Respuesta :

irspow
x^2+2x-35=0 is of the form ax^2+bx+c

To factor you want to find two values, j and k, which satisfy two conditions:

ac=jk=-35 and b=j+k=2, so j and k must be 7 and -5.

Now replace bx with jx and kx in the original equation to get:

x^2+7x-5x-35=0  now factor 1st and 2nd pair of terms...

x(x+7)-5(x+7)=0 which is equal to:

(x-5)(x+7)=0  

so x=-7 and 5
zebsi0n

[tex]x^2+2x-35=0[/tex]
There are a few ways to solve this: factoring, completing the square, quadratic formula, among others.
I prefer to factor in most cases.
What factors of -35 add up to 2?
That would be -5 and 7. So, we can set the equation up like this:
[tex](x-5)(x+7)=0[/tex]
Now we can solve each of the expressions.
[tex]x-5=0[/tex]
[tex]x=5[/tex]
and
[tex]x+7=0[/tex]
[tex]x=-7[/tex]

Hope this helps!

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