f(x) = 2x^3 - 5x^2 + ax + 18 where a is a constant - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 4

Now that we know a = -9, we can rewrite f(x):

$f(x) = 2x^3 - 5x^2 - 9x + 18$

To factor f(x), we will use synthetic division by (x - 3):

1. Performing synthetic division:

• 3 | 2 -5 -9 18 | 6 3 -18

  2 1 -6 0

This gives us:

$f(x) = (x - 3)(2x^2 + x - 6)$

2. Now we need to factor the quadratic 2x^2 + x - 6:

Using the quadratic formula, x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}:

• Here a = 2, b = 1, c = -6.
• Discriminant: b^2 - 4ac = 1^2 - 4(2)(-6) = 1 + 48 = 49,
• Roots are:

x = \frac{6}{4} = \frac{3}{2},\ x = \frac{-8}{4} = -2$$ Thus, we factor 2x^2 + x - 6 as: $$f(x) = (x - 3)(2x + 5)(x + 2)$$

Given: g(y) = 2(3^y) - 5(3^{2y}) - 9(3^y) + 18

To solve for g(y) = 0, we can substitute 3^y with z:

= -5z^2 - 7z + 18$$ Setting g(z) = 0 yields: $$-5z^2 - 7z + 18 = 0\ 5z^2 + 7z - 18 = 0$$ Using the quadratic formula: $$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4(5)(-18)}}{2(5)}\ = \frac{-7 \pm \sqrt{49 + 360}}{10}\ = \frac{-7 \pm \sqrt{409}}{10}\ = \frac{-7 \pm 20.223}{10}$$ So we find: 1. $$z_1 = \frac{13.223}{10} = 1.3223$$ 2. $$z_2 = \frac{-27.223}{10} = -2.7223$$ Returning to our variable, recalling that z = 3^y: 1. For $z_1$: $$3^y = 1.3223\y = \log_3(1.3223) \approx 0.374 (to 2 decimal places)$$ 2. For $z_2$: $$3^y = -2.7223\ y = \text{undefined since there are no real solutions.}$$ Thus the valid solution is: $$y \approx 0.37$$