How do you factor #x^5+5x^4+6x^3#?

First note that all of the terms are divisible by #x^3#, so we can separate that out as a factor first:

#x^5+5x^4+6x^3 = x^3(x^2+5x+6)#

To factor the remaining quadratic, we want to find two numbers whose sum is #5# and whose product is #6#, since in general we have:

#(x+p)(x+q) = x^2+(p+q)x+pqx#

The numbers #2, 3# work, so we have:

#x^2+5x+6 = (x+2)(x+3)#

Putting it all together:

#x^5+5x^4+6x^3 = x^3(x+2)(x+3)#

Factorization is accomplished by taking a common factor ,completing the square, applying the polynomial identities ,or using the quadratic formula .

#x^5+5x^4+6x^3#
#=color(blue)(x^3)color(brown)((x^2+5x+6)#

The expression #x^2+5x+6# can be factorized using the trial and error method, in other words, the idea is to find two integers whose sum is #color(red) 5# and their product is #color(violet) 6#.

#X^2+color(red)SX+color(violet)P#

For the expression #x^2+5x+6# , the two integers satisfying the #color(red)S and color(violet)P# requirements are #3 and 2#.

#color(brown)(x^2+5x+6=(x+3)(x+2))#

For the whole expression#\ x^5+5x^4+6x^3\ #, the factorized form is:

#color(blue)(x^3)color(brown)((x+3)(x+2)#

Therefore,

#x^5+5x^4+6x^3=x^3(x+3)(x+2)#