Question

For how many integer values of x^2+ax+6 a can be factored? What are they?

Answer 1

i need help on this

Answer 2

\[\mathrm{x^2+ax+6=(x+m)(x+n)
\\x^2+ax+6=x^2+(m+n)x+mn}\]
\[\mathrm{m+n=a\\
mn=6}\]
I hope this helps.

Answer 3

it does alittle

Answer 4

xD

Answer 5

\[\mathrm{m(a-m)=6=2\times3=6\times1}\]

Answer 6

You have to consider the negative integers :p

Answer 7

so is the awnser -7 -5 5 7

Answer 8

4 integer values; -7, -5, 5, and 7

Answer 9

Yes :p

Answer 10

i dont know what the work looks like

Answer 11

anyone there

Answer 12

X^2 + ax + 6 can be factored either in the form (x - b)(x - c) or (x + b)(x + c). b and c can any pair of the factors of 6: 4 integer values; -7, -5, 5, and 7 i hope this help..

Answer 13

\[m(a-m)=6\\
a=\frac{6}{m}+m,\quad m\neq0\\
\]
Integer+rational=rational

If m is integer, then 6/m is integer (if m=2,3,6) or rational.
If m is rational, then 6/m is integer. Then a is rational

hence these are all the solutions (what you said before)