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Sagot :
To find all the zeros of the polynomial function [tex]\( f(x) = 2x^3 + 6x^2 + 2x - 10 \)[/tex], follow these steps:
1. Verify the Zero: We are told that [tex]\( f(1) = 0 \)[/tex]. Verification involves substituting [tex]\( x = 1 \)[/tex] into the polynomial:
[tex]\[ f(1) = 2(1)^3 + 6(1)^2 + 2(1) - 10 = 2 + 6 + 2 - 10 = 0 \][/tex]
Therefore, [tex]\( x = 1 \)[/tex] is indeed a zero of the polynomial.
2. Factor out [tex]\((x - 1)\)[/tex]: Use polynomial division or synthetic division to divide [tex]\( f(x) \)[/tex] by [tex]\( (x - 1) \)[/tex].
Performing synthetic division with the zero [tex]\( x = 1 \)[/tex]:
- Coefficients of [tex]\( f(x) \)[/tex] are: [tex]\( 2, 6, 2, -10 \)[/tex]
[tex]\[ \begin{array}{r|rrrr} 1 & 2 & 6 & 2 & -10 \\ \hline & & 2 & 8 & 10 \\ \hline & 2 & 8 & 10 & 0 \\ \end{array} \][/tex]
- Bring down the 2.
- Multiply 1 by 2 and write it under the next coefficient 6, result is 8.
- Add 6 and 2, results is 8.
- Multiply 1 by 8 and write it under the next coefficient 2, result is 10.
- Add 2 and 8, result is 10.
- Multiply 1 by 10 and write it under the next coefficient -10, result is 0.
- Add -10 and 10, result is 0, which confirms the remainder is zero.
This means the quotient is:
[tex]\[ 2x^2 + 8x + 10 \][/tex]
Thus, [tex]\( f(x) \)[/tex] can be expressed as:
[tex]\[ f(x) = (x - 1)(2x^2 + 8x + 10) \][/tex]
3. Solve the Quadratic Equation: To find the remaining zeros, solve the quadratic equation [tex]\( 2x^2 + 8x + 10 = 0 \)[/tex].
- Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = 10 \)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 2 \cdot 10}}{2 \cdot 2} = \frac{-8 \pm \sqrt{64 - 80}}{4} = \frac{-8 \pm \sqrt{-16}}{4} = \frac{-8 \pm 4i}{4} = -2 \pm i \][/tex]
4. List All Zeros: The zeros of [tex]\( f(x) \)[/tex] are:
- [tex]\( x = 1 \)[/tex]
- [tex]\( x = -2 + i \)[/tex]
- [tex]\( x = -2 - i \)[/tex]
Therefore, the zeros of the polynomial function [tex]\( f(x) = 2x^3 + 6x^2 + 2x - 10 \)[/tex] are:
[tex]\[ x = 1, \quad x = -2 + i, \quad \text{and} \quad x = -2 - i \][/tex]
1. Verify the Zero: We are told that [tex]\( f(1) = 0 \)[/tex]. Verification involves substituting [tex]\( x = 1 \)[/tex] into the polynomial:
[tex]\[ f(1) = 2(1)^3 + 6(1)^2 + 2(1) - 10 = 2 + 6 + 2 - 10 = 0 \][/tex]
Therefore, [tex]\( x = 1 \)[/tex] is indeed a zero of the polynomial.
2. Factor out [tex]\((x - 1)\)[/tex]: Use polynomial division or synthetic division to divide [tex]\( f(x) \)[/tex] by [tex]\( (x - 1) \)[/tex].
Performing synthetic division with the zero [tex]\( x = 1 \)[/tex]:
- Coefficients of [tex]\( f(x) \)[/tex] are: [tex]\( 2, 6, 2, -10 \)[/tex]
[tex]\[ \begin{array}{r|rrrr} 1 & 2 & 6 & 2 & -10 \\ \hline & & 2 & 8 & 10 \\ \hline & 2 & 8 & 10 & 0 \\ \end{array} \][/tex]
- Bring down the 2.
- Multiply 1 by 2 and write it under the next coefficient 6, result is 8.
- Add 6 and 2, results is 8.
- Multiply 1 by 8 and write it under the next coefficient 2, result is 10.
- Add 2 and 8, result is 10.
- Multiply 1 by 10 and write it under the next coefficient -10, result is 0.
- Add -10 and 10, result is 0, which confirms the remainder is zero.
This means the quotient is:
[tex]\[ 2x^2 + 8x + 10 \][/tex]
Thus, [tex]\( f(x) \)[/tex] can be expressed as:
[tex]\[ f(x) = (x - 1)(2x^2 + 8x + 10) \][/tex]
3. Solve the Quadratic Equation: To find the remaining zeros, solve the quadratic equation [tex]\( 2x^2 + 8x + 10 = 0 \)[/tex].
- Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = 10 \)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 2 \cdot 10}}{2 \cdot 2} = \frac{-8 \pm \sqrt{64 - 80}}{4} = \frac{-8 \pm \sqrt{-16}}{4} = \frac{-8 \pm 4i}{4} = -2 \pm i \][/tex]
4. List All Zeros: The zeros of [tex]\( f(x) \)[/tex] are:
- [tex]\( x = 1 \)[/tex]
- [tex]\( x = -2 + i \)[/tex]
- [tex]\( x = -2 - i \)[/tex]
Therefore, the zeros of the polynomial function [tex]\( f(x) = 2x^3 + 6x^2 + 2x - 10 \)[/tex] are:
[tex]\[ x = 1, \quad x = -2 + i, \quad \text{and} \quad x = -2 - i \][/tex]
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