factoring trinomials with leading coefficient ax 2 + bx + c to (ax + b)(cx + d)

FACTORING TRINOMIALSwith leading coefficient

ax2 + bx + c

to

(ax + b)(cx + d)

REMEMBERThere are many different methods to use when

factoring. To be consistent, we will continue to use the factor by grouping method.

First, use parentheses to group terms with common factors.

Be sure middle sign is +If it is -, change it to +(-)

Next, factor the GCF from each grouping.

Now, Distributive Property…. Group both GCF’s. and bring down one of the other ( ) since they’re both the same.

Using grouping with trinomials

Multiply the first and last terms. 21c2(-4) =-84c2

Check your signs... ‘c’ is – so subtract

factors. ‘b’ is – so larger factor

will be negative.

Replace the middle term with these factors..

Then group.

21) 21 5 4c c

(1)( 84)(2)( 42)(3)( 28)(4)( 21)(6)( 14)(7)( 12) 7 12 5

7 12c c 221 7 12 4c c c 221 12 7 4c c c

Switch fa

ctors to

make middle sign +

( ) ( )3c(7 4)c 1 (7 4)c

(3 1)c (7 4)c

Using grouping with trinomials

Multiply the first and last terms. 25y2(9) = 225y2

Check your signs... ‘c’ is + so add factors. ‘b’ is – so both factors

will be negative.

Replace the middle term with these factors..

Then group.

22) 25 30 9y y

( 1)( 225) ( 5)( 45) ( 15)( 15) 15 15 30

15 15y y 225 15 15 9y y y ( ) ( )

5y(5 3)y 3 (5 3)y (5 3)y (5 3)y

2(5 3)y

If the first term in the ( ) is negative, You will factor out a negative number.

Using grouping with trinomials

Multiply the first and last terms. 9x2(2) = 18x2

Check your signs... ‘c’ is + so add factors. ‘b’ is – so both factors

will be negative.

Replace the middle term with these factors..

Then group.

23) 9 9 2x x

( 1)( 18) ( 2)( 9) ( 3)( 6) 3 6 9

3 6x x 29 3 6 2x x x ( ) ( )

3x (3 1)x 2 (3 1)x (3 2)x (3 1)x

If the first term in the ( ) is negative, You will factor out a negative number.

Using grouping with trinomials

Multiply the first and last terms. 12x2(5) = 60x2

Check your signs... ‘c’ is + so add factors. ‘b’ is + so both factors

will be positive.

Replace the middle term with these factors..

Then group.

24) 12 19 5x x

(1)(60)(2)(30)(3)(20)(4)(15) 4 15 19

4 15x x 212 4 15 5x x x ( ) ( )

4x(3 1)x 5 (3 1)x (4 5)x (3 1)x

25) 20 10 33x x

220 33 10x x

(1)(200)(2)(100)(4)(50)(5)(40)

8 25 33 (8)(25)

8 25x x 220 8 25 10x x x ( ) ( )

4x(5 2)x 5 (5 2)x (4 5)x (5 2)x

26) 20 2 3x x

220 3 2x x

( 1)( 40) ( 2)( 20) ( 4)( 10)

5 8 3 ( 5)( 8)

5 8x x 220 5 8 2x x x ( ) ( )

5x(4 1)x 2 (4 1)x (5 2)x (4 1)x

Using grouping with trinomials

Multiply the first and last terms. 4r2(7) = 28r2

Check your signs... ‘c’ is + so add factors. ‘b’ is - so both factors will

be negative.

None of the factors will ADD to give -1

27) 4 7r r

( 1)( 28) ( 2)( 14) ( 4)( 7) ( 7)( 4)

same factors

different order

PRIME

Using grouping with trinomials

Multiply the first and last terms. 3x2(-5) = -15x2

Check your signs... ‘c’ is - so subtract factors. ‘b’ is + so larger factor

will be positive.

None of the factors will SUBTRACT to give +7

28) 3 7 5x x

( 1)( 15) ( 3)( 5) ( 5)( 3)

same factors

different order

PRIME

Solutions to Trinomials

Now that we have the factors of each trinomial, we can carry it to the next step and find the SOLUTIONS for each trinomial.

Remember to set your factors equal to zero then solve for the variable…. Like this….

21) 21 5 4c c (3 1)c (7 4)c

3 1 0c 7 4 0c 1 1

3 1c 4 4

7 4c 1

3c

4

7c

22) 25 30 9y y 2(5 3)y

3

5y

23) 9 9 2x x (3 2)x (3 1)x

2

3x 1

3x