Question

How do you use synthetic substitution to find the real zeroes for `P(x)=x^4-6x^3+11x^2-8x+2`?

Answer 1

The real zeroes of `P(x) = x^4-6x^3+11x^2-8x+2` are `color(red)(1)` and `color(red)(1)` and `color(red)(2+sqrt2)` and `color(red)(2-sqrt2)`.

Explanation:

According to the rational root theorem, the rational roots of `f(x) = 0` must all be of the form `p/q` with `p` a divisor of `2` and `q` a divisor of `1`.

So the only possible rational roots are `1` and `2`.

We have to test both possibilities.

The only one that works is.

`1|1" "-6" "11" " "-8" " "color(white)(1)2`
`color(white)(1)|" " " "color(white)(1)1color(white)(1)-5" "color(white)(1)6-2`
`" "stackrel("————————————————————)`
`" "color(white)(1)1" "-5" "color(white)(1)6color(white)(1)-2 " " "color(red)(0)`

So `1` is a real zero of the polynomial.

That means that `x-1` and `x^3-5x^2+6x-2` are factors of `P(x)`.

A fourth degree polynomial must have an even number of real zeroes.

The real zeroes of `x^3-5x^2+6x-2` must all be of the form `p/q` with `p` a divisor of `-2` and `q` a divisor of `1`.

The possible real zeroes are `±1` and `±2`.

The only one that works is

`1|1" "-5" "color(white)(1)6" " "-2`
`color(white)(1)|" " " "color(white)(1)1color(white)(1)-4" " " "2`
`" "stackrel("————————————————————)`
`" "color(white)(1)1" "-4" "color(white)(1)2" " " "color(red)(0)`

So `1` is a second real zero.

The remaining factor is `x^2-4x+2`.

Whose zeros can be found by using the quadratic formula.

`x^2-4x+2=0`

`x = (-(-4)+-sqrt((-4)^2-4(1)(2)))/(2(1))`

`= (4 +-sqrt8)/2`

`= 2 +- sqrt2`

The real zeroes of `P(x)` are `1` and `1` and `2+sqrt2` and `2-sqrt2`.