Iv got an exam on this in a few days and need know how to do it!

Factoring a quadratic of the form ax^2+bx+c requires that you find two values, j and k, such that jk=ac and j+k=b.  Then you replace bx with jx and kx in the original equation.  Then you factor the first and second pair of terms.

So in your case...

x^2+6x+8

ac=jk=8 and b=j+k=6  so j and k must be 2 and 4 so your equation becomes:

x^2+2x+4x+8  now factor 1st and 2nd pair of terms...

x(x+2)+4(x+2) which is equal to:

(x+4)(x+2)



x^2-3x-18, ac=jk=-18, b=j+k=-3, so j and k must be -6 and 3 so

x^2+3x-6x-18

x(x+3)-6(x+3)

(x-6)(x+3)

...

x^2-7x+12, jk=12, j+k=-7, so j and k must be -3 and -4 so

x^2-3x-4x+12

x(x-3)-4(x-3)

(x-4)(x-3)

The question of the answer is found in the pic