I need help factoring....?

Question

I have 50 problems but 2 have stumped me umm they are 3x^2-11x-4 and 8x^2+2x+3

I'm sorry the question is 8x^2+2x-3

Answer 1

3x² - 11x - 4 = 0

The middle term is - 11x

Find the sum of the middle term

Multiply the first term 3 times the last term -4 equals - 12 and factor

factors of - 12

1 x - 12. . .←. .use these factors
2 x - 6
3 x - 4

- 12 and + 1 satisfy the sum of the middle term

Insert - 12x and + 1x into the equation

3x² - 11x - 4 = 0

3x² - 12x + 1x - 4 = 0

Group factor

(3x² - 12x) + (1x - 4) = 0

3x(x - 4) + 1(x - 4) = 0

(3x + 1)(x - 4) = 0

- - - - - - - -s-

Answer 2

The easiest way to factor polynomials in which the number in front of the squared term is different from 1 is called the AC method. First of all, remember the general form of a polynomial is:

ax^2 + bx + c

So what we want to do is MULTIPLY the a-term by the c-term and factor that. So in example 1:

3x^2 - 11x - 4
a = 3 b = -11 c = -4

a*c = 3*4 [we can ignore the negative for now]
a*c = 12

Now we want to factor 12 into pairs of factors. We get:

12 => (1*12), (2*6), (3*4)

After doing that, we want to pick the pair that can be added/subtracted to equal the b-term in the expression (11) in our case. We see that 12 - 1 is 11, so the pair we want to pick is (1*12).

From there, we want to rewrite the expression substituting our two new factors in for the old b-term. Here's where we pick the sign up. Since we want -11x, we want to use -12x + 1x. So:

3x^2 - 11x - 4 => 3x^2 -12x + 1x - 4

Since we've rewritten our three-term polynomial as a four-term polynomial, we can use factor by grouping which means we want to factor the first two terms, factor the second two terms and combine. So:

3x^2 -12x + 1x - 4
3x(x - 4) + 1(x - 4)

And we see that we've gotten an (x-4) in common. So if we pull that out, we're left with a (3x+1) remaining. So our factored expression is:

(x-4)(3x+1)


That sounds really long and really complicated but once you get the hang of it, it's simple. For instance, here's the second one:

8x^2+2x-3
(8*3) = 24 => (24*1), (12*2), (8*3), (6*4)
Since we have +2x, we want to use +6x - 4x
8x^2+2x+3 => 8x^2 +6x - 4x -3
2x(4x + 3) -1(4x+ 3)
(4x+3)(2x-1)

Does that make sense? Try a few more using that method to make sure you have the hang of it. It's much more reliable than the way most algebra teachers teach it; most of them teach "guess and check" which is only partially guaranteed AND it's very time consuming. This is like clockwork and it's guaranteed to get the right answer anytime we're dealing with a polynomial of degree 2.

Hope that helps. If you have more questions, email me at: [e-mail].

Answer 3

(3x+_)(x+_) since 3 is prime.
(3x-4)(x+1)
(3x-1)(x+4)
(3x-2)(x+2)
(3x+4)(x-1)
(3x+1)(x-4)
(3x+2)(x-2)
Are the only posible integer solutions.
(3x+1)(x-4) is the correct solution.

if you have learned the quadratic formula then you can see that for the second problem:
SQRT(b^2 - 4*a*c)
SQRT(4-4*8*3)
SQRT(-92)
Since you are looking for the square root of a negative number the problem doesn't have any real roots or any integer roots. It does have complex roots. I don't think you are learning about complex numbers right now so it is most likely that you've either misread the problem, or your teacher gave you a problem that he/she didn't prepare you to be able to solve.

Answer 4

3x^2-11x-4
(3x+1)(x-4)

8x^2+2x+3
Not factorable

Answer 5

the first one is (3x+1)(x-4)
the second is no factorable as is, so you have to complete the sqare... you get 8(x-1/8)^2 + 189/64

Answer 6

3x^2 - 11x - 4

(3x + 1) (x - 4)

Answer 7

twig131, man you really did it this time! Wow, you have time to write big notes for someone you don't know and for something trivial!!

Keep it up, its a good spirit you have there with you!