How do you factor #6t^ { 4} + 55t ^ { 3} - 29t ^ { 2}# by grouping?

1 Answer
Sridhar V.
Dec 6, 2017

By grouping the trinomial expression given, we get,
#color(blue)(t^2(3t + 29)(2t-1)# as it's factors.

Explanation:

We are given the polynomial #color(red)(6t^4 + 55t^3 - 29t^2)#. We note that this is a polynomial with three terms, also called a trinomial.

Please note that the degree of the given trinomial is #color(blue)(3)# since 3 is the highest exponent of the individual terms (also called monomials) .

We observe that the Greatest Common Factor ( GCF ) for our trinomial expression is #color(red)(t^2)#

Next, we will factor out the GCF and write our trinomial.

#color(blue)(t^2(6t^2+55t-29))# #.. color(red)(Expression.1)#

Next, consider the quadratic expression from #.. color(red)(Expression.1)# .

We will need our GCF when we write out our factors as the final answer. So, we will preserve the GCF for later use.

Our quadratic expression is

#color(blue)(6t^2+55t-29)# #.. color(red)(Expression.2)#

To factor this quadratic expression, we will follow the procedure given below:

#color(green)(Step.1)#

We must split the coefficient of middle term into two numbers , such that when we add them we get the middle term, and when we multiply them we must get the product of the coefficient of the #x^2 term# and the constant,

Note that the product of the coefficient of the #x^2 term# and the constant is #(-174)#,

#color(green)(Step.2)#

The two numbers are: #color(blue)(-3 and +58)#

When we add ( - 3) and ( +58 ) we get 55 and when we multiply the two values ( - 3) and ( +58 ) we get ( - 174 )

Now, we write our #.. color(red)(Expression.1)# as follows:

#color(blue)(6t^2-3t + 58t-29)# #.. color(red)(Expression.4)#

#color(green)(Step.3)#

In this step, we break our #.. color(red)(Expression.4)# into groups:

#color(blue)(rArr (6t^2 - 3t) + (58t - 29))#

Factor out #color(green)(3t)# from #color(blue)((6t^2 - 3t)# to obtain #color(blue)(rArr 3t*(2t - 1) )#

Factor out #color(green)(29)# from #color(blue)((58t - 29)# to obtain #color(blue)(rArr 29*(2t - 1) )#

#color(green)(Step.4)#

Using #color(green)(Step.3)# we can factor out the common term #color(blue)((2t-1)# and write the factors of our quadratic expression:

#color(blue)(rArr (3t + 29) * (2t - 1)#

We must also remember to include the GCF for our trinomial expression #color(red)(t^2)# as we write our final answer.

Hence, our final solution using #.. color(red)(Expression.1)# is

#color(blue)(t^2(3t + 29)(2t-1)#

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