If #x^2-5x+1=0# then what is the value of #x^4-2x^3-16x^2+13x+14# ?
3 Answers
Explanation:
From
Substituting again
Explanation:
Given:
#x^2-5x+1=0#
We find:
#x^4-2x^3-16x^2+13x+14 = (x^2-5x+1)(x^2+3x-2)+16#
#color(white)(x^4-2x^3-16x^2+13x+14) = 0(x^2+3x-2)+16#
#color(white)(x^4-2x^3-16x^2+13x+14) = 16#
Alternatively, use
#x^4-2x^3-16x^2+13x+14#
#= (5x-1)(5x-1)-2x(5x-1)-16(5x-1)+13x+14#
#= 25x^2-10x+1-10x^2+2x-80x+16+13x+14#
#= 15x^2-75x+31#
#= 15(5x-1)-75x+31#
#= 75x-15-75x+31#
#= 16#
Explanation:
Here's an approach using matrices.
Given:
#x^2-5x+1=0#
This quadratic equation is satisfied by a matrix called the companion matrix, taking the form:
#((0, color(blue)(-1)),(1, color(blue)(5)))#
Putting
We find:
#x^4-2x^3-16x^2+13x+14#
#= x(x(x(x-2)-16)+13)+14#
#= ((0, -1),(1, 5))(((0, -1),(1, 5))(((0, -1),(1, 5))(((0, -1),(1, 5))-((2,0),(0,2)))-((16,0),(0,16)))+((13,0),(0,13)))+((14,0),(0,14))#
#= ((0, -1),(1, 5))(((0, -1),(1, 5))(((0, -1),(1, 5))((-2, -1),(1,3))-((16,0),(0,16)))+((13,0),(0,13)))+((14,0),(0,14))#
#= ((0, -1),(1, 5))(((0, -1),(1, 5))(((-1, -3),(3, 14))-((16,0),(0,16)))+((13,0),(0,13)))+((14,0),(0,14))#
#= ((0, -1),(1, 5))(((0, -1),(1, 5))((-17, -3),(3, -2))+((13,0),(0,13)))+((14,0),(0,14))#
#= ((0, -1),(1, 5))(((-3, 2),(-2, -13))+((13,0),(0,13)))+((14,0),(0,14))#
#= ((0, -1),(1, 5))((10, 2),(-2, 0))+((14,0),(0,14))#
#= ((2, 0),(0, 2))+((14,0),(0,14))#
#= ((16,0),(0,16))#
#= 16#