McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Solving x^2+bx+c=0
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Exercise 1 Page 49

Start by identifying the values of a, b, and c.

(x+2)(x+12)

Practice makes perfect
To factor the quadratic expression, we will start by identifying the values of a, b, and c. x^2+14x+24 ⇔ 1x^2+ 14x+ 24 For our expression, we have that a= 1, b= 14, and c= 24. To factor a quadratic expression with leading coefficient a= 1, we need to find two factors of c= 24 whose sum is b= 14. Since 24 is a positive number, we will only consider factors with the same sign — both positive or both negative — so that their product is positive.
Factor Pair Product Sum
1 and 24 ^(1* 24) 24 1+24 25
- 1 and - 24 ^(- 1* (- 24)) 24 ^(- 1+(- 24)) - 25
2 and 12 ^(2* 12) 24 ^(2+12) 14
- 2 and - 12 ^(- 2* (-12)) 24 ^(- 2+(- 12)) - 14
3 and 8 ^(3* 8) 24 ^(3+8) 11
- 3 and - 8 ^(- 3* (- 8)) 24 ^(- 3+(- 8)) - 11
4 and 6 ^(4* 6) 24 ^(4+6) 10
- 4 and - 6 ^(- 4* (- 6)) 24 ^(- 4+(- 6)) - 10

The integers whose product is 24 and whose sum is 14 are 2 and 12. x^2+14x+24 ⇔ (x+2)(x+12) Let's use a graphing calculator to confirm our answer. To do so, we will graph the related functions in the same coordinate plane.

We see that only one graph appears. This means that both graphs coincide. Therefore, the expression has been factored correctly.