Question

If `x^2-5x+1=0` then what is the value of `x^4-2x^3-16x^2+13x+14` ?

Answer 1

`16`

Explanation:

From `x^2-5x+1=0` we have `x^2 = 5x-1` then

`x^4-2x^3-16x^2+13x+14 equiv (5 x-1)^2 - 2 x (5 x-1) - 16 (5 x-1) +13x+14 = 15 x^2-75x+31`

Substituting again

`15(5x-1)-75x+31 =16`

Answer 2

`x^4-2x^3-16x^2+13x+14 = 16`

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^2=5x-1` to find:

`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`

Answer 3

`x^4-2x^3-16x^2+13x+14 = 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 `x = ((0, -1),(1, 5))`

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`