Please factor 3z^2 +4z -4?

(z+2)(3z-2)

good luck

Factor using FOIL

3z^2 + 4z - 4

take first term 3 is prime so 3*1

then take third term 4 can 2*2 , 4*1 chose 2*2

use trial & error

(3z 2) ( z 2)

now the sign of middle term is +

also check the sign of the third term too -

(3z - 2) ( z + 2)

check to see to see if equal the middle term

6z -2z = +4z (check )

(3z - 2) ( z + 2) this is the answer

3z^2+4z-4= (3z-2)(z+2)

because:

- - - - - -

delta= 16+3*16=64 -> x1 = -2 and x2= 2/3 ;

the whole thing is 3(z+2)(z-2/3) = (3z-2)(z+2)

7

3z² +4z -4

(3z - 2)(z + 2)

it is extremely a case of (A^2 - B^2), after factoring out 2y^4 32y^4-2y^4z^4 2y^4(sixteen - z^4) 2y^4(4 - z^2)(4 + z^2) Oh seem! there is that (A^2 - B^2) back 2y^4(4 - z^2)(4 + z^2) 2y^4(2- z)(2+z)(4 + z^2)

z+9

( 3z - 2 ) ( z + 2 )

(z+2)(3z-2)

(3z-2)(z+2)