3. (a) (i) Show that $(x + 1)$ is a factor of $2x^3 - 9x^2 + 3x + 14$ - Scottish Highers Maths - Question 3 - 2016

Given that $(x + 1)$ is a factor, we can use polynomial division to factor the cubic equation:

1. Dividing $2x^3 - 9x^2 + 3x + 14$ by $(x + 1)$ results in:
   • The quotient is $2x^2 - 11x + 14$.
2. We can set this quadratic equation to zero: $2x^2 - 11x + 14 = 0$
3. Applying the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -11$, and $c = 14$: $x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 14}}{2 \cdot 2}$ $= \frac{11 \pm \sqrt{121 - 112}}{4}$ $= \frac{11 \pm 3}{4}$ This yields (x = 3.5) and (x = 2).
4. Therefore, the three roots are $x = -1$, $x = 2$, and $x = 3.5$.

The coordinates of point A, where the curve intersects the x-axis at $x = -1$, are $(-1, 0)$. The coordinates of point B, where the curve intersects at $x = 2$, are $(2, 0)$.

To find the shaded area under the curve between points A and B:

1. We integrate the function $y = 2x^3 - 9x^2 + 3x + 14$ from $x = -1$ to $x = 2$: $\int_{-1}^{2} (2x^3 - 9x^2 + 3x + 14) dx$.
2. Calculating this integral step-by-step:
   • The antiderivative is: $\frac{2}{4}x^4 - \frac{9}{3}x^3 + \frac{3}{2}x^2 + 14x$ $= \frac{1}{2}x^4 - 3x^3 + \frac{3}{2}x^2 + 14x$.
3. Evaluating this from $-1$ to $2$ results in: $\left[ \frac{1}{2}(2^4) - 3(2^3) + \frac{3}{2}(2^2) + 14(2) \right] - \left[ \frac{1}{2}(-1^4) - 3(-1^3) + \frac{3}{2}(-1^2) + 14(-1) \right].$
   • Calculating the definite values gives: $= \left[ 8 - 24 + 6 + 28 \right] - \left[ 0 + 3 + \frac{3}{2} - 14 \right]$ $= 18 - (-10.5) = 28.5.$
4. Therefore, the shaded area is 27 square units.