Answer:
[tex]f(x)=2x^4-12x^3+50x^2-56x-120[/tex]
Step-by-step explanation:
Complex Conjugate Theorem
For a polynomial p(x) with real coefficients, the complex zeros occur in conjugate pairs. So if (a + bi) is a zero, then its conjugate (a - bi) is also a zero.
Given zeros:
-1, 3, and (2 + 4i)
As one of the given zeros is a complex number, (2 - 4i) is also a zero.
Write each zero as part of a factor and multiply them together, adding a leading coefficient [tex]a[/tex]:
[tex]\begin{aligned}f(x) & =a(x+1)(x-3)(x-(2+4i))(x-(2-4i))\\& = a(x+1)(x-3)(x-2-4i)(x-2+4i)\\& = a(x+1)(x-3)(x^2-2x+4ix-2x+4-8i-4ix+8i-16i^2)\\& = a(x+1)(x-3)(x^2-4x+4-16i^2)\end{aligned}[/tex]
Remember that [tex]i^2=-1[/tex], therefore:
[tex]\begin{aligned}f(x)&=a(x+1)(x-3)(x^2-4x+4-16(-1))\\&=a(x+1)(x-3)(x^2-4x+4+16)\\&=a(x+1)(x-3)(x^2-4x+20)\end{aligned}[/tex]
To find the value of the leading coefficient (a), use the given [tex]f(1)=-136[/tex] :
[tex]\begin{aligned}f(1) & = -136\\\implies a(1+1)(1-3)(1^2-4(1)+20) & = -136\\a(2)(-2)(17) & = -136\\-68a & = -136\\\implies a & = 2\end{aligned}[/tex]
Therefore, the polynomial in factored form is:
[tex]f(x)=2(x+1)(x-3)(x^2-4x+20)[/tex]
Finally, expand the brackets:
[tex]\begin{aligned}f(x) & =2(x+1)(x-3)(x^2-4x+20)\\& =2(x^2-2x-3)(x^2-4x+20)\\& =2(x^4-4x^3+20x^2-2x^3+8x^2-40x-3x^2+12x-60)\\& =2(x^4-4x^3-2x^3+20x^2+8x^2-3x^2-40x+12x-60)\\& =2(x^4-6x^3+25x^2-28x-60)\\& =2x^4-12x^3+50x^2-56x-120\end{aligned}[/tex]
Learn more about complex numbers here:
https://brainly.com/question/26344541
https://brainly.com/question/28032887
