Question #ccd25

1 Answer
Jim G.
Jan 11, 2018

#x^4-4x^3-x^2+16x-12=0#

Explanation:

#"if a polynomial has roots say"#

#x=a,x=b,x=c" and "x=d#

#"then it's factors are"#

#(x-a),(x-b),(x-c)" and "(x-d)#

#"and the polynomial is the product of the factors"#

#"here "x=1,x=2,x=-2" and "x=3" are the roots"#

#"the factors are therefore"#

#(x-1),(x-2),(x-(-2))" and "(x-3)#

#rArrp(x)=(x-2)(x+2)(x-1)(x-3)#

#"expanding the factors using FOIL/distributive law"#

#=(x^2-4)(x^2-4x+3)#

#=x^4-4x^3+3x^2-4x^2+16x-12#

#=x^4-4x^3-x^2+16x-12#

#rArrx^4-4x^3-x^2+16x-12=0" is the equation"#
graph{x^4-4x^3-x^2+16x-12 [-10, 10, -5, 5]}

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