Factor w^2-11w+30 please?

Hi,

It's either(w - 6)(w - 5) or (6 - w)(5 - w), like you had,

Either one will work.

I hope that helps!! :-)

(w-5)(w-6) is the way I would have done it, but (5-w)(6-w) [which is exactly the same as what you've got] is correct as well.

w^2 +30 =11w w^2 -11w + 30 = 0 (carry each thing to a minimum of one part, equate to 0) (w - 5)(w - 6) = 0 -5 and -6 works because of the fact it has a sum of -11 (the b fee) and a manufactured from 30 (the ac fee)

( w - 6 ) ( w - 5 )

This quadratic equation has two real roots:

w = 5 and w = 6

so the factors are

(w - 5) and (w - 6)

Yup what you got is right. It can also be written like this:

(w - 5)(w - 6)

w^2 - 11w + 30
= w^2 - 5w - 6w + 30
= (w^2 - 5w) - (6w - 30)
= w(w - 5) - 6(w - 5)
= (w - 5)(w - 6)

(5 - w)(6 - w) is also a correct answer.

that would be correct 5x6=30 5x-1w=-5w+-1wx6=-6w so -5w+-6w=-11w and finally -1wx-1w=w^2

(W - 5) x (w - 6)

Well, you're technically correct, but it's easier to write like
(w - 5)(w - 6)

...and that's probably the answer that's written in the teacher's edition.