Respuesta :
Step-by-step explanation:
f ( 4 ) = 64 + 32 - 64 - 32 = 0
Thus x - 4 is a factor
Find other factors by synthetic division.
"""""""|1"""""2""""- 16"""""-32
"""""4|"""""""4""""""24""""""32
"""""""|1"""""6"" """""8""""""0
( x - 4 ) ( x ² + 6x + 8 ) = 0
( x - 4 ) ( x + 4 ) ( x + 2 ) = 0
x = - 2 , x = - 4, x = 4
Using the Factor Theorem, it is found that these following binomials are factors of the function:
[tex]x + 2[/tex]
[tex]x - 4[/tex]
[tex]x + 4[/tex]
The Factor Theorem states that if [tex]x_1, x_2, ..., x_n[/tex] are roots of a function f(x), then the binomials [tex](x - x_1), (x - x_2), ..., (x - x_3)[/tex] are factors of the function.
In this problem, the function is:
[tex]f(x) = x^3 + 2x^2 - 16x - 32[/tex]
One of the roots is x = -2, since [tex]f(-2) = 0[/tex], which means the binomial x + 2 is one factor, and the following equality holds true:
[tex](x + 2)(ax^2 + bx + c) = x^3 + 2x^2 - 16x - 32[/tex]
[tex]ax^3 + (2a + b)x^2 + (c + 2b)x + 2c = x^3 + 2x^2 - 16x - 32[/tex]
Thus:
[tex]a = 1[/tex]
[tex]2c = -32 \rightarrow c = -16[/tex]
[tex]b + 2a = 2 \rightarrow b = 0[/tex]
Thus, the other roots are given by:
[tex]x^2 - 16 = 0[/tex]
[tex]x^2 = 16[/tex]
[tex]x = \pm \sqrt{16}[/tex]
[tex]x = \pm 4[/tex]
Then, (x - 4) and (x + 4) are also binomial factors.
A similar problem is given at https://brainly.com/question/22340144
