(x - 1)(x - 14)
(x+7)(x+2)
(x - 14)(x - 2)
To factor the trinomial (7x^2 + 7x - 14), first factor out the common factor of 7: [ 7(x^2 + x - 2) ] Next, we can factor the quadratic (x^2 + x - 2) into ((x + 2)(x - 1)). Thus, the complete factorization of the original trinomial is: [ 7(x + 2)(x - 1) ]
Two numbers that multiply to give you +33 and add to give you 14 are 11 and 3. (x+11)(x+3)
That factors to (x - 14)(x - 2)
(x+7)(x+2)
-(x - 7)(x + 2)
(w + 7)(w - 2)
(2x + 7)(x - 2)
(x - 14)(x - 2)
3z2 + 45z + 42 = 3(z2 + 15z + 14) = 3(z + 1)(z + 14).
To factor the trinomial (7x^2 + 7x - 14), first factor out the common factor of 7: [ 7(x^2 + x - 2) ] Next, we can factor the quadratic (x^2 + x - 2) into ((x + 2)(x - 1)). Thus, the complete factorization of the original trinomial is: [ 7(x + 2)(x - 1) ]
x2+15x+14 = (x+1)(x+14)
Two numbers that multiply to give you +33 and add to give you 14 are 11 and 3. (x+11)(x+3)
I would do this. Multiply through by -1 -1 ( -X^2 +12X +28 ) = X^2 -12X -28 (X+2)(X-14)
That factors to (x - 14)(x - 2)
(x + 2)(x - 7)