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