Let's proceed step-by-step to solve the given problem.
### Step A: Finding the values of [tex]\( a, b \)[/tex], and [tex]\( c \)[/tex]
We are given that [tex]\( \left(x^2 - 4\right) \)[/tex] is a factor of the polynomial [tex]\( a x^3 + b x^2 - 4 x + c \)[/tex]. This factor can be written as [tex]\( (x + 2)(x - 2) \)[/tex]. Therefore, we can express the polynomial as:
[tex]\[ P(x) = (x + 2)(x - 2)(a x + k) \][/tex]
for some constant [tex]\( k \)[/tex].
For completeness:
[tex]\[ (x^2 - 4)(a x + k) = a x^3 + k x^2 - 4 a x - 4 k \][/tex]
We can compare this with [tex]\( a x^3 + b x^2 - 4 x + c \)[/tex] and get:
[tex]\[ a x^3 + k x^2 - 4 a x - 4 k = a x^3 + b x^2 - 4 x + c \][/tex]
Equating coefficients, we get:
[tex]\[ k = b \][/tex]
[tex]\[ -4a = -4 \Rightarrow a = 1 \][/tex]
[tex]\[ -4k = c \][/tex]
Thus, [tex]\( c = -4k \)[/tex].
Next, we use the information that [tex]\( P(x) \)[/tex] when divided by [tex]\((x+1)\)[/tex] gives a remainder of 6. Using the Remainder Theorem:
[tex]\[ P(-1) = 6 \][/tex]
Substituting [tex]\( x = -1 \)[/tex] in [tex]\( a x^3 + b x^2 - 4 x + c \)[/tex]:
[tex]\[ P(-1) = 1(-1)^3 + b(-1)^2 - 4(-1) + c \][/tex]
[tex]\[ 6 = -1 + b + 4 + c \][/tex]
[tex]\[ 6 = 3 + b + c \][/tex]
Using [tex]\( c = -4k \)[/tex] and [tex]\( b = k \)[/tex], we get:
[tex]\[ 6 = 3 + k - 4k \][/tex]
[tex]\[ 6 = 3 - 3k \][/tex]
[tex]\[ 3 = -3k \][/tex]
[tex]\[ k = -1 \][/tex]
Thus, [tex]\( k = b = -1 \)[/tex] and [tex]\( c = -4k = 4 \)[/tex].
So, the values are:
[tex]\[ a = 1, \, b = -1, \, c = 4 \][/tex]
### Step B: Factorize the polynomial
Using the values we found, the polynomial is:
[tex]\[ P(x) = x^3 - x^2 - 4x + 4 \][/tex]
We know [tex]\( x^2 - 4 = (x + 2)(x - 2) \)[/tex] is a factor. The remaining factor is:
[tex]\[ P(x) = (x^2 - 4)(x + k) \Rightarrow x^3 - x^2 - 4x + 4 = (x + 2)(x - 2)(x - 1) \][/tex]
Thus, the polynomial factorizes as:
[tex]\[ P(x) = (x + 2)(x - 2)(x - 1) \][/tex]
### Step C: Solve the equation [tex]\( P(x) = 0 \)[/tex]
Set the polynomial equal to zero:
[tex]\[ (x + 2)(x - 2)(x - 1) = 0 \][/tex]
So, the solutions are:
[tex]\[ x = -2, x = 2, x = -1 \][/tex]
### Final results
a) The values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are [tex]\(a = 1\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = 4\)[/tex].
b) The factorized form of the polynomial is:
[tex]\[ (x + 2)(x - 2)(x - 1) \][/tex]
c) The solutions to the equation [tex]\( ax^3 + bx^2 - 4x + c = 0 \)[/tex] are:
[tex]\[ x = -2, x = 2, \text{ and } x = -1 \][/tex]
