f(x) = 2x³ - 5x² + ax + 18 where a is a constant - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 4

Now we can factorise f(x) completely. We already know that (x - 3) is a factor.

By synthetic division, we have:

1. Divide f(x) by (x - 3)
2. The result of this division is: $2x^2 + x - 6$

Next, we can factor $2x^2 + x - 6$:

To factor, we need two numbers that multiply to $2 * (-6) = -12$ and add to 1. These numbers are 4 and -3.

This leads us to:

$$= (2x^2 - 3x) + (4x - 6) \\ = x(2x - 3) + 2(2x - 3) = (2x - 3)(x + 2).$$

Thus:

$f(x) = (x - 3)(2x - 3)(x + 2).$

To find the values of y that satisfy g(y) = 0, we set:

$$2(3^y) - 5(3^{2y}) - 9(3^y) + 18 = 0.$$

Rearranging gives:

$$-5(3^{2y}) + (2 - 9)(3^y) + 18 = 0 \\ -5(3^{2y}) - 7(3^y) + 18 = 0.$$

Letting $u = 3^y$, we convert this to:

$$-5u^2 - 7u + 18 = 0.$$

Multiplying through by -1 gives:

$$5u^2 + 7u - 18 = 0.$$

Using the quadratic formula:

$$u = \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}.

Calculating:

$$\sqrt{409} \approx 20.223 \\ = u = \frac{-7 + 20.223}{10} \approx 1.3223 \\ = u = \frac{-7 - 20.223}{10} \approx -2.7223 ext{ (not valid since } 3^y ext{ cannot be negative)} \\$$

Thus:

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Calculating:

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