The polynomial $P(x) = 5x^3 + ax^2 - x + b$, where $a$ and $b$ are constants is exactly divisible by $x^2 + 4x + 3$. Show that the value of $a = 16$ and find the value of $b$. Hence solve the equation $P(x) = 0$. Using a suitable substitution and your answers in (b), solve the equation $5x^6 + ax^4 - x^2 + b = 0$.

1. Factorise the quadratic divisor to find the roots.

Let the divisor be $D(x) = x^2 + 4x + 3$.
$x^2 + 4x + 3 = (x + 1)(x + 3)$

2. Apply the Factor Theorem using the first root.

Since $P(x)$ is exactly divisible by $(x + 1)(x + 3)$, by the Factor Theorem, $P(-1) = 0$.
$5(-1)^3 + a(-1)^2 - (-1) + b = 0$
$-5 + a + 1 + b = 0$
$a + b = 4 \quad \text{--- (1)}$

3. Apply the Factor Theorem using the second root.

Similarly, $P(-3) = 0$.
$5(-3)^3 + a(-3)^2 - (-3) + b = 0$
$-135 + 9a + 3 + b = 0$
$9a + b = 132 \quad \text{--- (2)}$

4. Solve the simultaneous equations to show $a = 16$ and find $b$.

Subtracting (1) from (2):
$(9a + b) - (a + b) = 132 - 4$
$8a = 128$
$a = 16 \quad \text{(shown)}$

Substitute $a = 16$ into (1):
$16 + b = 4$
$b = -12$

1. State the full polynomial using the values of $a$ and $b$ from part (a)

From part (a), $a = 16$ and $b = -12$.
$P(x) = 5x^3 + 16x^2 - x - 12$

2. Find the remaining linear factor of $P(x)$

Since $P(x)$ is exactly divisible by $x^2 + 4x + 3$, we can write:
$5x^3 + 16x^2 - x - 12 = (x^2 + 4x + 3)(cx + d)$
By comparing the coefficients of $x^3$ and the constant terms:
$c = 5$
$3d = -12 \implies d = -4$
So, $P(x) = (x^2 + 4x + 3)(5x - 4)$

3. Factorise the quadratic factor completely

$x^2 + 4x + 3 = (x + 1)(x + 3)$
Thus, $P(x) = (x + 1)(x + 3)(5x - 4)$

4. Solve the equation $P(x) = 0$

$(x + 1)(x + 3)(5x - 4) = 0$
$x = -1$, $x = -3$, or $x = \frac{4}{5}$

1. Identify the substitution to relate the new equation to $P(x) = 0$.

Let $y = x^2$.
The equation $5x^6 + ax^4 - x^2 + b = 0$ becomes $5y^3 + ay^2 - y + b = 0$, which is $P(y) = 0$.

2. Use the roots from part (b) to find the possible values of $y$ and solve for $x$.

From (b), the roots of $P(y) = 0$ are $y = -1$, $y = -3$, and $y = \frac{4}{5}$.
Since $y = x^2$, we have:
$x^2 = -1$ (no real solution)
$x^2 = -3$ (no real solution)
$x^2 = \frac{4}{5} \implies x = \pm \sqrt{\frac{4}{5}} = \pm \frac{2}{\sqrt{5}} = \pm \frac{2\sqrt{5}}{5}$