6. Solving x^2+bx+c=0
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Start by identifying the values of a, b, and c.
(x+2)(x+12)
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.