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## Sagot :

We evaluate each candidate factor, and if the remainder of the synthetic division is zero, then the candidate is a root of the polynomial. Let's walk through the synthetic division process for each valid root.

### Step-by-Step Synthetic Division Process

#### Testing [tex]\( x = -4 \)[/tex]:

1. Write down the coefficients of [tex]\( p(x) \)[/tex]: [tex]\( 1, 4, -7, -22, 24 \)[/tex].

2. Set up the synthetic division:

[tex]\[ \begin{array}{r|rrrrr} -4 & 1 & 4 & -7 & -22 & 24 \\ & & -4 & 0 & 28 & -24 \\ \hline & 1 & 0 & -7 & 6 & 0 \\ \end{array} \][/tex]

Since the remainder is [tex]\( 0 \)[/tex], [tex]\( x = -4 \)[/tex] is a root.

#### Testing [tex]\( x = -3 \)[/tex]:

1. Using the same coefficients [tex]\( 1, 4, -7, -22, 24 \)[/tex]:

2. Synthetic division setup:

[tex]\[ \begin{array}{r|rrrrr} -3 & 1 & 4 & -7 & -22 & 24 \\ & & -3 & -3 & 30 & -24 \\ \hline & 1 & 1 & -10 & 8 & 0 \\ \end{array} \][/tex]

The remainder [tex]\( 0 \)[/tex], so [tex]\( x = -3 \)[/tex] is a root.

#### Testing [tex]\( x = 1 \)[/tex]:

1. Using the coefficients [tex]\( 1, 4, -7, -22, 24 \)[/tex]:

2. Synthetic division setup:

[tex]\[ \begin{array}{r|rrrrr} 1 & 1 & 4 & -7 & -22 & 24 \\ & & 1 & 5 & -2 & -24 \\ \hline & 1 & 5 & -2 & -24 & 0 \\ \end{array} \][/tex]

The remainder [tex]\( 0 \)[/tex], so [tex]\( x = 1 \)[/tex] is a root.

#### Testing [tex]\( x = 2 \)[/tex]:

1. Using the coefficients [tex]\( 1, 4, -7, -22, 24 \)[/tex]:

2. Synthetic division setup:

[tex]\[ \begin{array}{r|rrrrr} 2 & 1 & 4 & -7 & -22 & 24 \\ & & 2 & 12 & 10 & -24 \\ \hline & 1 & 6 & 5 & -12 & 0 \\ \end{array} \][/tex]

The remainder [tex]\( 0 \)[/tex], so [tex]\( x = 2 \)[/tex] is a root.

### Conclusion

We have identified through synthetic division that the factors of the polynomial [tex]\( p(x) \)[/tex] are [tex]\( x = -4, -3, 1, 2 \)[/tex].

Thus, the correct roots using synthetic division are:

[tex]\[ \boxed{x = -4, x = -3, x = 1, x = 2} \][/tex]