Question

How do you factor the trinomial `4z^2 + 32z + 63`?

Answer 1

It is `[2z+7]*[2z+9]`

Explanation:

Rewrite this as follows

`4*(z^2+8z)+63=4*(z^2+2*4*z+16)+63-64= 4*(z+4)^2-1=(2*(z+4))^2-1=[2*(z+4)-1]*[2*(z+4)+1]= [2z+7]*[2z+9]`

Answer 2

y = (2x + 7)(2x + 9)

Explanation:

I use the systematic new AC Method to factor trinomials (Socratic Search)
`y = 4x^2 + 32x + 63 =` 4(x + p)(x + q)
Converted trinomial `y' = x^2 + 32 x + 252 =`(x + p')(x + q').
p' and q' have same sign because ac > 0.
Compose factor pairs of (ac = 252) --> ...(12, 21)(14, 18). This sum is (14 + 18 = 32 = b). Then p' = 14 and q' = 18.
Back to original trinomial: `p = (p')/a = 14/4 = 7/2` and `q = (q')/a = 18/4 = 9/2`.
Factored form: y = 4(x + 7/2)(x + 9/2) = (2x + 7)(2x + 9)