To factor the quadratic function \( f(x) = 2x^2 + 4x + 96 \) and find its roots, we can follow these steps:
a.) Factored form of \( f(x) \)
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First, we can factor out the common coefficient of 2 from the quadratic: \[ f(x) = 2(x^2 + 2x + 48) \]
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Next, we can factor the quadratic \( x^2 + 2x + 48 \). To do this, we can use the quadratic formula or check for factors, but since 48 has no pairs of factors that add to 2, the quadratic does not factor nicely into rational numbers.
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Therefore, \( x^2 + 2x + 48 \) does not have real roots, and we can use the quadratic formula to express it. The roots are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 2 \), and \( c = 48 \).
Calculate the discriminant: \[ b^2 - 4ac = 2^2 - 4 \times 1 \times 48 = 4 - 192 = -188 \]
Since the discriminant is negative, the factorization will involve complex numbers. Thus, we find: \[ x = \frac{-2 \pm \sqrt{-188}}{2} = \frac{-2 \pm \sqrt{188}i}{2} = -1 \pm \sqrt{47}i \]
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Hence, the factored form in terms of complex numbers, using the roots found, is: \[ f(x) = 2 \left( x - (-1 + \sqrt{47}i) \right) \left( x - (-1 - \sqrt{47}i) \right) \] or, more neatly, \[ f(x) = 2 \left( x + 1 - \sqrt{47}i \right) \left( x + 1 + \sqrt{47}i \right) \]
b.) Roots of \( f(x) \)
The roots of the function \( f(x) = 2x^2 + 4x + 96 \) are: \[ x = -1 + \sqrt{47}i \quad \text{and} \quad x = -1 - \sqrt{47}i \]
Thus, we can summarize:
- Factored Form: \( f(x) = 2(x + 1 - \sqrt{47}i)(x + 1 + \sqrt{47}i) \)
- Roots: \( x = -1 + \sqrt{47}i \) and \( x = -1 - \sqrt{47}i \)