Question #ba75e

1 Answer
Jim G.
Oct 24, 2017

#p(x)=1/5(x^4-7x^3+8x^2+16x)#

Explanation:

#"roots are "x=0" and "x=-1larr"multiplicity 1"#

#rArrx" and "(x+1)" are the factors"#

#"also root of multiplicity 2 at "x=4#

#rArr(x-4)^2" is the factor"#

#p(x)" is then the product of the factors"#

#rArrp(x)=ax(x+1)(x-4)^2larr" a is a multiplier"#

#"to find a substitute "(5,6)" into the equation"#

#6=a(5)(6)(1)=30a#

#rArra=6/30=1/5#

#rArrp(x)=1/5x(x+1)(x-4)^2#

#"expanding the factors gives"#

#p(x)=1/5(x^4-7x^3+8x^2+16x)#
graph{1/5(x^4-7x^3+8x^2+16x) [-10, 10, -5, 5]}

Related questions
Impact of this question
301 views around the world
You can reuse this answer
Creative Commons License