How do you find the cubic polynomial function with two of its zeros 2 and -3+√2 and a y-intercept of 7?

Question

4822 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-31T22:16:35

#f(x) = -1/2x^3-2x^2+5/2x+7#

If the cubic has rational coefficients, then its other zero must be the conjugate #-3-sqrt(2)# of #-3+sqrt(2)# and it will take the form:

#f(x) = a(x-2)(x+3-sqrt(2))(x+3+sqrt(2))#

#=a(x-2)((x+3)^2-2)#

#=a(x-2)(x^2+6x+7)#

#=a(x^3+4x^2-5x-14)#

#=ax^3+4ax^2-5ax-14a#

In order that the #y# intercept be #7#, we must have:

#f(0) = -14a=7#

So #a=-1/2# and our cubic function is:

#f(x) = -1/2x^3-2x^2+5/2x+7#

graph{-1/2x^3-2x^2+5/2x+7 [-20.34, 19.66, -9.08, 10.92]}

Note that if we do not require #f(x)# to have rational coefficients, then there is a whole family of other solutions.