How do you factor #3t^3-21t^2-12t#?

2 Answers
Jun 9, 2018

Answer:

#3t(t-1/2*(7-sqrt(65)))(t-1/2*(7+sqrt(65)))#

Explanation:

We can write
#3t(t^2-7t-4)#

to get the other factors we must solve the quadrat6ic

#t_{1,2}=7/2pmsqrt(49/4+16/4)#
this is

#t_{1,2}=7/2pmsqrt(65)/2#

Jun 9, 2018

Answer:

#3t^3-21t^2-12t = 3t(t-7/2-sqrt(65)/2)(t-7/2+sqrt(65)/2)#

Explanation:

Given:

#3t^3-21t^2-12t#

Note that all of the terms are divisible by #3t#. So we can separate that out as a monomial factor:

#3t^3-21t^2-12t = 3t(t^2-7t-4)#

The remaining quadratic is in the form:

#at^2+bt+c#

with #a=1#, #b=-7# and #c=-4#

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (color(blue)(-7))^2-4(color(blue)(1))(color(blue)(-4)) = 49+16 = 65#

Since #Delta > 0# this quadratic does have real zeros and hence linear factors with real coefficients, but since #Delta# is not a perfect square those zeros and coefficients are irrational.

We can use the quadratic formula to find them:

#t = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(t) = (-b+-sqrt(Delta))/(2a)#

#color(white)(t) = (7+-sqrt(65))/2#

Hence:

#t^2-7t-4 = (t-7/2-sqrt(65)/2)(t-7/2+sqrt(65)/2)#

Putting it all together:

#3t^3-21t^2-12t = 3t(t-7/2-sqrt(65)/2)(t-7/2+sqrt(65)/2)#