1. 8.6: FACTORING ax2 + bx + c where a ≠ 1
Factoring: A process used to break down any polynomial into simpler polynomials.

2. FACTORING ax2 + bx + c Procedure:
1) Always look for the GCF of all the terms
2) Factor the remaining terms – pay close attention to the value of coefficient a and follow the proper steps.
3) Re-write the original polynomial as a product of the polynomials that cannot be factored any further.

3. GOAL:

4. 5x2+11x+2? FACTORING: When AC is POSSIBLE.
Ex: What is the FACTORED form of:
5x2+11x+2?

5. 5x2+11x+2 ax2+bx+c  b= +11  ac =(5)(2)  ac = +10 : (1)(10), (2)(5)
FACTORING:
To factor a quadratic trinomial with a coefficient ≠ 1 in the x2, we must look at the b and ac coefficients:
5x2+11x+2
ax2+bx+c
 b= +11
 ac =(5)(2)
Look at the factors of ac:
 ac = +10 : (1)(10), (2)(5)
Take the pair that equals to b when adding the two integers.
In our case it is 1x10 since 1+10 =11= b

6. 5x2+11x+2 5x2 + 1x + 10x + 2 5x2 + 1x  x(5x + 1) 10x + 2  2(5x + 1)
Re-write using factors of ac that = b.
5x2+11x+2
5x2 + 1x + 10x + 2
Look at the GCF of the first two terms:
5x2 + 1x
 x(5x + 1)
Look at the GCF of the last two terms:
10x + 2
 2(5x + 1)
Look at the GCF of both:
 x(5x + 1) + 2(5x + 1)
Thus the factored form is: (5x+1) (x+2)

7. YOU TRY IT:
Ex: What is the FACTORED form of:
6x2+13x+5?

8. 6x2+13x+5 ax2+bx+c  b= +13  ac =(6)(5)
SOLUTION:
To factor a quadratic trinomial with a coefficient ≠ 1 in the x2, we must look at the b and ac coefficients:
6x2+13x+5
ax2+bx+c
 b= +13
 ac =(6)(5)
Look at the factors of C:
ac = +30 :(1)(30), (2)(15), (3)(10)
Take the pair that equals to b when adding the two integers.
In our case it is 3x10 since 3+10 =13= b

9. 6x2+13x+5  6x2 + 3x + 10x + 5 6x2 + 3x  3x(2x + 1) 10x + 5
Re-write using factors of ac that = b.
6x2+13x+5
 6x2 + 3x + 10x + 5
Look at the GCF of the first two terms:
6x2 + 3x
 3x(2x + 1)
Look at the GCF of the last two terms:
10x + 5
 5(2x + 1)
Look at the GCF of both:
 3x(2x + 1)+ 5(2x + 1)
Thus the factored form is:(3x+5)(2x+1)

10. CLASSWORK: Page :
Problems: 1, 2, 5, 8, , 11, 20.

11. 3x2+4x-15? FACTORING: When ac is Negative.
Ex: What is the FACTORED form of:
3x2+4x-15?

12.  ac = -45 : (-1)(45), (1)(-45) (-3)(15), (3)(-15) (-5)(9), (5)(-9)
FACTORING:
Since a ≠ 1, we still look at the b and ac coefficients:
3x2+4x-15
ax2+bx+c
 b= +4
 ac =(3)(-15)
Look at the factors of ac:
 ac = -45 : (-1)(45), (1)(-45) (-3)(15), (3)(-15) (-5)(9), (5)(-9)
Take the pair that equals to b when adding the two integers.
In our case it is (-5)(9)since -5+9 =+4 =b

13. 3x2+4x-15 3x2 -5x + 9x -15 3x2 - 5x  x(3x - 5) 9x -15  3(3x -5)
Re-write: using factors of ac that = b.
3x2+4x-15
3x2 -5x + 9x -15
Look at the GCF of the first two terms:
3x2 - 5x
 x(3x - 5)
Look at the GCF of the last two terms:
9x -15
 3(3x -5)
Look at the GCF of both:
 x(3x - 5) + 3(3x - 5)
Thus the factored form is: (3x-5) (x+3)

14. YOU TRY IT:
Ex: What is the FACTORED form of:
10x2+31x-14?

15. 10x2+31x-14 ax2+bx+c  b= +31  ac =(10)(-14)
FACTORING:
Since a ≠ 1, we still look at the b and ac coefficients:
10x2+31x-14
ax2+bx+c
 b= +31
 ac =(10)(-14)
Look at the factors of ac:
 ac = : (-1)(140), (1)(-140) (-2)(70), (2)(-70) (-4)(35), (4)(-35)
Take the pair that equals to b when adding the two integers.
This time it is(-4)(35)since -4+35=+31=b

16. 10x2+31x-14  10x2-4x + 35x -14 10x2 - 4x  2x(5x - 2) 35x -14
Re-write using factors of ac that = b.
10x2+31x-14
 10x2-4x + 35x -14
Look at the GCF of the first two terms:
10x2 - 4x
 2x(5x - 2)
Look at the GCF of the last two terms:
35x -14
 7(5x - 2)
Look at the GCF of both:
 2x(5x - 2)+ 7(5x - 2)
Thus the factored form is:(2x+7)(5x-2)

17. CLASSWORK: Page :
Problems: 3, 6, 14, 16, , 18, 34.

18. REAL-WORLD:
The area of a rectangular knitted blanket is 15x2-14x-8. What are the possible dimensions of the blanket?

19. 15x2-14x-8 ax2+bx+c  b= -14  ac =(15)(-8)
SOLUTION:
Since a ≠ 1, we still look at the b and ac coefficients:
15x2-14x-8
ax2+bx+c
 b= -14
 ac =(15)(-8)
Look at the factors of ac:
 ac = : (-1)(120), (1)(-120) (-2)(60), (2)(-60),(-3)(40), (3)(-40) (-4)(30), (4)(-30), (-5)(24), (5)(-24) (-6)(20), (6)(-20), (-8)(15), (8)(-15)
This time it is(6)(-20)since 6-20=-14=b

20. 15x2-14x-8  15x2 + 6x - 20x -8 15x2 + 6x  3x(5x + 2) -20x -8
Re-write using factors of ac that = b.
15x2-14x-8
 15x2 + 6x - 20x -8
Look at the GCF of the first two terms:
15x2 + 6x
 3x(5x + 2)
Look at the GCF of the last two terms:
-20x -8
-4 (5x + 2)
Look at the GCF of both:
 3x(5x + 2)- 4(5x + 2)
Thus the factored form is:(3x-4)(5x+2)

21. VIDEOS: Factoring Quadratics Factoring when a ≠ 1:

22. CLASSWORK: Page :
Problems: 4, 7, 15, 21, , 36, 37.