The given function is:
$ f(x) = 2a x^3 + x^2 - bx + 3a, $
where $a$ and $b$ are integers, and $(x + 2)$ and $(x - 1)$ are factors of $f(x)$.
(a) Show that $a = 2$ and find the value of $b$
Since $(x + 2)$ and $(x - 1)$ are factors of $f(x)$, substituting $x = -2$ and $x = 1$ into $f(x)$ must result in 0.
Substitute $x = -2$:
$ f(-2) = 2a(-2)^3 + (-2)^2 - b(-2) + 3a = 0. $
Simplify:
$ f(-2) = 2a(-8) + 4 + 2b + 3a = 0. $
$ f(-2) = -16a + 4 + 2b + 3a = 0. $
$ f(-2) = -13a + 4 + 2b = 0. $
Rearrange:
$ 2b = 13a - 4. \tag{1} $
Substitute $x = 1$:
$ f(1) = 2a(1)^3 + (1)^2 - b(1) + 3a = 0. $
Simplify:
$ f(1) = 2a + 1 - b + 3a = 0. $
$ f(1) = 5a + 1 - b = 0. $
Rearrange:
$ b = 5a + 1. \tag{2} $
Solve the equations for $a$ and $b$:
From equation $(2)$, substitute $b = 5a + 1$ into equation $(1)$:
$ 2(5a + 1) = 13a - 4. $
Simplify:
$ 10a + 2 = 13a - 4. $
$ 10a - 13a = -4 - 2. $
$ -3a = -6 \implies a = 2. $
Substitute $a = 2$ into $b = 5a + 1$:
$ b = 5(2) + 1 = 10 + 1 = 11. $
Final Results for (a):
$ a = 2, \quad b = 11. $
(b) Factorise $f(x)$ completely
Substitute $a = 2$ and $b = 11$ into $f(x)$:
$ f(x) = 2(2)x^3 + x^2 - 11x + 3(2). $
$ f(x) = 4x^3 + x^2 - 11x + 6. $
We know that $(x + 2)$ and $(x - 1)$ are factors of $f(x)$. Therefore, divide $f(x)$ by $(x + 2)(x - 1)$.
$ (x + 2)(x - 1) = x^2 + x - 2. $
Perform synthetic or polynomial division:
$ \frac{4x^3 + x^2 - 11x + 6}{x^2 + x - 2}. $
The quotient is:
$ 4x - 3. $
Thus, $f(x)$ factorises as:
$ f(x) = (x + 2)(x - 1)(4x - 3). $
(c) Solve $h(y) = 0$
The function $h(y)$ is:
$ h(y) = 2^{3y + 2} + 2^{2y} - 11(2^y) + 6. $
Let $z = 2^y$, so:
$ 2^{3y + 2} = 4z^3, \quad 2^{2y} = z^2, \quad 2^y = z. $
Substitute into $h(y)$:
$ h(y) = 4z^3 + z^2 - 11z + 6. $
Solve:
$ 4z^3 + z^2 - 11z + 6 = 0. $
Using trial and error or synthetic division, we find that $z = 1$ is a root. Factorise:
$ 4z^3 + z^2 - 11z + 6 = (z - 1)(4z^2 + 5z - 6). $
Factorise $4z^2 + 5z - 6$:
$ 4z^2 + 5z - 6 = (4z - 3)(z + 2). $
Thus:
$ h(y) = (z - 1)(4z - 3)(z + 2). $
Substitute back $z = 2^y$:
$ (2^y - 1)(4(2^y) - 3)(2^y + 2) = 0. $
Solve each factor:
- $2^y - 1 = 0 \implies 2^y = 1 \implies y = 0.$
- $4(2^y) - 3 = 0 \implies 2^y = \frac{3}{4} \implies y = \log_2\left(\frac{3}{4}\right) \approx -0.415.$
- $2^y + 2 = 0 \implies 2^y = -2$ (not possible as $2^y > 0$).
Final Answers:
- (a) $a = 2, \, b = 11.$
- (b) $f(x) = (x + 2)(x - 1)(4x - 3).$
- (c) $y = 0$ or $y \approx -0.415.$
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