How do you factor #5.4x^9+2.4x^8-3x^2+4.2x#?

1 Answer
Jan 17, 2017

Answer:

#5.4x^9+2.4x^8-3x^2+4.2x#

#= 1/5 x(27x^8+12x^7-15x+21)#

#~~ 1/5x(x^2+2.0369x+1.2066)(x^2+0.774536x+1.04923)(x^2-0.789118x+0.869413)(x^2-1.57788x+0.706631)#

Explanation:

#f(x) = 5.4x^9+2.4x^8-3x^2+4.2x#

Note that all of the terms are divisible by #x# and that if we multiply all of the terms by #5# then we get integral coefficients. So let's start with:

#5.4x^9+2.4x^8-3x^2+4.2x = 1/5 x(27x^8+12x^7-15x+21)#

The remaining octic has no zeros expressible in terms of elementary functions, including #n#th roots, trigonometric, exponential or logarithmic functions.

In theory, it has four quadratic factors with real coefficients, but those coefficients can only be approximated using numerical methods.

The zeros of the octic are approximately:

#x_(1,2) ~~ -1.01845+-0.411536i#

#x_(3,4) ~~ -0.387268+-0.948291i#

#x_(5,6) ~~ 0.394559+-0.844829i#

#x_(7,8) ~~ 0.78894+-0.290181i#

Hence, taking these zeros in complex conjugate pairs, the real quadratic factors are approximately:

#(x-x_1)(x-x_2) ~~ x^2+2.0369x+1.2066#

#(x-x_3)(x-x_4) ~~ x^2+0.774536x+1.04923#

#(x-x_5)(x-x_6) ~~ x^2-0.789118x+0.869413#

#(x-x_7)(x-x_8) ~~ x^2-1.57788x+0.706631#