Question

Factor completely: 2x^2 + 14x + 12

(x + 1)(x + 6)
2(x + 2)(x + 3)
2(x + 1)(x + 6)
2(x + 2)(x + 6)

Factor completely: 3x^2 – 21x + 30

(x – 2)(x – 5)
2(x – 3)(x – 10)
(3x – 2)(x – 15)
3(x – 2)(x – 5

Factor completely: 2x^3 − 32x

2(x + 4)(x − 4)
x(x − 4)(x − 4)
2x(x − 16)
2x(x + 4)(x − 4)

Factor completely: 16x^4 − 1

(4x^2 + 1)(4x2^ − 1)
(2x + 1)3^(2x − 1)
(4x^2 + 1)(2x + 1)(2x − 1)
(4x + 1)(4x − 1)

Factor completely: 6x^2 − 12x − 18

6(x + 1)(x − 3)
6(x − 1)(x − 3)
3(2x + 3)(x − 1)
(3x + 3)(3x − 3)

Answer 1

Got any ideas?

Answer 2

no clue :(

Answer 3

Lets start with the first one
Look at the terms:
2x^2, 14x, and 12
What is their greatest common factor?

Answer 4

2?

Answer 5

Correct.
So your first step would be to factor out that 2 they have in common.
2(x^2 +7x+6)
Do you know how to factor what's in the parantheses?

Answer 6

no

Answer 7

To factor what is in the parentheses, you have to find 2 numbers that when multiplied = 6 and when added = 7.

Answer 8

The only 2 numbers multiplied together to make 6 are 2 and 3..that doesn't equal 7 though..

Answer 9

There are 2 other numbers though.

Answer 10

Oh! 6 and 1!!

Answer 11

Very Good !!

Answer 12

So what would it be now..

Answer 13

So it'll be
(x+_)(x+_)
We want to find the two blanks
You should know the standardized form ax^2 + bx+c=0
When factoring, you want to find two factors of ac
Such that when added b

In your problem, a=1, b=7, c=6
ac=6
and some factors of 6 are 1,6 and 2,3
The only pair that would work is 1,6 becuz 1+6=7
So the two numbers that go in the blank are 1 and 6
So it's (x+1)(x+6)
And don't forget the 2 you factored in the beginning
So the entire thing is
2(x+6)(x+1)

Answer 14

Oh, ok. Thank you two so much for explaining this to me! Now I can do the rest on my on! :)

Answer 15

You are welcome Diane - and I'll admit I got here much later than Nikato did.

Answer 16

Your welcome! And you did a great job explaining wolf

Answer 17

Thanks nikato