Question

How do you factor 2x squared + 20x + 50?

Answer 1

2x^2 + 20x + 50

2(x^2 + 10x + 25)

Answer 2

now you need to factor x^2 + 10x + 25

Answer 3

thanks

Answer 4

np

Answer 5

so is that now 2 (x - 5) (x-5)

Answer 6

close, but (x-5)(x-5) = x^2 - 10x + 25

Answer 7

we want the 10x to be positive

Answer 8

so then it should be 2 (x + 5) (x+5)

Answer 9

better

Answer 10

the answer is either

\[\Large 2(x+5)(x+5)\]

or

\[\Large 2(x+5)^2\]

Answer 11

so the actual problem was to simplify 2x squared + 20x + 50 over x squared - 25. The answer I came up with was 2 (x+5) over (x-5) and then I turned that into 2x + 10 over x-5

Answer 12

did I do anything wrong?

Answer 13

no you did that perfectly, nice work

Answer 14

you can leave the answer as

\[\Large \frac{2(x+5)}{x-5}\]

or you can distribute to get

\[\Large \frac{2x+10}{x-5}\]

Answer 15

either one works in my opinion

Answer 16

and then ix 2x cubed - 128 this is factored form?

2 (x - 4) (x squared + 4x + 16)

Answer 17

2x^3 - 128

2(x^3 - 64)

2(x-4)(x^2 + 4x + 16) ... use difference of cubes factoring formula

I'm getting the same

Answer 18

so I'm trying to solve for x in this equation: 10x + 3 over 5x + 2 = 1\2. I got rid of the 5x and turned the 10x into 2x and now have 2x over x + 2 = 1\2. is it all good so far?

Answer 19

the equation is

\[\Large \frac{10x+3}{5x+2} = \frac{1}{2}\]

right?

Answer 20

wait...it should be 5x + 6

Answer 21

so it's

\[\Large \frac{10x+3}{5x+6} = \frac{1}{2}\]

??

Answer 22

correct

Answer 23

You would cross multiply to get


\[\Large \frac{10x+3}{5x+6} = \frac{1}{2}\]


\[\Large 2(10x+3) = 1(5x+6)\]


\[\Large 20x+6 = 5x+6\]


I'll let you finish

Answer 24

I got 15x = 0

Answer 25

keep going

Answer 26

then wouldn't x = 0 ?

Answer 27

yep

Answer 28

I'm trying to solve for y in this problem: 6y squared - 13y - 8 and I factored it out to (2y - 1) (3y+8) but now I'm stuck

Answer 29

you have the signs mixed up

Answer 30

it should factor to (2y+1)(3y-8)

Answer 31

oh woops

Answer 32

so if

6y^2-13y-8 = 0

then

(2y+1)(3y-8) = 0

which allows us to say

2y+1=0 or 3y-8=0

Answer 33

solve those two equations for y

Answer 34

so y = - 1\2 and y = 8\3

Answer 35

yep

y = -1/2 or y = 8/3

Answer 36

one more problem and I did part of it already. I'm solving for x and the equation is sqrt(5-x) - 3 = 0

Answer 37

I multiplied both sides by the square root of 5 - x and got 5 - x = 3 times the square root of 5 - x and then tried multiplying the side with 3 by the square root of 5 - x to get rid of the square root, but something went wrong

Answer 38

sqrt(5-x) - 3 = 0

sqrt(5-x) = 3

( sqrt(5-x) )^2 = 3^2

5-x = 9

keep going

Answer 39

make sure to check the answer

Answer 40

cuz I ended up with 5 - x = 3 - 5 - x which is 3 = 0x

Answer 41

ohhhh

Answer 42

- 4 = x

Answer 43

yep x = -4

Answer 44

ok I promise this is the last one. I tried solving this with my friend but neither of us could solve it for some reason. I'm adding 3 over x - 2 + 5 over x squared + 5x + 6 and I then got 3 over x - 2 + 5 over (x + 2) (x + 3)

Answer 45

is there anything else I need to do?

Answer 46

the expression is

\[\Large \frac{3}{x-2} + \frac{5}{x^2+5x+6}\]

correct?

Answer 47

yes

Answer 48

ok one sec

Answer 49

hmm are you sure that it's not 3 over x+2 ?

and could the +5x be -5x ?

Answer 50

yes I'm sure

Answer 51

ok just checking

Answer 52

I"m just struggling so much cuz I haven't taken a math class since the spring

Answer 53

oh so it's been nearly a year since you've taken math?

Answer 54

ya...I was in the pastry program and was just focusing on that

Answer 55

i gotcha, that's definitely a long time to be out of practice, but here's how you simplify

Answer 56

but then I switched my major this quarter

Answer 57

\[\Large \frac{3}{x-2} + \frac{5}{x^2+5x+6}\]


\[\Large \frac{3}{x-2} + \frac{5}{(x+2)(x+3)}\]


\[\Large \frac{3(x+2)(x+3)}{(x-2)(x+2)(x+3)} + \frac{5}{(x+2)(x+3)}\]


\[\Large \frac{3(x+2)(x+3)}{(x-2)(x+2)(x+3)} + \frac{5(x-2)}{(x-2)(x+2)(x+3)}\]


\[\Large \frac{3(x+2)(x+3)+5(x-2)}{(x-2)(x+2)(x+3)}\]


\[\Large \frac{3(x^2+5x+6)+5(x-2)}{(x-2)(x+2)(x+3)}\]


\[\Large \frac{3x^2+15x+18+5x-10}{(x-2)(x+2)(x+3)}\]


\[\Large \frac{3x^2+20x+8}{(x-2)(x+2)(x+3)}\]


---------------------------------------------------------------------------------

So

\[\Large \frac{3}{x-2} + \frac{5}{x^2+5x+6}\]

simplifies to

\[\Large \frac{3x^2+20x+8}{(x-2)(x+2)(x+3)}\]

Answer 58

being a chef wasn't that exciting anymore?

Answer 59

no. It's what I decided to do when I was in 5th grade and then when it came down to choosing, it was either being a pastry chef or going into a music degree, so I tried the chef thing, but I wasn't committed enough and music is kinda my true passion

Answer 60

i gotcha, it's probably a tough field

and if you don't love it fully, then no point doing it

Answer 61

but does that simplification make sense?

Answer 62

it's really competitive and I had lots of problems with the teachers, so it wasn't like I was happy to be there every day

Answer 63

it does, but it's just a really long process

Answer 64

well maybe it was just the teachers? you never know, you might have loved the people in the actual workforce and the job itself

Answer 65

yeah it's a lot, it would be simpler if the x-2 was x+2, but oh well

Answer 66

or it would be simpler if the x^2 + 5x + 6 was x^2 - 5x + 6

Answer 67

I got through half the program which was the basics of it...the next half is just more advanced things

Answer 68

well you got the basics down at least, so you can use that in some settings (like making your own desserts at home? that's useful I guess)

Answer 69

can you factor 2x to the 4th power - 24x any other way besides 3x (x cubed - 8)

Answer 70

ya what I learned will be useful

Answer 71

2x^4 - 24x

2x(x^3 - 8)

2x(x^3 - 2^3)

2x(x - 2)(x^2 + 2x + 4) ... use the difference of cubes factoring rule

Answer 72

*3x to the 4th power I mean

Answer 73

yeah and you can decide to go back to it if you wanted to later down the road now that you have the basics under your belt

Answer 74

oh 3x^4

Answer 75

oh wait, lol made a serious typo above, ignore that...

Answer 76

ok

Answer 77

3x^4 - 24x

3x(x^3 - 8)

3x(x^3 - 2^3)

3x(x - 2)(x^2 + 2x + 4) ... use the difference of cubes factoring rule

Answer 78

there's the correct version

Answer 79

but how did you get the 2 cubed?

Answer 80

8 = 2^3

Answer 81

2^3 = 2*2*2 = 8

Answer 82

oh right

Answer 83

and if I wanted to solve for x, what would I do next?

Answer 84

if 3x^4 - 24x = 0

then 3x(x - 2)(x^2 + 2x + 4) = 0

Answer 85

3x(x - 2)(x^2 + 2x + 4) = 0

turns into

3x=0, or x-2=0 or x^2+2x+4=0

Answer 86

you'll have to use the quadratic formula to solve x^2+2x+4=0 for x (you'll get 2 solutions here)

Answer 87

and the quadratic formula is b = + - b times the square root of b squared - 4ac over 2a?

Answer 88

close, the quadratic formula is

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 89

so it should be - 2 + - the square root of 2 squared - 4 times 1 times 4 over 2 x 1, which turns into -2 + - the square root of -12 over 2

Answer 90

and now I have to find the square root of 12 which is is the square root of 4 times 3?

Answer 91

good, you should get

\[\Large x = \frac{-2\pm\sqrt{-12}}{2}\]

and that turns into


\[\Large x = \frac{-2+\sqrt{-12}}{2} \ \text{or} \ x = \frac{-2-\sqrt{-12}}{2}\]


\[\Large x = \frac{-2+2i\sqrt{3}}{2} \ \text{or} \ x = \frac{-2-2i\sqrt{3}}{2}\]


\[\Large x = -1+i\sqrt{3} \ \text{or} \ x = -1-i\sqrt{3}\]

Answer 92

yes

\[\large \sqrt{-12} = \sqrt{-1*4*3}\]

\[\large \sqrt{-12} = \sqrt{-1}*\sqrt{4}*\sqrt{3}\]

\[\large \sqrt{-12} = i*2*\sqrt{3}\]

\[\large \sqrt{-12} = 2i*\sqrt{3}\]

Answer 93

I finished all of my problems. Thanks so much for your help :)

Answer 94

glad to be of help

Answer 95

what did you major in, btw?

Answer 96

math lol

Answer 97

well math education really, learning to become a teacher

Answer 98

would you want to teach high school or middle school?