Determine the zeros of f(x) = x^3 - 3x^2 - 16x + 48
show work please

Question

Determine the zeros of f(x) = x^3 - 3x^2 - 16x + 48
show work please

apoorvk · Answer

$$=x^2(x-3) - 16(x-3)$$

Can you proceed now?

anonymous · Answer

what is that?

anonymous · Answer

could you plug it in please

apoorvk · Answer

Just proceed like you do it for a normal binomial-binomial factorisation.

anonymous · Answer

factoring by grouping

take largest term you can factor from left side (x^2) and largest term you can factor from right side (16), and factor them out.

anonymous · Answer

zack, look at what @apoorvk 's last line... you have common factors of (x-3)... factor those out...

anonymous · Answer

what now?

anonymous · Answer

=x2(x−3)−16(x−3) what after this? is the answer -4?

anonymous · Answer

You see, you factor by grouping first:
f(x) = x^3 - 3x^2 - 16x + 48
f(x) = x^2(x - 3) - 16(x - 3)
f(x) = (x^2 - 16)(x - 3)
Now, there is a difference in squares pattern in there.
f(x) = (x^2 - 16)(x - 3)
f(x) = (x + 4)(x - 4)(x - 3)
Set it equal to zero.
0 = (x + 4)(x - 4)(x - 3)
Now solve for x in each of these binomials by setting it equal to zero.
x + 4 = 0
x = -4

x - 4 = 0
x = 4

x - 3 = 0
x = 3

Therefore, the roots are x = -4, 4, 3.

anonymous · Answer

thanks

anonymous · Answer

You're welcome. Do you understand it?