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

##### 1 Answer

#### 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:

Note that all of the terms are divisible by

#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

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#