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

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

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Answer 1 · 2017-12-06T11:40:41

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

Explanation:

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

Please note that the degree of the given trinomial is $3$ $color(blue)(3)$ 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 $t^{2}$ $color(red)(t^2)$

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

$t^{2} (6 t^{2} + 55 t - 29)$ $color(blue)(t^2(6t^2+55t-29))$ $. . E x p r e s s i o n .1$ $.. color(red)(Expression.1)$

Next, consider the quadratic expression from $. . E x p r e s s i o n .1$ $.. color(red)(Expression.1)$ .

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

$6 t^{2} + 55 t - 29$ $color(blue)(6t^2+55t-29)$ $. . E x p r e s s i o n .2$ $.. color(red)(Expression.2)$

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

$S t e p .1$ $color(green)(Step.1)$

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 $x^{2} t e r m$ $x^2 term$ and the constant,

Note that the product of the coefficient of the $x^{2} t e r m$ $x^2 term$ and the constant is $(- 174)$ $(-174)$ ,

$S t e p .2$ $color(green)(Step.2)$

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

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 $. . E x p r e s s i o n .1$ $.. color(red)(Expression.1)$ as follows:

$6 t^{2} - 3 t + 58 t - 29$ $color(blue)(6t^2-3t + 58t-29)$ $. . E x p r e s s i o n .4$ $.. color(red)(Expression.4)$

$S t e p .3$ $color(green)(Step.3)$

In this step, we break our $. . E x p r e s s i o n .4$ $.. color(red)(Expression.4)$ into groups:

$\Rightarrow (6 t^{2} - 3 t) + (58 t - 29)$ $color(blue)(rArr (6t^2 - 3t) + (58t - 29))$

Factor out $3 t$ $color(green)(3t)$ from $(6 t^{2} - 3 t)$ $color(blue)((6t^2 - 3t)$ to obtain $\Rightarrow 3 t \cdot (2 t - 1)$ $color(blue)(rArr 3t*(2t - 1) )$

Factor out $29$ $color(green)(29)$ from $(58 t - 29)$ $color(blue)((58t - 29)$ to obtain $\Rightarrow 29 \cdot (2 t - 1)$ $color(blue)(rArr 29*(2t - 1) )$

$S t e p .4$ $color(green)(Step.4)$

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

$\Rightarrow (3 t + 29) \cdot (2 t - 1)$ $color(blue)(rArr (3t + 29) * (2t - 1)$

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

Hence, our final solution using $. . E x p r e s s i o n .1$ $.. color(red)(Expression.1)$ is

$t^{2} (3 t + 29) (2 t - 1)$ $color(blue)(t^2(3t + 29)(2t-1)$

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

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