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
Explanation:
Given:
#f(x) = x^4+9x^3+29x^2+81x+180#
By the rational roots theorem, any rational zeros of
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
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+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 = 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)#