Flowchart to factor
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Flowchart to factor. Factor out the Great Common Factor. Ex 1. Can be the expression written as A 2 – B 2 or A+ B 3 or A 3 – B 3 ?. YES. Ex 2. Factor. Two terms ?. YES. NO. END. NO. Can the expression be written as ax 2 +bx+c?. YES. YES. perfect square?. Ex 3.

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Flowchart to factor

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Flowchart to factor

Flowchart to factor

Factor out the Great

Common Factor

Ex 1

Can be the expression

written as A2 – B2 or

A+ B3 or A3 – B3 ?

YES

Ex 2

Factor

Two terms?

YES

NO

END

NO

Can the expression be

written as ax2+bx+c?

YES

YES

perfect square?

Ex 3

Three Terms?

YES

NO

NO

product of two

binomials?

YES

END

NO

Ex 4

NO

END

Can you factorize the

expression by grouping ?

Four or More

Terms?

YES

YES

Ex 5

NO

END


Ex 1 greatest common factor gcf

Ex 1a: Factor 6ab+8ac+4a

Look for the GCF 6ab+8ac+4a = 2 ·3 ab + 2 · 4ac + 2 ·2a

Take out the GCF = 2a (     )

Working on the parenthesis = 2a ( 3 b + 4 c + 2 )

Answer:6ab+8ac+4a =2a ( 3b + 4c + 2)

Ex 1b: Factor 5x2y(2a–3) –15xy(2a-3)

Look for the GCF 5x2y(2a–3)–15xy(2a-3) = 5xxy(2a-3)– 3•5xy(2a-3)

Take out the GCF = 5xy(2a-3)(  )

Working on the parenthesis = 5xy(2a-3)(x - 3 )

Answer: 5x2y(2a – 3) –15xy(2a - 3) = 5xy(2a - 3)( x - 3 )

Ex 1: Greatest Common Factor (GCF)

Return to Flowchart


Ex 2 factoring a binomial

Ex 2a: Factor 3x3 – 12xy2

Taking the GCF out 3x3 – 12xy2= 3x ( x2 – 4y2 )

Factoring as difference of squares

Ex 2b: Factor 16x4 + 54x2

Taking the GCF out 16x5 + 54x2 = 2x2 (8x3 + 27)

Working on the parenthesis = 2x2( [2x]3 + [3]3 )

Ex 2: Factoring a Binomial

= 3x (x+2y)(x -2y)

Answer:3x3 – 12xy2 =3x (x+2y(x -2y)

Factoring as difference of cubes

= 2x2(2x+ 3)[ (2x)2 –(2x)3 +(3)2]

Simplifying = 2x2 (2x+ 3 )(4x2 - 6x + 9)

Answer :16x5 – 54x2 = 2x2(2x + 3)(4x2 – 6x + 9)

Return to Flowchart


Ex 3 factoring as perfect square

Ex 3a: Factor 63x4 – 210x3 + 175x2

Taking GCF out 63x4 – 210x3 + 175x2 = 7x2(9x2 – 30x + 25)

Check for perfect square (    )2

Take square root first and last terms = 7x2(3x – 5 )2

Answer:63x4 – 30x3 + 25x2 =7x2 (3x – 5 )2

Ex 3b:Factor 3x3y2 +18x2y +27x

Taking the GCF out 3x3y2 + 18x2y +27x = 3x (x2y2 + 6xy + 9)

Check for perfect square (    )2

Take square root first and last terms = 3x ( xy + 3 )2

Ex 3: Factoring as Perfect Square

Middle term’s check 2(3x)5 = 30x OK!

Middle term’s check 2(xy)3 = 6xy OK!

Answer:3x3y2 +18x2y + 27 =3x (xy + 5)2

Return to Flowchart


Ex 4 factor as product of two binomials simple case

Ex 4: Factor as Product of Two Binomials Simple Case

Ex 4a :Factor 3x3 – 21x2 + 30x

Taking GCF out 3x3 – 21x2 + 30x = 3x (x2 – 7x+ 10)

Trinomial is not a perfect square

Let’s try to factor as product of two binomials = 3x(x + a )(x + b )

Where a·b = 10 & a + b = - 7. Since a·b is positive, a & b should have same sign, and since ab = - 7 both should be negative.

Possibilities for a, b are 1,10 or 2,5. Let’s check

x - 1

x - 10

- x

- 10x

-11x

x - 2

x - 5

-2x

-5x

-7x

NO!

OK!

So, a = -2 and b = - 5 3x3 – 21x2 + 30x = 3x (x -2)(x –5)

Answer:3x3 – 21x2 + 30x = 3x (x - 2)(x - 5)

Return to Flowchart


Ex 4 simple case continued

Ex 4b: Factor 4x5 – 20x3 + 16x

Taking the GCF out 4x5 – 20x3 + 16x = 4x (x4 – 5x2 + 4 )

Trinomial is not a perfect square !  

Factoring as product of two binomials = 4x(x2 – 1) (x2 – 4)

Factoring as difference of squares  

=4x(x+1)(x–1)(x+2)(x–2)

Answer:4x5– 20x3+16x = 4x (x+1)(x–1)(x+2)(x–2)

Ex 4c: Factor 128 x7 – 2x

Taking the GCF out … 128 x7 – 2x = 2x (64x6 – 1)

Working on the parenthesis = 2x ( (8x3 )2– 1)

 

Factoring as difference of squares = 2x (8x3 + 1) (8x3– 1)

Working on the parenthesis … = 2x ( (2x)3+ 1 ) ((2x)3 –1)

  

Addition and difference of cubes =2x(2x+1)(4x2 –2x+1)(2x–1)(4x2+2x+1)

Answer:128 x7–162x =2x(2x+1)(4x2–2x+1)(2x-1)(4x2+2x+1)

Ex 4: Simple Case Continued …


Ex 4 general case

Ex 4d: Factor 30 x2 – 21x – 36

Taking the GCF out 30x2 – 21x – 36 = 3(10x2 – 7x – 12)

Check if it factors as = 3(ax + p)(bx + q)

Since ab = 10, possible values for a,b are 1&10, or 2&5

Since pq = - 12, p & q has different sign and all possible ways of getting 12 are 12(1) , 6(2), 4(3).

The following tables summarize the situation for pairs 1&10,2&5

Ex 4: General Case

7

Each number on the last row is the adding of cross multiplication of the blue column by each one of the red columns. For example: 1 (-1) + 10 (12) = 119 …

The pair 1 &10 doesn’t work ! ( since 7 didn’t appear on the last row).

  • Let’s try now with p=2 & q=5. Filling the blanks on third row …

Since we need – 7x as the middle term, so we pick p = -2 and q = 5. (Do theCheck)

Answer:30x2 – 21x – 36 = 3(2x - 3)(5x +4)

Return to Flowchart


Ex 5 factor by grouping

Ex 5a:Factor3x3 – 6x2 + 5x –10

Splitting in two groups 3x3 – 6x2 + 5x –10 = 3x3 – 6x2+ 5x –10

Taking the GCF out from both groups = 3x2(x– 2) +5(x – 2)

Taking the GCF( x – 2 ) out (to the right)

= ( 3x2+ 5)

Answer:3x3 – 6x2 + 5x –10 = ( x – 2)( 3x2 + 5)

Ex 5b: Factor4x2 – 40x + 100 – 4y2

Taking the GCF out 4x2 – 40x + 100 – 4y2 = 4(x2 – 10x + 25 –y2)

Split expression in parenthesis into two groups = 4 ([x2–10x+25]– y2 )

Factor the first group as perfect square = 4 ([ x - 5]2 – y2 )

Factor as difference of squares

Ex 5: Factor by grouping

( x – 2)

= 4( [x – 5 + y][x – 5 – y] )

Answer: 4x2 – 40x + 100 –4y2 = 4(x + y – 5)(x – y – 5)

Return to Flowchart


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