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Sagot :
Sure, let's go through the detailed steps to solve the division problem: [tex]\((3x^3 + 2x^2 - 5x + 1) \div (3x + 1)\)[/tex].
### Step-by-Step Solution:
1. Setup the Division:
We need to divide the polynomial [tex]\(3x^3 + 2x^2 - 5x + 1\)[/tex] by the polynomial [tex]\(3x + 1\)[/tex].
2. First Division:
- Take the leading term of the numerator [tex]\(3x^3\)[/tex] and the leading term of the denominator [tex]\(3x\)[/tex].
- Divide [tex]\(3x^3\)[/tex] by [tex]\(3x\)[/tex]:
[tex]\[ \frac{3x^3}{3x} = x^2 \][/tex]
- Multiply [tex]\(x^2\)[/tex] by [tex]\(3x + 1\)[/tex]:
[tex]\[ x^2 \cdot (3x + 1) = 3x^3 + x^2 \][/tex]
- Subtract this result from the original polynomial:
[tex]\[ 3x^3 + 2x^2 - 5x + 1 - (3x^3 + x^2) = (2x^2 - x^2) - 5x + 1 = x^2 - 5x + 1 \][/tex]
3. Second Division:
- Take the new numerator [tex]\(x^2 - 5x + 1\)[/tex] and the leading term of the denominator [tex]\(3x\)[/tex].
- Divide [tex]\(x^2\)[/tex] by [tex]\(3x\)[/tex]:
[tex]\[ \frac{x^2}{3x} = \frac{1}{3}x \][/tex]
- Multiply [tex]\(\frac{1}{3}x\)[/tex] by [tex]\(3x + 1\)[/tex]:
[tex]\[ \frac{1}{3}x \cdot (3x + 1) = x^2 + \frac{1}{3}x \][/tex]
- Subtract this result from the current numerator:
[tex]\[ x^2 - 5x + 1 - (x^2 + \frac{1}{3}x) = -5x - \frac{1}{3}x + 1 = \left(-\frac{15}{3}x - \frac{1}{3}x\right) + 1 = -\frac{16}{3}x + 1 \][/tex]
4. Third Division:
- Take the new numerator [tex]\( -\frac{16}{3}x + 1 \)[/tex] and the leading term of the denominator [tex]\(3x\)[/tex].
- Divide [tex]\( -\frac{16}{3}x\)[/tex] by [tex]\(3x\)[/tex]:
[tex]\[ \frac{-\frac{16}{3}x}{3x} = -\frac{16}{9} \][/tex]
- Multiply [tex]\(-\frac{16}{9}\)[/tex] by [tex]\(3x + 1\)[/tex]:
[tex]\[ -\frac{16}{9} \cdot (3x + 1) = -\frac{48}{9}x - \frac{16}{9} \][/tex]
- Subtract this result from the current numerator:
[tex]\[ -\frac{16}{3}x + 1 - \left( -\frac{48}{9}x - \frac{16}{9} \right) = -\frac{16}{3} x - (-\frac{48}{9}x) + 1 - (-\frac{16}{9}) = \frac{-16}{3}x + \frac{48}{9}x + 1 + \frac{16}{9} \][/tex]
- Simplify:
[tex]\[ \left( -\frac{48}{9} + \frac{48}{9} \right) x + \left( 1 + \frac{16}{9} \right) = \frac{25}{9} \][/tex]
Thus, the quotient is:
[tex]\[ x^2 + \frac{1}{3}x - \frac{16}{9} \][/tex]
And the remainder is:
[tex]\[ \frac{25}{9} \][/tex]
### Final Result:
The quotient of the division [tex]\((3x^3 + 2x^2 - 5x + 1) \div (3x + 1)\)[/tex] is:
[tex]\[ x^2 + \frac{x}{3} - \frac{16}{9} \][/tex]
And the remainder is:
[tex]\[ \frac{25}{9} \][/tex]
### Step-by-Step Solution:
1. Setup the Division:
We need to divide the polynomial [tex]\(3x^3 + 2x^2 - 5x + 1\)[/tex] by the polynomial [tex]\(3x + 1\)[/tex].
2. First Division:
- Take the leading term of the numerator [tex]\(3x^3\)[/tex] and the leading term of the denominator [tex]\(3x\)[/tex].
- Divide [tex]\(3x^3\)[/tex] by [tex]\(3x\)[/tex]:
[tex]\[ \frac{3x^3}{3x} = x^2 \][/tex]
- Multiply [tex]\(x^2\)[/tex] by [tex]\(3x + 1\)[/tex]:
[tex]\[ x^2 \cdot (3x + 1) = 3x^3 + x^2 \][/tex]
- Subtract this result from the original polynomial:
[tex]\[ 3x^3 + 2x^2 - 5x + 1 - (3x^3 + x^2) = (2x^2 - x^2) - 5x + 1 = x^2 - 5x + 1 \][/tex]
3. Second Division:
- Take the new numerator [tex]\(x^2 - 5x + 1\)[/tex] and the leading term of the denominator [tex]\(3x\)[/tex].
- Divide [tex]\(x^2\)[/tex] by [tex]\(3x\)[/tex]:
[tex]\[ \frac{x^2}{3x} = \frac{1}{3}x \][/tex]
- Multiply [tex]\(\frac{1}{3}x\)[/tex] by [tex]\(3x + 1\)[/tex]:
[tex]\[ \frac{1}{3}x \cdot (3x + 1) = x^2 + \frac{1}{3}x \][/tex]
- Subtract this result from the current numerator:
[tex]\[ x^2 - 5x + 1 - (x^2 + \frac{1}{3}x) = -5x - \frac{1}{3}x + 1 = \left(-\frac{15}{3}x - \frac{1}{3}x\right) + 1 = -\frac{16}{3}x + 1 \][/tex]
4. Third Division:
- Take the new numerator [tex]\( -\frac{16}{3}x + 1 \)[/tex] and the leading term of the denominator [tex]\(3x\)[/tex].
- Divide [tex]\( -\frac{16}{3}x\)[/tex] by [tex]\(3x\)[/tex]:
[tex]\[ \frac{-\frac{16}{3}x}{3x} = -\frac{16}{9} \][/tex]
- Multiply [tex]\(-\frac{16}{9}\)[/tex] by [tex]\(3x + 1\)[/tex]:
[tex]\[ -\frac{16}{9} \cdot (3x + 1) = -\frac{48}{9}x - \frac{16}{9} \][/tex]
- Subtract this result from the current numerator:
[tex]\[ -\frac{16}{3}x + 1 - \left( -\frac{48}{9}x - \frac{16}{9} \right) = -\frac{16}{3} x - (-\frac{48}{9}x) + 1 - (-\frac{16}{9}) = \frac{-16}{3}x + \frac{48}{9}x + 1 + \frac{16}{9} \][/tex]
- Simplify:
[tex]\[ \left( -\frac{48}{9} + \frac{48}{9} \right) x + \left( 1 + \frac{16}{9} \right) = \frac{25}{9} \][/tex]
Thus, the quotient is:
[tex]\[ x^2 + \frac{1}{3}x - \frac{16}{9} \][/tex]
And the remainder is:
[tex]\[ \frac{25}{9} \][/tex]
### Final Result:
The quotient of the division [tex]\((3x^3 + 2x^2 - 5x + 1) \div (3x + 1)\)[/tex] is:
[tex]\[ x^2 + \frac{x}{3} - \frac{16}{9} \][/tex]
And the remainder is:
[tex]\[ \frac{25}{9} \][/tex]
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