Factorise fully: (x^2-16) - (x-4)(3x+5)

Thank you

Knowed answer but not how to do it.

Re: Factorise fully: (x^2-16) - (x-4)(3x+5)

OK...this one isn't too difficult.....note that the first thing is just a difference of two squares, so we can write it as (x+4) (x-4)   

And we have

(x+4) (x-4)  - (x-4) (3x-5)

Since (x-4) is common to both terms, we can factor it out, and this gives us

(x-4) [(x+4) - (3x -5)]  Now, simplifying the stuff in the brackets, we have

(x-4) (9 - 2x)

And that's it !!!

OK...this one isn't too difficult.....note that the first thing is just a difference of two squares, so we can write it as (x+4) (x-4)   

And we have

(x+4) (x-4)  - (x-4) (3x-5)

Since (x-4) is common to both terms, we can factor it out, and this gives us

(x-4) [(x+4) - (3x -5)]  Now, simplifying the stuff in the brackets, we have

(x-4) (9 - 2x)

Above solution is incorrect

Correct is solution

Thanks, Anonymous...I made a slight error in transcribing the problem.....sorry!!

Let me correct this

(x+4) (x-4)  - (x-4) (3x+5)

Since (x-4) is common to both terms, we can factor it out, and this gives us

(x-4) [(x+4) - (3x + 5)]  Now, simplifying the stuff in the brackets, we have

(x-4) (-1 - 2x) =

(x - 4) (1 + 2x) (-1) =

(-1) (x - 4) (1 + 2x)    And multiplying (-1) across (x-4) we end up with

(4 - x) (1 + 2x)

And that's it !!!