# Determine the value(s) of K that makes 9w^2-kw+36 factorable.Please explain how you answer this question fully thank you.

### 4 Answers | Add Yours

`P = 9w^2-kw+36`

For P to be factorable `Delta >= 0 `

For a function of `ax^2+bx+c` Delta is defined as;

`Delta = b^2-4ac`

So for this question;

`Delta = (-k)^2-4*9*36`

`Delta >=0`

`(-k)^2-4*9*36 >=0`

`k^2 - 36^2 >=0`

`(k+36)(k-36) >=0`

When `k<-36` suppoose k = -40 then `(k+36)(k-36) >0`

When `k = -36` then (k+36)(k-36) = 0

When` -36<k<36` suppoose k = 0 then `(k+36)(k-36) <0`

When `k = 36` then `(k+36)(k-36) = 0`

When `k>36` suppoose k = 40 then `(k+36)(k-36) >0`

So `(k+36)(k-36) >=0 ` only when ;

k<=-36 and k>=36

In advanced notation we could write this as;

`k in (-oo,-36]U[36,+oo)`

*So `9w^2-kw+36` factorable when k<=-36 and k>=36.*

** So ` ` factorable when** `k in (-oo,-36]U[36,+oo)`

**Sources:**

Determine the values of k so that `9w^2-kw+36` is factorable:

There are multiple answers to this question depending on the set of numbers you are factoring over:

(1) **Complex numbers**: this will always factor. The fundamental theorem of algebra guarantees that every polynomial factors into linear factors.

(2) **Real numbers**: the analysis by jeew-m is correct. Any value of k where the determinant is greater than or equal to zero will allow factoring over the reals.

(3) **Rationals**: Let p and q be numbers so that their product is 324 . Then any k=p+q will be factorable over the rationals.

For example, let p=27 and q=12. Then k=39.

`9w^2-39w+36=9w^2-27w-12w+36=9w(w-3)-12(w-3)=(w-3)(9w-12)`

Here the roots are 3 and `4/3`

The possibilities are 325,164,111,85,60,45,39,36

(4) **Integers**: Here the discriminant must not only be positive, but also a perfect square number. k=45 or k=36 are the possibilities.

k=45: `9w^2-45w+36=9(w^2-5w+4)=9(w-1)(w-4)`

k=36: `9w^2-36w+36=9(w^2-4w+4)=9(w-2)^2`

This type of sum is not thought in class 7.

i mean **taught.**