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

1 Answer
Aug 5, 2015

Answer:

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#.