6. f(x) = -3x³ + 8x² - 9x + 10, x ∈ ℝ (a) (i) Calculate f(2) (ii) Write f(x) as a product of two algebraic factors - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 2

To factor $f(x)$, we can identify the polynomial form:

$f(x) = -3x^3 + 8x^2 - 9x + 10$

We know that $x = 2$ is a root since $f(2) = 0$. Thus, $(x - 2)$ is a factor. We can use polynomial long division to divide:

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

Next, we can further factor the second polynomial:

We need to find two numbers that multiply to $-15$ (the product of -3 and -5) and add to $2$. These numbers are $5$ and $-3$. Therefore, we can rewrite it:

$-3x^2 + 5x - 3x - 5$

Factoring by grouping gives:

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

Thus,

$f(x) = -(x - 2)(3x - 5)(x - 2)$

The final factorization is:

$f(x) = -(x - 2)^2(3x - 5)$.

Using the result from part (a)(ii), we can analyze:

$y = 2$ is a double root from the factor $(y - 2)^2$. We can find the other root by solving:

$3y - 5 = 0$

which simplifies to:

y = rac{5}{3}.

This means the equation has real solutions at $y = 2$ (with multiplicity 2) and y = rac{5}{3}. Since both of these are real, there are exactly two distinct real solutions.

To solve the equation:

$3 an^2θ - 8 anθ + 9 anθ - 10 = 0$,

we first rewrite it as:

$3 an^2θ - 8 anθ - 10 = 0$.

Using the quadratic formula:

anθ = rac{-b ext{±} ext{√}(b^2 - 4ac)}{2a},

where $a = 3$, $b = -8$, and $c = -10$:

1. Calculate the discriminant:

$D = (-8)^2 - 4(3)(-10) = 64 + 120 = 184$,

Since $D > 0$, there are two distinct real solutions.

2. Each solution for $anθ$ corresponds to periodic solutions. The general solution for $anθ = k$ occurs every rac{π}{n} where $n$ is the period, thus from $7π$ to $10π$, which covers a range of 3π, yielding:

$2 ext{ solutions per period of } an$,

Therefore,

$ext{ total solutions } = 2 imes 3 = 6.$

Thus, there are 6 real solutions in the specified interval.