How do you find the polynomial function with roots #-1/2, 1/4, 1, 2#?

1 Answer
mason m
Nov 23, 2015

#8x^4-22x^3+9x^2+7x-2#

Explanation:

Consider the equation #(x-2)(2x+3)=0#.

To figure out what #x# could be, we set each term being multiplied equal to #0#.

#{(x-2=0rarrx=2),(2x+3=0rarrx=-3/2):}#

To answer your question, we use this process in reverse.

If we know one of the roots of the polynomial is #-1/2#, we know #x=-1/2#. Therefore, #x+1/2=0#, but having it's easier if we don't have fractions. We can multiply everything by #2# to see that #2x+1=0#. From this, we know that one of the multiplied terms in our polynomial will be #(2x+1)#.

We can use the same logic for each of the roots:

#{(-1/2rarr(2x+1)),(1/4rarr(4x-1)),(1rarr(x-1)),(2rarr(x-2)):}#

Then, we multiply them:

#f(x)=(2x+1)(4x-1)(x-1)(x-2)#

If we foil, we find that #f(x)=8x^4-22x^3+9x^2+7x-2#.

Related questions
See all questions in Zeros
Impact of this question
2381 views around the world
You can reuse this answer
Creative Commons License