Question #4a1e9

2 Answers
NickTheTurtle
Nov 6, 2017

#y=-1/272(x+5)(x^2+4x+13)#

Explanation:

All cubic equations can be written in the form #y=a(x-r_1)(x-r_2)(x-r_3)#, where #a# is a constant and #r_1,r_2,r_3# are roots of the equation.

Since we know the roots, we can write the cubic equation as #y=a(x+5)(x+2+3i)(x+2-3i)# (the complex roots of all real polynomials occur in conjugate pairs; thus since one of the roots is #-2-3i#, another root is #-2+3i#).

Simplifying this gives #y=a(x+5)(x^2+4x+13)#.

It is given that #y(3)=-1#, then #-1=a(3+5)((3)^2+4*3+13)=272a#, or #a=-1/272#.

The final answer is #y=-1/272(x+5)(x^2+4x+13)#.

Anjali G
Nov 6, 2017

Equation for #f(x)#:
#f(x) = -1/112(x+5)(x^2+5)#

Explanation:

Because one of the zeros is irrational in a way that it includes #i#, there must be another zero that includes the conjugate of that term.

#f(x) = a(x+5)(x-(-3i-2))(x+(-3i-2))#

#f(x) = a(x+5)(x+3i+2)(x-3i-2)#

#f(x) = a(x+5)(x^2 cancel(-3ix)cancel(-2x) cancel(+3ix)+9 cancel(-6i) cancel(+2x) cancel(-6i) -4)#

#f(x) = a(x+5)(x^2+5)#

Now use the given to find #a#: #f(3)=-1#
#-1 = a(3+5)(3^2+5)#

#-1=8*14*a#

#a=-1/(112)#

Equation for #f(x)#:
#f(x) = -1/112(x+5)(x^2+5)#

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