Factorise $x^3 - 125$. The expression $-2x^3 + 7x^2 + ax + b$, where $a$ and $b$ are integers, has a factor of $x - 5$ and a remainder of $-11$ when divided by $2x + 1$. Find the value of $a$ and of $b$. Hence, or otherwise, show that the equation $x^3 - 125 = -2x^3 + 7x^2 + ax + b$ has only one real root.

1. Recognise the expression as a difference of two cubes

$x^3 - 125 = x^3 - 5^3$

2. Apply the difference of cubes formula

$x^3 - 5^3 = (x - 5)(x^2 + 5x + 5^2)$
$= (x - 5)(x^2 + 5x + 25)$

1. Apply the Factor Theorem to form the first equation in $a$ and $b$.

Let $f(x) = -2x^3 + 7x^2 + ax + b$.
Since $x - 5$ is a factor, $f(5) = 0$.
$-2(5)^3 + 7(5)^2 + a(5) + b = 0$
$-2(125) + 7(25) + 5a + b = 0$
$-250 + 175 + 5a + b = 0$
$5a + b = 75 \quad \text{--- (1)}$

2. Apply the Remainder Theorem to form the second equation in $a$ and $b$.

Since the remainder is $-11$ when divided by $2x + 1$, $f\left(-\frac{1}{2}\right) = -11$.
$-2\left(-\frac{1}{2}\right)^3 + 7\left(-\frac{1}{2}\right)^2 + a\left(-\frac{1}{2}\right) + b = -11$
$-2\left(-\frac{1}{8}\right) + 7\left(\frac{1}{4}\right) - \frac{1}{2}a + b = -11$
$\frac{1}{4} + \frac{7}{4} - \frac{1}{2}a + b = -11$
$2 - \frac{1}{2}a + b = -11$
$-\frac{1}{2}a + b = -13$
$-a + 2b = -26 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find $a$ and $b$.

From (1), $b = 75 - 5a$.
Substitute into (2):
$-a + 2(75 - 5a) = -26$
$-a + 150 - 10a = -26$
$-11a = -176$
$a = 16$
Substitute $a = 16$ back into $b = 75 - 5a$:
$b = 75 - 5(16) = 75 - 80 = -5$

1. Find the values of $a$ and $b$ from part (b).

Let $P(x) = -2x^3 + 7x^2 + ax + b$.
By the Factor Theorem, $P(5) = 0$:
$-2(125) + 7(25) + 5a + b = 0 \implies 5a + b = 75$
By the Remainder Theorem, $P(-\frac{1}{2}) = -11$:
$-2(-\frac{1}{8}) + 7(\frac{1}{4}) - \frac{1}{2}a + b = -11 \implies -\frac{1}{2}a + b = -13 \implies a - 2b = 26$
Solving simultaneously gives $a = 16$ and $b = -5$.

2. Factorise the right-hand side of the equation.

The right-hand side is $-2x^3 + 7x^2 + 16x - 5$.
Since $(x - 5)$ is a factor, by inspection or long division:
$-2x^3 + 7x^2 + 16x - 5 = (x - 5)(-2x^2 - 3x + 1)$

3. Equate both sides using the factorised form from part (a) and rearrange.

From part (a), $x^3 - 125 = (x - 5)(x^2 + 5x + 25)$.
The equation becomes:
$(x - 5)(x^2 + 5x + 25) = (x - 5)(-2x^2 - 3x + 1)$
$(x - 5)(x^2 + 5x + 25) - (x - 5)(-2x^2 - 3x + 1) = 0$
$(x - 5)[(x^2 + 5x + 25) - (-2x^2 - 3x + 1)] = 0$
$(x - 5)(3x^2 + 8x + 24) = 0$

4. Analyse the roots of the resulting equation.

From $(x - 5)(3x^2 + 8x + 24) = 0$, one real root is $x = 5$.
For the quadratic factor $3x^2 + 8x + 24 = 0$, we check the discriminant:
$\Delta = b^2 - 4ac = 8^2 - 4(3)(24) = 64 - 288 = -224$
Since $\Delta < 0$, the equation $3x^2 + 8x + 24 = 0$ has no real roots.
Therefore, $x = 5$ is the only real root.