What is a cubic polynomial function in standard form with zeros 3, -4, and 5?

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Answer 1 · 2018-02-16T22:53:05

#y=x^3-4x^2-17x+60#

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The standard form of a cubic polynomial function is:

#y=ax^3+bx^2+cx+d#

If we know the zeros as being #alpha#, #beta#, and #theta# then we know its factored form to be:

#y=(x-alpha)(x-beta)(x-theta)#

simply because you would have had to set the expression within each set of parentheses equal to #0# and solve for a root.

Therefore, we can write the polynomial in this problem as:

#y=(x-3)(x+4)(x-5)#

Now, if we multiply these through and remove the parentheses we will have the standard form of it:

#y=(x-3)(x^2-x-20)#

#y=x^3-x^2-20x-3x^2+3x+60#

#y=x^3-4x^2-17x+60#

Answer 2 · 2018-02-16T22:53:19

#x^3-4x^2-17x+60#

There is only one cubic polynomial in standard form (i.e. monic) with three given distinct zeros, #a, b# and #c#.
It is given by the product:

#(x-a)(x-b)(x-c)#

In your case, it is

#(x-3)(x+4)(x-5)=(x^2+x-12)(x-5)#
#=x^3-4x^2-17x+60#