First one, 12q^2+34q-28 divide whole thing by 2 6q^2+17q-14 use trial and error and get (2x+7)(3x-2) factored out form is (2)(2x+7)(3x-2)
2. divide by 3 6h^2+5h-6 trial and error and get (2x+3)(3x-2) so the factored form is (3)(2x+2)(3x-2)
3. divide by 2 6p^2-11x-10 use trial and error and get (2x-5)(3x+2) the factored out form is (2)(2x-5)(3x+2)
4.divide by 4 2z^2+5z-12 use trial and error and get (x+4)(2x-3) the factored form is (4)(x+4)(2x-3)
to factor the basic thing is ax^2+bx+c b=x+y ac=xy solve for x and y so exg 2z^2+5z-12 a=2 b=5 c=-12 2 times -12=-24
then factctor -24 and find factors that add up to 5 -1, 24 -2, 12 -3, 8 -4,6 now add -1+24=23 -2+12=10 -3+8=5 check -4+6=2 the numbers are -3 and 8 split it into such
2z^2+8z-3z-12 group (2z^2+8z)+(-3z-12) undistribute using distributive property ab+ac=a(b+c) (2z)(z+4)+(-3)(z+4) now reverse distribute again ((2z)+(-3))(z+4) (2z-3)(z+4)