How can you put this in factored form? Can you give me step by step, as well, please?

#f(x)=x^4+9x^3+29x^2+81x+180#

1 Answer
Feb 6, 2018

#f(x) = (x+4)(x+5)(x^2+9) = (x+4)(x+5)(x-3i)(x+3i)#

Explanation:

Given:

#f(x) = x^4+9x^3+29x^2+81x+180#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #180# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-5, +-6, +-9, +-10, +-12, +-15, +-18, +-20, +-30, +-36, +-45, +-60, +-90, +-180#

In addition note that the signs of the coefficients are in the pattern #+ + + + +#. With no change of sign, Descartes' Rule of Signs tells us that this quartic has no positive real zeros.

So the only possible rational zeros are:

#-1, -2, -3, -4, -5, -6, -9, -10, -12, -15, -18, -20, -30, -36, -45, -60, -90, -180#

Trying each in turn, we first find:

#f(-4) = (-4)^4+9(-4)^3+29(-4)^2+81(-4)+180#

#color(white)(f(-4)) = 256-576+464-324+180 = 0#

So #x=-4# is a zero and #(x+4)# a factor:

#x^4+9x^3+29x^2+81x+180 = (x+4)(x^3+5x^2+9x+45)#

Next note 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:

#x^3+5x^2+9x+45 = (x^3+5x^2)+(9x+45)#

#color(white)(x^3+5x^2+9x+45) = x^2(x+5)+9(x+5)#

#color(white)(x^3+5x^2+9x+45) = (x^2+9)(x+5)#

Finally note that #x^2+9 > 0# for all real values of #x#. It has non-real complex zeros #+-3i# ...

#x^2+9 = x^2+3^2 = x^2-(3i)^2 = (x-3i)(x+3i)#

Putting it all together:

#f(x) = (x+4)(x+5)(x^2+9) = (x+4)(x+5)(x-3i)(x+3i)#