How do you write a polynomial with zeros: -3,1,5,6?

1 Answer
David G.
Feb 4, 2016

For a zero, when #x# has that value the overall bracket (parenthesis) needs to sum to zero. For example #(x+3)=(-3+3)=0#. That means the solution is #(x+3)(x-1)(x-5)(x-6)#.

Explanation:

A 'zero' or 'root' of an equation is a point at which it crosses the #x#-axis: that is, the line #y=0# (or #f(x)=0#).

For that to happen, one of the factors must go to zero. If we take the example of #x=-3#, we need a factor that, when #x# has that value, will equal zero. If it's in the form #(x+a)# and we put in the value #x=-3#, the value of #a# required is #3#.

For a positive value of #x# such as #x=5#, the factor needs to be #(x-5)# so that it becomes zero when #x=5#.

Using this approach, over all we get:

#(x+3)(x-1)(x-5)(x-6)#

We could multiply this through to get an expression in #x^4# and all lower values, but it's probably best left in this factorized form.

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