Non - Pst 42127

FACTORING

NON- PERFECT SQUARE TRINOMIAL (NONPST) of the form + bx + c and  

of the form+ bx + c

• Objectives   Recall • FOIL Method / Pattern in Multiplying Binomials • Factoring by Grouping Method • PST or not PST Factor • Non-Perfect Square Trinomial (Non-PST) of the form + bx + c • Non-Perfect Square Trinomial (Non-PST) of the form + bx + c

 

(c + d)  

Activity: Recall FOIL Method / Pattern   c + bd

+ 3x + 2x + 6   + 5x + 6 (x – 4 ) (x – 5)   – 5x – 4x + 20   – 9x + 20   (x + 6) (x – 7) – 7x + 6x – 42   – x – 42   (x – 8) (x + 9) + 9x – 8x – 72   + x – 72 (x + 2) (x + 3)

 

(c + d)

Activity: Recall FOIL Method / Pattern   c + bd

 

(x + 2) (3x + 4) + 4x + 6x + 8   + 10x + 8 (5x – 6 ) (7x – 8)   – 40x – 42x + 48   – 82x + 48 (9x + 10) (11x – 12)

 

– 108x + 110x – 120

 

+ 2x – 120

Activity: Recall Factoring By Grouping c + bd

   

 

c + bd) =  

 

c + d) = (c + d)

  Factor 3x + 6y + + 2xy =   (3x + 6y) + + 2xy) = 3(x + 2y) + x(x + 2y) = (x + 2y)(3 + x) or (3 + x) (x + 2y)

LET’S GO BACK AND FACTOR TH Do you those that are Non-PST in the previous activity, PST or Not? We are going to factor some of them.

a. a2 + 2a + 1a. PST (a + 1)2 2 2 b. PST (1 + 3c) b. 1 + 6c + 9c 2 c. PST (x + 2) c. x + 4x + 4 2

d. a2 – 5a + 6 d. Not PST e. x2 + 4x + 3e. Not PST

* *

LET’S GO BACK AND FACTOR TH Do you those that are Non-PST in the previous activity, PST or Not? We are going to factor some of them.

f. 9x2 – 12xy + 4yf.2 PST (3x – 2y)2 g. x2 – 7x – 18 g. Not PST h.

a2 – 8a + 16

h. PST (a – 4)2

*

Prime; Non-factorable i. a2 + 10a + 100 i. Not PST j. x2 + 8x + 15 j. Not PST *

ACTIVITY: EXAMPLE ANALYSIS 1. x2 + 5x + 4 = x2 + x + 4x + 4 By grouping we have, (x2 + x) + (4x + 4) = x(x + 1) + 4(x + 1) = (x + 1)(x + 4) 2. x2 – 8x + 12 = x2 – 2x – 6x + 12 = (x2 – 2x) – (6x – 12) = x(x – 2) – 6 (x – 2) = (x – 2)(x – 6) 3. y2 – y – 12 = y2 – 4y + 3y – 12 = (y2 – 4y) + (3y – 12) = y(y – 4) + 3(y – 4) = (y – 4)(y +3) 4. y2 + 6y – 27 = y2 + 9y –3y – 27 = (y2 + 9y) – (3y + 27) = y(y + 9) – 3(y + 9) = (y + 9)(y – 3) 5. 2n2 + 7n + 3 = 2n2 + 6n + n + 3 = (2n2 + 6n) + (n + 3) Activity 3: Examples Analysis (Pair Work) = 2n(n + 3) + (n + 3) = (n + 3)(2n + 1) 6. 3n2 + 2n – 5 = 3n2 – 3n + 5n – 5 = (3n2 – 3n ) + (5n – 5) = 3n(n – 1) + 5(n – 1) = (n – 1)(3n + 5) 7. 6m2 – 17m + 10 = 6m2 –12m – 5m + 10 = (6m2 – 12m ) – (5m – 10) = 6m(m – 2) – 5(m – 2) = (m – 2)(6m –5) 8. 4m2 – 5m – 6 = 4m2 –8m + 3m – 6 = (4m2 – 8m ) + (3m –

Factoring Non  PST of the form + bx + c A. Using Grouping Method Factor x2 + 4x + 3

1 and Think of factors of 3 that sum up to 4. 3 Rewrite the trinomial into its equivalent multinomial.

x2 + 4x = + x32 + x + 3x + 3

Factor the resulting multinomial by grouping. 2 x2 + x + 3x =+ (x3 + x) + (3x + 3) = x(x + 1) + 3(x + 1) x2 + 4x ++ 3 1)(x + 3) or (x + 3)(x + = (x

Factoring Non  PST of the form + bx + c

B. Using Criss Cross Method   Think of factors of . Factor x2 + 4x + 3 of factors of 3 that sum up to x + 3 +Think 3x x

+

1

Cross multiply.

+x Determine the correct signs to satisfy + 4xthe middle term of the trinomial.

Therefore, x2 + 4x + 3 = (x + 3)(x + 1)

Factoring Non  PST of the form + bx + c

C. Using Trial and Error/Success Method Factor x2 + 4x = + (3 x +3 ) (x+1   Think of factors of .

3x

)

+ x

ink of factors of 3 that sum up to 4.4x

etermine the correct signs to satisfy the middle term of the trinomial.

Therefore, x2 + 4x + 3 = (x + 3)(x + 1)

Using any of the three methods factor the following: Non-PST of the form + bx + c

 

1. x2 + 8x +=15 (x + 3)(x + 5) 2. a2 – 5a + = 6 (a – 3)(a – 2) 3.

x2 – 7x – = 18(x – 9)(x + 2)

Factoring Non-PST   of the form a+ bx + c A. Using AC / Grouping Method Factor 2x2 + 5xa+=32; b = 5, c = 3

et the product of a and acc.= 2(3) = Middle 6 term = 2 + 3 = factors of ac that sum up to the coefficient of the mi the trinomial into its equivalent multinomial, then fac

2x2 + 5x=+2x 3 2 + 2x + 3x + 3 = (2x2 + 2x) + (3x + 3) = 2x(x + 1) + 3(x + 1) (2x + 3)(x + 1) = (x + 1)(2x or + 3)

Factoring Non-PST   of the form a+ bx + c B. Using Criss Cross Method   Think of factors of .

1. 2x + 5x + 3 2x + 3 + 3xThink of 2

x

+

1

factors of 3.

Cross multiply. + 2x

Determine the correct positions of the factors of 3 and the +5x corresponding signs 2 to satisfy the middle term of Therefore, 2x + 5x + 3 = (2x + 3)(x + 1 the trinomial.

Factoring Non-PST   of the form a+ bx + c

C. Using Trial and Error/Success Method 1. 2x2 + 5x= +(3 2x+3) (x+1   Think of factors of .

Think of factors of 3.

)

3x

+ 2x 5x

ne the correct positions of the factors of the 3 rd term onding signs to satisfy the middle term of the trinomia

Therefore, 2x2 + 5x + 3 = (2x + 3)(x + 1

Factoring Non-PST   of the form a+ bx + c D. Using Bottoms Up Method 1. 2x2 + 5x + 3 a = 2; b = 5, c = 3

ac = 2(3) = 6 Get the product of a and c. rite the trinomial in the form of x 2 + bx + ac, then fa

x2 + 5x = + (x 6 + 2)(x + 3)

Sinceby a =a.2 in the give de each constant in the binomial factor Simplify whenever possible. If not,(x  + )( x + ) write the denominator as a coefficient of the 1st term Therefore, 2x2 + 5x + 3 = (x + 1)(2x and the numerator as the constant

Let’s Go Back and Chec • Go back to the examples in the activity, Example Analysis, and check their factors using other methods. Assignment: • Study illustrative examples in your Math text book, pages 35-39, and answer selected items on pages 44-45, and write your solutions and answers on one whole sheet of paper.

Assignment : • Review all the methods of factoring Non- PST through the PowerPoint Presentation shared through Edmodo.