Respuesta :
Try out some factors of 72 to see which are roots of f(x).
f(3) = 27 - 45 - 54 + 72 = 0 and f(-3) = -27 - 45 + 54 + 72 = 54,
so 3 is a root and -3 is not. For confirmation, we have f(-4) = 0
and f(6) = 0, so (x - 3)(x + 4)(x - 6)
f(3) = 27 - 45 - 54 + 72 = 0 and f(-3) = -27 - 45 + 54 + 72 = 54,
so 3 is a root and -3 is not. For confirmation, we have f(-4) = 0
and f(6) = 0, so (x - 3)(x + 4)(x - 6)
Answer: The given polynomial can be expressed as (x-3)(x+4)(x-6).
Step-by-step explanation: We are given to express the following cubic polynomial as a product of linear factors :
[tex]f(x)=x^3-5x^2-18x+72.[/tex]
We have the factor theorem that states as follows :
FACTOR THEOREM : If x = a is a zero of the polynomial p(x), then (x - a) is a factor of the polynomial p(x).
If x = 1, then
[tex]f(x)=1^3-5\times1^2-18\times1+72=1-5-18+72=50\neq 0.[/tex]
If x = 2, then
[tex]f(x)=2^3-5\times2^2-18\times2+72=8-20-36+72=24\neq 0.[/tex]
If x = 3, then
[tex]f(x)=3^3-5\times3^2-18\times3+72=27-45-54+72=0.[/tex]
So, x = 3 is a zero of f(x), implies that (x - 3) is a factor of f(x).
Therefore, we have
[tex]f(x)\\\\=x^3-5x^2-18x+72\\\\=x^2(x-3)-2x(x-3)-24(x-3)\\\\=(x-3)(x^2-2x-24)\\\\=(x-3)(x^2-6x+4x-24)\\\\=(x-3)\{x(x-6)+4(x-6)\}\\\\=(x-3)(x+4)(x-6).[/tex]
Thus, the given polynomial can be expressed as (x-3)(x+4)(x-6).