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Sagot :
Given:
The expression is
[tex]x^2+bx+6[/tex]
To find:
The values of b so that the given expression can be factored into binomials factors.
Solution:
An expression is [tex]ax^2+bx+c[/tex] factorable if b is the sum of possible factors of ac.
We have,
[tex]x^2+bx+6[/tex]
Here, [tex]a=1,b=b,c=6[/tex].
[tex]ac=(1)(6)[/tex]
[tex]ac=6[/tex]
Some, factor forms of 6 are (1×6) and (2×3).
[tex]1+6=7[/tex]
[tex]2+3=5[/tex]
For b=7,
[tex]x^2+7x+6=x^2+x+6x+6[/tex]
[tex]x^2+7x+6=x(x+1)+6(x+1)[/tex]
[tex]x^2+7x+6=(x+1)(x+6)[/tex]
For b=5,
[tex]x^2+5x+6=x^2+2x+3x+6[/tex]
[tex]x^2+5x+6=x(x+2)+3(x+2)[/tex]
[tex]x^2+5x+6=(x+2)(x+3)[/tex]
Therefore, the two possible values of b are 7 and 5.
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