Math help!!?

Question

Could someone please please!! help me I'm so stuck on these.. Thank you

1. Factor:
2x^2 + 11x + 12
Which of the following is one of the factors?

A) (2x + 4)
B) (x + 4)
C) (x + 3)
D) prime

2. Factor:
3y^2 - 4y - 15
Which of the following is one of the factors?

A) (3y - 15)
B) (y + 3)
C) (3y + 5)
D) prime

3. Factor completely:
8x^4 + 4x^3 - 24x^2

A) 4x^2(2x^2 + x – 6)
B) 4x^2(2x – 3)(x + 2)
C) 4(2x^4 + x^3 - 6x^2)
D) prime


4. Factor completely:
4x^4 - 64

A) 4(x^4 - 16)
B) 4(x^2 + 4)(x^2 - 4)
C) 4(x^2 + 4)(x + 2)(x – 2)
D) prime

Answer 1

Here's the answers

1. Factor:
2x^2 + 11x + 12
(2x+3)(x+4)

B) (x + 4)
---------------------------
2. Factor:
3y^2 - 4y - 15
(3y+5)(y-3)

C) (3y + 5)
-----------------------
3. Factor completely:
8x^4 + 4x^3 - 24x^2
4x^2(2x^2+x-6) <---- THE DIRECTION SAID FACTOR COMPLETELY....
SO THE ANSWER IS 4x^2(2x-3)(x+2)

B) 4x^2(2x – 3)(x + 2)
--------------------------------------
4. Factor completely:
4x^4 - 64
4(x^4-16)
4(x^2+4)((x^2 - 4)
4(x^2+4)(x + 2)(x – 2)

C) 4(x^2 + 4)(x + 2)(x – 2)

YOU HAVE TO READ THE DIRECTION CAREFULLY
WHEN ITS SAID FACTOR COMPLETELY YOU HAVE TO FACTOR ALL !!!

Answer 2

1. Factor:
2x^2 + 11x + 12

In factoring the middle term is the sum of the factors of the last term. Since the first term has a coefficient we must take that into account as well.

First we see if that coefficient has more than one way to factor it. It does not. The only factors are 1 and 2. Therefore, we know that our answer will be (2x ___) (x ___).

Now what are the factors of twelve? 1 and 12, 2 and 6, and 3 and 4. Which of these placed in the blanks will get us 11 when we multiply the number in the last blank by 2 and then add it to the number in the first blank.

3 and 4. 4 x 2= 8 and 8+3=11. Therefore our answer is (2x+3)(x+4). Which means that B is the correct answer.

Now for one with a 4th power.
3. Factor completely:
8x^4 + 4x^3 - 24x^2

First off I would factor out the largest common variable I can. That is x^2. So I would be left with (x^2)(8x^2+4x-24).

Next I would factor out the largest common coefficient I can which is 4. We get: (4x^2)(2x^2+x-6)

Now let's factor the second parentheses like we did the first problem:
We know again that 2 has only one way to factor so the answer must be (2x____)(x____). The factors of 6 are: 1 and 6, and 2 and 3. Now one of these must be negative in order for it to become -6. If we put -3 in the first blank and 2 in the second blank we get 4-3 which equals 1. So our answer is: (2x-3)(x+2).

Our final answer is: (4x^2)(2x-3)(x+2). Therfore, answer B is correct.

I hope that helps and that you are able to complete the other two problems!

Answer 3

1. 2 factors 1,2
12 factors 2,6.. 1,12..3,4
so what combination of 1,2 and 3,4 gets us 11
(2x+3) (x+4)

2. Same theory 1,3 and 3,5
(3y+5)(y-3) because the middle factor is negative

3divide by 4x^2 and you get A..but it says factor completely
B!

4. divide by 4 and factor out - B