# Is it possible to factor #y=6x^4-11x^3-51x^2+98x-27 #? If so, what are the factors?

##### 1 Answer

#### Explanation:

I suspect a typo in the question.

Consider the quartic:

#f(x) = 6x^4-11x^3-51x^2+color(red)(99)x-27#

By the rational roots theorem, any *rational* zeros of

That means that the only possible *rational* zeros of

#+-1/6, +-1/3, +-1/2, +-1, +-3/2, +-3, +-9/2, +-9, +-27/2, +-27#

Trying each in turn, the first zero we come across is:

#f(1/3) = 6(1/81)-11(1/27)-51(1/9)+99(1/3)-27 = (2-11-153+891-729)/27 = 0#

So

#6x^4-11x^3-51x^2+99x-27 = (3x-1)(2x^3-3x^2-18x+27)#

Notice that in the remaining cubic, the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

#2x^3-3x^2-18x+27 = (2x^3-3x^2)-(18x-27)#

#color(white)(2x^3-3x^2-18x+27) = x^2(2x-3)-9(2x-3)#

#color(white)(2x^3-3x^2-18x+27) = (x^2-9)(2x-3)#

#color(white)(2x^3-3x^2-18x+27) = (x^2-3^2)(2x-3)#

#color(white)(2x^3-3x^2-18x+27) = (x-3)(x+3)(2x-3)#

So we find:

#6x^4-11x^3-51x^2+99x-27 = (3x-1)(x-3)(x+3)(2x-3)#

**Footnote**

The quartic as specified "will" factor, but only with very complicated coefficients. Here's one of them:

#11/24-1/24 sqrt(937+(10376 3^(2/3))/(1/2 (37557+i sqrt(24771372627)))^(1/3)+4 2^(2/3) (3 (37557+i sqrt(24771372627)))^(1/3))-1/2 sqrt(937/72-(1/2 (37557+i sqrt(24771372627)))^(1/3)/(6 3^(2/3))-1297/(3 2^(2/3) (3 (37557+i sqrt(24771372627)))^(1/3))+13429/(72 sqrt(937+(10376 3^(2/3))/(1/2 (37557+i sqrt(24771372627)))^(1/3)+4 2^(2/3) (3 (37557+i sqrt(24771372627)))^(1/3))))#