1. Determine the zeros of f(x) = x3 – 3x2 – 16x + 48

Question

1.	Determine the zeros of f(x) = x3 – 3x2 – 16x + 48

anonymous · Answer

soryy it took so long @ganeshie8

ganeshie8 · Answer

aha its okay :)

ganeshie8 · Answer

$ f(x) = x^3 – 3x^2 – 16x + 48 $

ganeshie8 · Answer

you have four terms.

ganeshie8 · Answer

look at first two terms. we can pull out x^2  ?

anonymous · Answer

yes so x²(x-3) ?

ganeshie8 · Answer

thats right ! next, look at last two terms, can we pull out -16 ?

ganeshie8 · Answer

$ f(x) = x^3 – 3x^2 – 16x + 48 $
         $ = x^2(x-3) - 16(x - 3)$

anonymous · Answer

okk , I get that part.

ganeshie8 · Answer

good :) next, see (x-3) is common

ganeshie8 · Answer

pull it out

ganeshie8 · Answer

$ f(x) = x^3 – 3x^2 – 16x + 48 $

         $ = x^2(x-3) - 16(x - 3)$

         $ = (x-3)(x^2 - 16)$

anonymous · Answer

(x²-16)(x-3)

ganeshie8 · Answer

Now, to get the zeroes set it to 0

anonymous · Answer

sooo 4 and 3 ?

ganeshie8 · Answer

$(x-3)(x^2 - 16) = 0 $

 $(x-3) = 0 ,   \ \   (x^2 - 16) = 0 $

 $x  = 3 ,   \ \   x^2  = 16 $

 $x  = 3 ,   \ \   x  =  \pm\sqrt{16} $

 $x  = 3 ,   \ \   x  =  4, \ \ \  x = -4 $

ganeshie8 · Answer

3 zeroes, since it is a 3rd degree polynomial

anonymous · Answer

ohhh okk thankyouu veryy much, i understand it now (: . okk im going to try and do the next 3 and ill tag you in them (;

ganeshie8 · Answer

brilliant :)