Question

What is the factored form of 2x^2 + 20x + 50

Answer 1

So far, I took out 2

Answer 2

Okay, let's see what you got.

Answer 3

And that left me with 2(x^2 + 10 x + 25)

Answer 4

so its 2(x+5)(x+5)

Answer 5

This can be done without the quadratic formula.
And @IrisL please don't give away answers,

Answer 6

I thought the answer was 2(x +5)(x + 5) too

Answer 7

However, it isn't one of the listed answers

Answer 8

yea exactly, its 2(x+5)(x+5)

Answer 9

maybe your textbook solution is wrong, you don't need the quadratic formula when it is factorable.

Answer 10

Hmm, what are your solutions listed? Just curious.

Answer 11

(x + 5)(2x − 8)

(x − 5)(2x + 8)

(x + 5)(2x + 10)

(x + 5)(2x − 10)

Answer 12

Oh, simple, just distribute that 5 back in.

Answer 13

Ugh, stop giving away answers!

Answer 14

How would I get to that answer though? Distribute which 5?

Answer 15

oops, sorry I meant 2.
Okay, so we have 2(x+5)(x+5)
Multiply one of them with the 2, so [2(x+5)](x+5)
Do what's in the brackets.

Answer 16

Oh! Okay. So it would be (2x+10)(x+5)

Answer 17

Cools XD

Answer 18

Yup :P

Answer 19

What about polynomials of the third degree?

Answer 20

64^3 +343, to be exact

Answer 21

Anything you can factor out right off the bat?

Answer 22

2 isn't a factor, 3 isn't a factor, 4 isn't a factor... I'm not sure.

Answer 23

7 is a factor of 343, but not 64

Answer 24

Just as a notice, this has one solution..

Answer 25

How do you know?

Answer 26

I graphed it :)

Answer 27

lol cool

Answer 28

So, how would I go about it without any factors? I notice they are both perfect cubes

Answer 29

I think

Answer 30

Oh, I feel dumb now.
What we have to do is solve it as a normal equation. So set it equal to zero, and move everything away from x.

Answer 31

\[64x = \sqrt{-343}\]

Answer 32

But wouldn't that be bringing in imaginary numbers?

Answer 33

I meant to put a three as the power

Answer 34

Hold your horses there, you forgot tht 64

Answer 35

*that

Answer 36

lol \[x = \frac{ \sqrt[3]{343} }{ 64 }\]

Answer 37

ANd I meant negative -343

Answer 38

x = 7i/64

Answer 39

Factor out 2, 2(x^2 + 10x + 25)
= 2 (x +5)(x+5)
or you can write it as 2(x+5)^2.
Done.

Answer 40

You can take the cube root of a negative number. It's only even powers that get into the imaginary stuff.

Answer 41

Close:
\[64x^3+343=0\]
\[64x^3=-343\]
\[x^3=-\frac{343}{64}\]
\[\LARGE \sqrt[3]{-\frac{343}{64}}\]
Now solve it. And when you are taking the cubic root if anything, there are no imaginary numbers, it's just negative.
Because \[\LARGE -x^3=-x*-x*-x\]

Answer 42

(Well, negative numbers have complex cube roots too. But I assume you care more about the real one.)

Answer 43

Okay. So - 343/64 = 5.35?

Answer 44

Take the cubic root of each, first.

Answer 45

Oh. so 7/4

Answer 46

*-7/4
but yessssssss

Answer 47

lol okay. So -1.75?

Answer 48

Yea, and that's your one solution.

Answer 49

4x − 7

16x2− 28x + 49

16x2+ 56x + 49

None of the above

Answer 50

So the first one?

Answer 51

Remember what I taught you about working backwards from a solution?
x=-7/4

Answer 52

Yes, so far 4x + 7?

Answer 53

Oh. So none of them?

Answer 54

Looks like it.
Because it factors into:
\[\LARGE (4x+7)(64x^2-112x+196)\]

Answer 55

Wait, no ignore that.

Answer 56

lol k

Answer 57

Hmm, you know what.. I'm stuck >.>
Open a new question and see if anyone else can answer it. Sorry.
If I get anything I'll let you know.

Answer 58

After distributing, I got that the factors were (4x+7)(16x^2-28x+49)