Answer :
Given that [tex]\((x-3)^2\)[/tex] is a factor of the polynomial [tex]\(x^3 + px^2 + 3x + q\)[/tex], let's start by expressing the polynomial in terms of its factors.
Since [tex]\((x-3)^2\)[/tex] is a factor, we know that the polynomial can be written as:
[tex]\[ P(x) = (x-3)^2 (x - a) \][/tex]
where [tex]\(a\)[/tex] is another root of the polynomial. To determine [tex]\(p\)[/tex] and [tex]\(q\)[/tex], we need to expand this product and match the coefficients with the given polynomial [tex]\(x^3 + px^2 + 3x + q\)[/tex].
First, we'll expand [tex]\((x-3)^2\)[/tex]:
[tex]\[ (x-3)^2 = (x-3)(x-3) = x^2 - 6x + 9 \][/tex]
Now multiply [tex]\((x-3)^2\)[/tex] by [tex]\((x - a)\)[/tex]:
[tex]\[ P(x) = (x^2 - 6x + 9)(x - a) \][/tex]
Expanding this, we get:
[tex]\[ P(x) = x^3 - ax^2 - 6x^2 + 6ax + 9x - 9a \][/tex]
[tex]\[ P(x) = x^3 + (-a - 6)x^2 + (6a + 9)x - 9a \][/tex]
We can now compare this expanded polynomial to the given polynomial [tex]\(x^3 + px^2 + 3x + q\)[/tex]:
[tex]\[ x^3 + (-a - 6)x^2 + (6a + 9)x - 9a = x^3 + px^2 + 3x + q \][/tex]
By comparing the coefficients, we obtain the following system of equations:
1. Coefficient of [tex]\(x^2\)[/tex]:
[tex]\[ p = -a - 6 \][/tex]
2. Coefficient of [tex]\(x\)[/tex]:
[tex]\[ 6a + 9 = 3 \][/tex]
3. Constant term:
[tex]\[ q = -9a \][/tex]
Let's solve these equations step by step.
From the second equation:
[tex]\[ 6a + 9 = 3 \][/tex]
[tex]\[ 6a = 3 - 9 \][/tex]
[tex]\[ 6a = -6 \][/tex]
[tex]\[ a = -1 \][/tex]
Using [tex]\(a = -1\)[/tex] in the first equation:
[tex]\[ p = -a - 6 \][/tex]
[tex]\[ p = -(-1) - 6 \][/tex]
[tex]\[ p = 1 - 6 \][/tex]
[tex]\[ p = -5 \][/tex]
Using [tex]\(a = -1\)[/tex] in the third equation:
[tex]\[ q = -9a \][/tex]
[tex]\[ q = -9(-1) \][/tex]
[tex]\[ q = 9 \][/tex]
Therefore, the values of the constants are:
[tex]\[ p = -5 \quad \text{and} \quad q = 9 \][/tex]
Now, we will find the third factor. We already know that the polynomial can be written as:
[tex]\[ P(x) = (x-3)^2(x + 1) \][/tex]
So the third factor is:
[tex]\[ x + 1 \][/tex]
Thus, the polynomial [tex]\(x^3 + px^2 + 3x + q\)[/tex] can be factorized as:
[tex]\[ P(x) = (x-3)^2(x + 1) \][/tex]
In conclusion, the values of the constants are [tex]\(p = -5\)[/tex] and [tex]\(q = 9\)[/tex], and the third factor is [tex]\(x + 1\)[/tex].
Since [tex]\((x-3)^2\)[/tex] is a factor, we know that the polynomial can be written as:
[tex]\[ P(x) = (x-3)^2 (x - a) \][/tex]
where [tex]\(a\)[/tex] is another root of the polynomial. To determine [tex]\(p\)[/tex] and [tex]\(q\)[/tex], we need to expand this product and match the coefficients with the given polynomial [tex]\(x^3 + px^2 + 3x + q\)[/tex].
First, we'll expand [tex]\((x-3)^2\)[/tex]:
[tex]\[ (x-3)^2 = (x-3)(x-3) = x^2 - 6x + 9 \][/tex]
Now multiply [tex]\((x-3)^2\)[/tex] by [tex]\((x - a)\)[/tex]:
[tex]\[ P(x) = (x^2 - 6x + 9)(x - a) \][/tex]
Expanding this, we get:
[tex]\[ P(x) = x^3 - ax^2 - 6x^2 + 6ax + 9x - 9a \][/tex]
[tex]\[ P(x) = x^3 + (-a - 6)x^2 + (6a + 9)x - 9a \][/tex]
We can now compare this expanded polynomial to the given polynomial [tex]\(x^3 + px^2 + 3x + q\)[/tex]:
[tex]\[ x^3 + (-a - 6)x^2 + (6a + 9)x - 9a = x^3 + px^2 + 3x + q \][/tex]
By comparing the coefficients, we obtain the following system of equations:
1. Coefficient of [tex]\(x^2\)[/tex]:
[tex]\[ p = -a - 6 \][/tex]
2. Coefficient of [tex]\(x\)[/tex]:
[tex]\[ 6a + 9 = 3 \][/tex]
3. Constant term:
[tex]\[ q = -9a \][/tex]
Let's solve these equations step by step.
From the second equation:
[tex]\[ 6a + 9 = 3 \][/tex]
[tex]\[ 6a = 3 - 9 \][/tex]
[tex]\[ 6a = -6 \][/tex]
[tex]\[ a = -1 \][/tex]
Using [tex]\(a = -1\)[/tex] in the first equation:
[tex]\[ p = -a - 6 \][/tex]
[tex]\[ p = -(-1) - 6 \][/tex]
[tex]\[ p = 1 - 6 \][/tex]
[tex]\[ p = -5 \][/tex]
Using [tex]\(a = -1\)[/tex] in the third equation:
[tex]\[ q = -9a \][/tex]
[tex]\[ q = -9(-1) \][/tex]
[tex]\[ q = 9 \][/tex]
Therefore, the values of the constants are:
[tex]\[ p = -5 \quad \text{and} \quad q = 9 \][/tex]
Now, we will find the third factor. We already know that the polynomial can be written as:
[tex]\[ P(x) = (x-3)^2(x + 1) \][/tex]
So the third factor is:
[tex]\[ x + 1 \][/tex]
Thus, the polynomial [tex]\(x^3 + px^2 + 3x + q\)[/tex] can be factorized as:
[tex]\[ P(x) = (x-3)^2(x + 1) \][/tex]
In conclusion, the values of the constants are [tex]\(p = -5\)[/tex] and [tex]\(q = 9\)[/tex], and the third factor is [tex]\(x + 1\)[/tex].