ 1 2 ( ๐ฅ 2 โ9) 3 ๐ฅ โ 1 2 (๐ฅโ3)(๐ฅ+3) 3 ๐ โ ๐ ๐ ( ๐ฅ ๐ ๐ โ โ๐ ๐ โ9 ๐ฅ โ ๐ ๐ โ โ๐ ๐ )

4(x โ 5)4 โ 6(x โ 5)3 2(x โ 5)3 [2(x โ 5) โ 3] 4(x โ 5)4 โ 6(x โ 5)3 2(x โ 5)3 [2x โ 10 โ 3] 2(x โ 5)3 [2(x โ 5)4-3 โ 3(x โ 5)3-3] 2(x โ 5)3 (2x โ 13) 2(x โ 5)3 [2(x โ 5)1 โ 3(x โ 5)0]

6 ๐ฅโ2 โ5 โ24 ๐ฅโ2 โ6 6 ๐โ๐ โ5 โ24 ๐โ๐ โ6 6 ๐โ๐ โ6 [ ๐ฅโ2 โ5โ โ6 โ4 ๐ฅโ2 โ6โ โ6 ] 6 ๐โ๐ โ6 [ ๐ฅโ2 1 โ4 ๐ฅโ2 0 ] 6 ๐โ๐ โ6 [๐ฅโ2โ4] 6 ๐ฅโ2 โ6 (๐ฅโ6)

8 ๐ฅ+6 โ 1 2 โ6 ๐ฅ+6 โ 3 2 8 ๐+๐ โ 1 2 โ6 ๐+๐ โ 3 2 2 ๐+๐ โ 3 2 [4 ๐+๐ โ 1 2 โ โ3 2 โ3 ๐+๐ โ 3 2 โ โ3 2 ] 2 ๐+๐ โ 3 2 [4 ๐+๐ โ3 ๐+๐ 0 ] 2 ๐+๐ โ 3 2 [4 ๐+๐ 1 โ3 ๐+๐ 0 ] 2 ๐+๐ โ 3 2 [4๐ฅ+24โ3] 2 ๐ฅ+6 โ 3 2 [4๐ฅ+21]

Factor by Decomposition Example
6x2 โ 11x + 3 6 ๐ฅ 2 โ11๐ฅ+3 6 ๐ฅ 2 +2๐ฅโ9๐ฅ+3 ๐ร๐=18 2รโ9=18 2+(โ9)=โ7 2๐ฅ(3๐ฅ+1)โ3(3๐ฅ+1) (2๐ฅโ3)(3๐ฅ+1)

Quadratic Formula ax2 + bx + c = 0
๐= โ๐ยฑ ๐ ๐ โ๐๐๐ ๐๐

Solve for x 3x2 โ 2x โ 4 = 0 ๐ฅ= โ(โ2)ยฑ (โ2) 2 โ4(3)(โ4) 2(3)
๐ฅ= โ(โ2)ยฑ (โ2) 2 โ4(3)(โ4) 2(3) ๐ฅ= 2ยฑ ๐ฅ= 2ยฑ 4ร13 6 ๐ฅ= 2ยฑ ๐ฅ= 2ยฑ ๐ฅ= 1ยฑ Answers in simplest and exact radical form Approximate decimal answers to nearest hundredth. ๐ฅ=โ0.95 ๐๐๐ ๐ฅ=1.54

๐ฅ= โ(โ3)ยฑ (โ3) 2 โ4(5)(10) 2(5) ๐ฅ= 3ยฑ 9โ200 10 ๐ฅ= 3ยฑ โ191 6
ย Solve for x x2 โ 3x + 10 = 0 ๐ฅ= โ(โ3)ยฑ (โ3) 2 โ4(5)(10) 2(5) ๐ฅ= 3ยฑ 9โ ๐ฅ= 3ยฑ โ191 6 Non-real answer.

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

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

Divide x3 + 3x2 โ 5 by x + 2 SYNTHETIC DIVISION ๐๐ + ๐๐๐ + ๐๐ โ๐
Method I: SUBTRACTION โ5 โ4 โ โ1 Quotient is x2 + x โ 2 Remainder is โ1 NOTE: x3 + 3x2 โ 5 (โ2)3 + 3(โ2)2 โ 5 = โ1

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

Divide x3 โ 8 by x โ 2 SYNTHETIC DIVISION ๐๐ +๐ ๐ ๐ +๐๐โ๐
Method I: SUBTRACTION โ โ8 โ2 โ4 โ8 Quotient is x2 + 2x + 4 Remainder is 0 NOTE: x3โ 8 (2)3 โ 8 = 0

Difference of Cubes Formula
a3 โ b3 = (a โ b)(a2 + ab + b2) Factor x3 โ 8 ๐=๐ฅ ๐=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. (๐ฅโ2)( ๐ฅ 2 +2๐ฅ+ 2 2 ) (๐ฅโ2)( ๐ฅ 2 +2๐ฅ+4) Factor 27x3 โ 64 ๐=3๐ฅ ๐=4 3๐ฅโ4 [ 3๐ฅ 2 +(3๐ฅ)(4)+ 4 2 ] 3๐ฅโ4 (9 ๐ฅ 2 +12๐ฅ+16)

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

Sum of Cubes Formula a3 + b3 = (a + b)(a2 โ ab + b2)
Factor x3 + 27 ๐=๐ฅ ๐=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. (๐ฅ+3 )( ๐ฅ 2 โ3๐ฅ+ 3 2 ) (๐ฅ+3)( ๐ฅ 2 โ3๐ฅ+9) Factor ๐ฅ3+125 ๐ฆ 3 Factor 27๐ฅ3โ64 ๐=๐ฅ ๐=5๐ฆ ๐=3๐ฅ ๐=4 3๐ฅโ4 [ 3๐ฅ 2 +(3๐ฅ)(4)+ (4) 2 ] ๐ฅ+5๐ฆ [ ๐ฅ 2 โ(๐ฅ)(5๐ฆ)+ (5๐ฆ) 2 ] 3๐ฅโ4 (9 ๐ฅ 2 +12๐ฅ+16) ๐ฅ+5๐ฆ ( ๐ฅ 2 โ5๐ฅ๐ฆ+25 ๐ฆ 2 )