Respuesta :

Answer:

[tex](3x-4)(x+1)[/tex]

Step-by-step explanation:

This is in the form [tex]ax^2+bx+c[/tex].

If your wish is to factor by grouping, then you goal is too look for two numbers that multiply to be [tex]ac[/tex] and adds up to be [tex]b[/tex].

Then once you find those numbers you replace b with those numbers.  Then the factor by grouping can be done.

So you have [tex]a=3,b=-1,c=-4[/tex]

[tex]ac=3(-4)[/tex]

[tex]b=-1[/tex]

The numbers that we need are already present since -4+3 is -1.

So replace -1 in

[tex]3x^2-1x-4[/tex]

with (-4+3)

[tex]3x^2-4x+3x-4[/tex]

Now group the first two terms together and group the last two terms together:

[tex](3x^2-4x)+(3x-4)[/tex]

Factor what you can from both pairs:

[tex]x(3x-4)+1(3x-4)[/tex]

Notice you have two terms: x(3x-4)  and 1(3x-4).  These terms have a common factor of (3x-4) so factor that out of our expression like so:

[tex](3x-4)(x+1)[/tex]

Check with foil if you like:

First: 3x(x)=3x^2

Outer: 3x(1)=3x

Inner: -4(x)=-4x

Last: -4(1)=-4

----------------------Add together:

3x^2-x-4

luisejr77

Answer: [tex]=(3x-4)(x+1)[/tex]

Step-by-step explanation:

Given the polynomial [tex]3x^2- x -4[/tex]

We can observe that it is written in this form:

[tex]ax^2+bx+c[/tex]

To factor completely, we need to follow these steps:

- Rewrite the term "b" as the sum of two terms whose product be [tex](3)(-4)=-12[/tex] and whose sum be -1:

[tex]=3x^2+(-4+3)x-4[/tex]

- Applying Distributive property, we get:

[tex]=3x^2-4x+3x-4[/tex]

- Make two groups of two terms each:

[tex]=(3x^2-4x)+(3x-4)[/tex]

- Factor out "x" from the first group and 1 from the second group:

 [tex]=x(3x-4)+1(3x-4)[/tex]

- Factoring out [tex](3x-4)[/tex], we get:

 [tex]=(3x-4)(x+1)[/tex]