Certainly! Let's address the problem step-by-step.
Part (a): Show that [tex]\((x+3)\)[/tex] is a factor of [tex]\(f(x)\)[/tex] using the factor theorem.
The factor theorem states that [tex]\((x + a)\)[/tex] is a factor of a polynomial [tex]\(f(x)\)[/tex] if and only if [tex]\(f(-a) = 0\)[/tex]. In our case, we need to show that [tex]\((x + 3)\)[/tex] is a factor by proving that [tex]\(f(-3) = 0\)[/tex].
The polynomial given is:
[tex]\[ f(x) = -6x^3 - 7x^2 + 40x + 21 \][/tex]
Substitute [tex]\(x = -3\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(-3) = -6(-3)^3 - 7(-3)^2 + 40(-3) + 21 \][/tex]
Let's evaluate this step by step:
[tex]\[ (-3)^3 = -27 \][/tex]
[tex]\[ -6(-27) = -6 \cdot -27 = 162 \][/tex]
[tex]\[ (-3)^2 = 9 \][/tex]
[tex]\[ -7(9) = -63 \][/tex]
[tex]\[ 40(-3) = -120 \][/tex]
Therefore:
[tex]\[ f(-3) = 162 - 63 - 120 + 21 \][/tex]
Combine the terms:
[tex]\[ 162 - 63 = 99 \][/tex]
[tex]\[ 99 - 120 = -21 \][/tex]
[tex]\[ -21 + 21 = 0 \][/tex]
Since [tex]\(f(-3) = 0\)[/tex], by the factor theorem, [tex]\((x + 3)\)[/tex] is indeed a factor of [tex]\(f(x)\)[/tex].
Part (b): Factorize [tex]\(f(x)\)[/tex] completely.
Given that [tex]\((x + 3)\)[/tex] is a factor, we can use this to factorize the polynomial further. The polynomial can be expressed as:
[tex]\[ f(x) = (x + 3) \cdot q(x) \][/tex]
where [tex]\(q(x)\)[/tex] is another polynomial. To find [tex]\(q(x)\)[/tex], we need to perform polynomial division or factorization.
Since the factored form of the polynomial is already given, we get:
[tex]\[ f(x) = -(x + 3)(2x + 1)(3x - 7) \][/tex]
So the complete factorization of [tex]\(f(x)\)[/tex] is:
[tex]\[ f(x) = -(x + 3)(2x + 1)(3x - 7) \][/tex]
Putting it all together, we have shown that:
1. Using the factor theorem, [tex]\((x + 3)\)[/tex] is a factor of [tex]\(f(x)\)[/tex].
2. The complete factorization of [tex]\(f(x)\)[/tex] is:
[tex]\[ f(x) = -(x + 3)(2x + 1)(3x - 7) \][/tex]
This concludes our solution to the problem.
