Find two values of "b" so that x^2 + b x + 6 can be factored into binomial factors.
1 answer:
Given:
The expression is
![x^2+bx+6](https://tex.z-dn.net/?f=x%5E2%2Bbx%2B6)
To find:
The values of b so that the given expression can be factored into binomials factors.
Solution:
An expression is
factorable if b is the sum of possible factors of ac.
We have,
![x^2+bx+6](https://tex.z-dn.net/?f=x%5E2%2Bbx%2B6)
Here,
.
![ac=(1)(6)](https://tex.z-dn.net/?f=ac%3D%281%29%286%29)
![ac=6](https://tex.z-dn.net/?f=ac%3D6)
Some, factor forms of 6 are (1×6) and (2×3).
![1+6=7](https://tex.z-dn.net/?f=1%2B6%3D7)
![2+3=5](https://tex.z-dn.net/?f=2%2B3%3D5)
For b=7,
![x^2+7x+6=x^2+x+6x+6](https://tex.z-dn.net/?f=x%5E2%2B7x%2B6%3Dx%5E2%2Bx%2B6x%2B6)
![x^2+7x+6=x(x+1)+6(x+1)](https://tex.z-dn.net/?f=x%5E2%2B7x%2B6%3Dx%28x%2B1%29%2B6%28x%2B1%29)
![x^2+7x+6=(x+1)(x+6)](https://tex.z-dn.net/?f=x%5E2%2B7x%2B6%3D%28x%2B1%29%28x%2B6%29)
For b=5,
![x^2+5x+6=x^2+2x+3x+6](https://tex.z-dn.net/?f=x%5E2%2B5x%2B6%3Dx%5E2%2B2x%2B3x%2B6)
![x^2+5x+6=x(x+2)+3(x+2)](https://tex.z-dn.net/?f=x%5E2%2B5x%2B6%3Dx%28x%2B2%29%2B3%28x%2B2%29)
![x^2+5x+6=(x+2)(x+3)](https://tex.z-dn.net/?f=x%5E2%2B5x%2B6%3D%28x%2B2%29%28x%2B3%29)
Therefore, the two possible values of b are 7 and 5.
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