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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!

Exercice n°2: On donne C = -3 × (4 - 5x) + (4 - 5x)(-3 - 4x). 1) Développer et réduire C. 2) Factoriser C. 3) Calculer C pour: a) x = -2, b) x = 2/3.

In this exercise, expand, factor, and calculate the quadratic expression C = -3 × (4 - 5x) + (4 - 5x)(-3 - 4x). Detailed steps include simplifying the expression, factoring it, and finding C for x = -2 and x = 2/3. Suitable for students in Grades 9-12.

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