Find an nth degree polynomial function with real coefficients satisfying the given conditions..
n=3; -4 and 6+5i are zeros; f(1)=250
(if you know how to do this but dont feel like doing it for me; could you just inform me on how i's work while factoring, i. e. what do they affect, etc.

we know that if we have a polynomial with real coefients and one root is a+bi, another root is a-bi

so

some roots are

3,-4,6+5i, 6-5i

for roots, r1 and r2

teh factors are

(x-r1)(x-r2)

so

we can put them in and

(x-3)(x+4)(x-6-5i)(x-6+5i)

(x^2+x-12)(2x^2-12x+61)

x^4-11x^3+37x^2+205x-732

to keep the same coefients, we multiply whole thing by c

c(x^4-11x^3+37x^2+205x-732)

we don't know what c is

we know that

f(1)=250

f(1)=c(x^4-11x^3+37x^2+205x-732)

find c

f(1)=c(1^4-11(1^3)+37(1^2)+205(1)-732)=250

f(1)=c(-500)=250

-500c=250

divide both sides by -500

c=-1/2

the polynomial (factored with real coefients) is

f(x)=-0.5(x-3)(x+4)(2x^2-12x+61)

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation:

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