Question

Will medal, Factor the trinomial
36x^2+24x+4
A: (36x+4)^2
B: (18x+2)^2
C: (6x+2^2
D: (6x+4)^2
E: none

Answer 1

take the square root of the first and last term

Answer 2

36x and 2?

Answer 3

Factor out a 4. \[4(9x^2+6x+1)\]\[= 4(x^2+6x+9)\]\[= 4(x+3)(x+3)\]\[= 4\left(x+\frac{3}{9}\right)\left(x+\frac{3}{9}\right)\]\[= 4\left(x+\frac{1}{3}\right)\left(x+\frac{1}{3}\right)\]\[= 4(3x+1)(3x+1)\]

Answer 4

So what I did was:

1. Factor out the LCM 4
2. factored what was inside the parenthesis.
3. Found 2 numbers that multiplied to give me 9, added to give me 6.
4. Because I had originally multiplied the 9 over to the constant to give me a simpler quadratic equation to solve for, I had to readd it back into my equation, so I divided all my terms by it.
5. simplified the fraction.
6. Multiplied the denominator over to the variable to simplify the function over a common denominator.

Answer 5

okay i understand that

Answer 6

Oh you are the first!`

Answer 7

really?

Answer 8

jhanny its the ancient arabic method that idiotic modern men have found on a tablet in a cave, and claimed it as their own

Answer 9

Yes, what most people do not understand is that when you take a complex quadratic function, such as \(9x^2+6x+1\) you can simplify it by multiplying the leading coefficient, \(9\) to the constant at the end, \(1\). This allows for easier factoring.

because you're trying to keep to the ORIGINAL equation, you must divide by the same leading coefficient \(9\) when attaining the factored form.

Answer 10

Yeah i didnt know that untill you explained it but i understood

Answer 11

in sorry what letter was that?

Answer 12

What letter?

Answer 13

i had a-e, and sorry for typeing funny i cup open my hand

Answer 14

cut*

Answer 15

Im not seeing the answer choice there.

Answer 16

A: (36x+4)^2
B: (18x+2)^2
C: (6x+2^2
D: (6x+4)^2
E: none

Answer 17

@Jhannybean
are all these steps correct/intentional which you wrote
4( 9x^2 + 6x + 1)
= 4( x^2 + 6x + 9)
= 4(x+3)(x+3)
= 4( x + 3/9) (x+3/9)
=4( x + 1/3) (x+1/3)
= 4 ( 3x+1)(3x+1)

Answer 18

this method is very clever :)
I agree with Jhanny, i dont see answer choice

Answer 19

thank you both can you help with another?

Answer 20

ok

Answer 21

Nevermind, I am wrong, it is there.

Answer 22

o.o?

Answer 23

how so?

Answer 24

4 ( 3x+1)^2 = 2 (3x+1) * 2(3x+1)

Answer 25

4 ( 3x+1)^2 = (6x + 2) * (6x+2)

Answer 26

but the factorization was correct

Answer 27

I did not know how to explain it.

Answer 28

Thats okay

Answer 29

4( 9x^2 + 6x + 1)
= 4( x^2 + 6x + 9)
= 4(x+3)(x+3)
= 4( x + 3/9)(x+3/9)
=4(x + 1/3)(x+1/3)
= 4(3x+1)(3x+1)
=2*2 (3x+1)*(3x+1)
= 2(3x+1) * 2(3x+1)
= (6x + 2)( 6x+2)
=(6x+2)^2

Answer 30

Oh now that makes sense. \[4(3x+1)^2\]\[2^2(3x+1)^2\]\[\left[2(3x+1)\right]^2\]\[2(3x+1)\cdot 2(3x+1)\]

Answer 31

\[=(6x+2)(6x+2)\]\[=(6x+2)^2\]

Answer 32

Thank you both so much this really helps, could i get help with this one
x^2-17x+16
A:(x-2)(x-8)
B:(x-16)(x-1)
C:(x-16)(x+1)
D:(x-4)(x+4)
E: none

Answer 33

What are two numbers that multiply to give you 16, add to give you -17?

Answer 34

umm 8*2 is 16 4*4 is 16

Answer 35

Think bigger, and take the factors of 17 into consideration.

Answer 36

answer C would be Ok for the first one - it looks like an end bracket is missing

Answer 37

it should be (6x + 2)^2

Answer 38

Im really sorry i have to go get lunch and dinner for my family thank you for all the help @Jhannybean

Answer 39

No problem :)

Answer 40

Hi jhannybean - you have made 2 fans here!

Answer 41

I tried to message you back but could not

Answer 42

Oh, because I have been getting spammed quite a bit lately.