Let's solve the given exercise step by step.

We are given:

$C = -3 \times (4 - 5x) + (4 - 5x) \times (-3 - 4x)$

1) Développer et réduire C (Expand and simplify $C$):

We will first expand the terms and then simplify.

$C = -3(4 - 5x) + (4 - 5x)(-3 - 4x)$

Expanding:

• First term: $-3(4 - 5x)$: $-3(4 - 5x) = -12 + 15x$

• Second term: $(4 - 5x)(-3 - 4x)$ (distribute both terms): $(4 - 5x)(-3 - 4x) = 4(-3) + 4(-4x) + (-5x)(-3) + (-5x)(-4x)$ $= -12 - 16x + 15x + 20x^2$

Now, combine the expanded terms: $C = (-12 + 15x) + (-12 - 16x + 15x + 20x^2)$ $C = -12 + 15x - 12 - 16x + 15x + 20x^2$

Simplify: $C = 20x^2 + (15x - 16x + 15x) - 12 - 12$ $C = 20x^2 + 14x - 24$

So, the simplified form of $C$ is: $C = 20x^2 + 14x - 24$

2) Factoriser $C$ (Factor $C$):

We have the quadratic expression: $C = 20x^2 + 14x - 24$

To factor this, we can first factor out the greatest common factor (GCF) of the coefficients, which is 2: $C = 2(10x^2 + 7x - 12)$

Now, we will factor the quadratic $10x^2 + 7x - 12$ using the splitting middle term method.

We need to find two numbers that multiply to $10 \times (-12) = -120$ and add to 7. The numbers are 15 and -8, since: $15 \times (-8) = -120 \quad \text{and} \quad 15 + (-8) = 7$

We split the middle term: $10x^2 + 15x - 8x - 12$

Group the terms: $(10x^2 + 15x) - (8x + 12)$

Factor each group: $5x(2x + 3) - 4(2x + 3)$

Now, factor out the common binomial factor $(2x + 3)$: $(2x + 3)(5x - 4)$

Thus, the fully factored form of $C$ is: $C = 2(2x + 3)(5x - 4)$

3) Calculer $C$ pour $x = -2$ et $x = \frac{2}{3}$:

a) When $x = -2$:

Substitute $x = -2$ into the factored form: $C = 2(2(-2) + 3)(5(-2) - 4)$ $= 2((-4) + 3)(-10 - 4)$ $= 2(-1)(-14) = 2(14) = 28$

So, $C = 28$ when $x = -2$.

b) When $x = \frac{2}{3}$:

Substitute $x = \frac{2}{3}$ into the factored form: $C = 2\left(2\left(\frac{2}{3}\right) + 3\right)\left(5\left(\frac{2}{3}\right) - 4\right)$ $= 2\left(\frac{4}{3} + 3\right)\left(\frac{10}{3} - 4\right)$ $= 2\left(\frac{4}{3} + \frac{9}{3}\right)\left(\frac{10}{3} - \frac{12}{3}\right)$ $= 2\left(\frac{13}{3}\right)\left(\frac{-2}{3}\right)$ $= 2 \times \frac{-26}{9} = \frac{-52}{9}$

So, $C = \frac{-52}{9}$ when $x = \frac{2}{3}$.

Would you like further details on any of these steps, or do you have questions? Here are some related problems:

1. Expand and factor: $C = 3(2x + 5) - 4(3x - 2)$.
2. Solve $2x^2 + 5x - 12 = 0$ using the quadratic formula.
3. Calculate $C = 5x^2 - 7x + 3$ for $x = 1$.
4. Factor $15x^2 - 14x + 3$.
5. Expand $(2x + 3)(4x - 5)$.

Tip: Always check for a greatest common factor (GCF) before factoring quadratic equations!