Let's solve the two questions in the image step by step.

Question 1: Find factors of $x^3 - 7x - 6$

We want to factor $x^3 - 7x - 6$. To do this, let's first attempt synthetic division or trial and error using rational roots.

Step 1: Check Rational Roots

Possible rational roots are factors of $-6$ (constant term) over factors of 1 (leading coefficient of $x^3$), i.e., $\pm 1, \pm 2, \pm 3, \pm 6$.

We'll try substituting these values into the equation to check if they are roots.

• Try $x = -1$: $(-1)^3 - 7(-1) - 6 = -1 + 7 - 6 = 0$ So, $x = -1$ is a root.

Step 2: Polynomial Division

Since $x = -1$ is a root, one factor is $x + 1$. Now, divide $x^3 - 7x - 6$ by $x + 1$ using synthetic or polynomial division:

Dividing $x^3 - 7x - 6$ by $x + 1$, we get $x^2 - x - 6$.

Step 3: Factor the Quadratic

Now, factor $x^2 - x - 6$: $x^2 - x - 6 = (x - 3)(x + 2)$

So, the full factorization is: $x^3 - 7x - 6 = (x + 1)(x - 3)(x + 2)$

The correct option for Question 1 is C: $(x - 3)(x + 2)(x + 1)$.

Question 2: Find factors of $x^3 + 12x^2 - 19x = -20$

First, rewrite the equation as: $x^3 + 12x^2 - 19x + 20 = 0$

Step 1: Check Rational Roots

The possible rational roots are factors of $20$ over factors of $1$, i.e., $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$.

• Try $x = 1$: $1^3 + 12(1)^2 - 19(1) + 20 = 1 + 12 - 19 + 20 = 14 \quad (\text{not a root})$

• Try $x = -1$: $(-1)^3 + 12(-1)^2 - 19(-1) + 20 = -1 + 12 + 19 + 20 = 50 \quad (\text{not a root})$

• Try $x = 5$: $5^3 + 12(5)^2 - 19(5) + 20 = 125 + 300 - 95 + 20 = 350 \quad (\text{not a root})$

• Try $x = -5$: [ (-5)^3 + 12(-5)^2 - 19(-5) + 20 = -125 + 300 + 95 + 20 = 290 \quad (\text{wrong})