Question

(x^2+8x+13) divide (x+5)

Answer is quotient and remainder.
Help please! I've given my best tries to the others...but I can't get this one...

Answer 1

take the first term of the dividend (the thing being divided), and divide it by the first term of the divisor (the thing doing the dividing), what do you get?

Answer 2

5x^2? how do you divide an x?
an x with an exponent

Answer 3

\[\frac{x^2}{x} = \]

Answer 4

2x^2? or just x?

Answer 5

pick one. then test your answer by multiplying it by \(x\) and see if you get \(x^2\) as the result.

\[\frac{a}{b} =c\] only if \[c*b = a\]

Answer 6

x works!

Answer 7

yay learning :)

Answer 8

very good.

\[x^2 = x*x\]\[\frac{x^2}{x} =\frac{x*x}{x} = \frac{\cancel{x*}x}{\cancel{x}} = x\]

Or
\[\frac{x^2}{x} = \frac{x^2}{x^1} = x^{2-1} = x^1 = x\]

Answer 9

okay, so we do this like long division...we take the first term of the thing being divided, and divide it by the first term of the thing doing the dividing. we did that, and we got \(x\) as the first term of our quotient.

Now, we multiply \(x\) with the entire divisor (the thing doing the dividing), and subtract the resulting quantity from what we started with:

\[x(x+5) =\]

Answer 10

x^2 + 5x

Answer 11

good. what is \[x^2+8x+13 - (x^2+5x) = \]

Answer 12

do I simplify the first part first? like have 8x^3 + 13 - (blah blah)

Answer 13

that seems like the thing I should do?

Answer 14

am I wrong?

Answer 15

Uh, could you simplify my expression again?

Answer 16

hmmm? what do you mean?

Answer 17

8x^3 + 13 and then I multiply across right?

Answer 18

is your internet connection bad?

Answer 19

no, I'm just stunned at some of the things I'm seeing :-)

Answer 20

\[x^2+8x+13 - (x^2+5x) = x^2+8x+13-1(x^2+5x)\]Could you use the distributive property on the stuff in the parentheses, please?

Answer 21

you're adding and subtracting, so there's no way you can invent a new exponent, such as \(x^3\)...

Answer 22

ohhh sorry...I'm really bad at math...I'm a writer...this would explain a lot I'm sure

Answer 23

(surprising things? good things I hope? and not my math fail?)

Answer 24

alas, I must dash your hopes on the rocks :-)

Answer 25

x^4 + 5x^3 + 8x^3 + 40x^2 + 13x^2 +65x -x^2 -5x

Answer 26

yes?

Answer 27

distributive property:

\[a(b+c) = a*b + a*c\]\[a(b-c) = a*b-a*c\]
this means that \[-(b+c) = -1(b+c) = -1b -1c = -b-c\]

the -( stuff )
construct trips up a lot of people, in my experience. Sometimes, writing it as the equivalent -1( stuff ) serves as enough of a visual reminder, but not always.

Answer 28

Anyhow, we are just doing addition and subtraction of like terms here. We aren't able to create new exponents, so you only get the same exponents as you started with.

\[x^2+8x+13 - (x^2+5x) = x^2+8x+13-1(x^2+5x)\]\[\qquad = x^2+8x+13 -1x^2 -1*5x\]\[\qquad = x^2 + 8x + 13 - x^2 -5x\]\[\qquad = x^2 - x^2 + 8x - 5x + 13\]\[\qquad = 3x+13\]

any questions about that?

We are NOT multiplying, just subtracting/adding

Answer 29

If we draw this problem like a long division problem, this is what we have (so far):

x
-----------------
x + 5 | x^2 + 8x + 13
- (x^2 + 5x) |
----------------
3x + 13

Answer 30

nope, that makes perfect sense (I will attempt to not be such a fail...but no guarantees)

Answer 31

does that make sense?

Answer 32

it does

Answer 33

Now, we repeat the process. We divide the first term of \(3x+13\) by the first term of \(x+5\); what is the result?

Answer 34

\[\frac{3x}{x} = \]

Answer 35

3

Answer 36

Good. Now multiply 3 * the divisor; what do you get?

Answer 37

3(x^2 + 8x + 13) = 3x^2 + 24 + 39

Answer 38

24x

Answer 39

my bad

Answer 40

No, that's the dividend. The divisor is \(x+5\)

\[3(x+5) = 3x+15\]


x + 3
-----------------
x + 5 | x^2 + 8x + 13
- (x^2 + 5x) |
----------------
3x + 13
- (3x + 15)
-----------
-2

What does this mean?

Answer 41

is this the remainder? and 3x +15 was the quotient?

Answer 42

yes, no

Answer 43

x+3 is quotient

Answer 44

ohhhhhh, geez I feel dumb

Answer 45

is that "geez I feel dumb" because you understand it now, or because you still don't understand it? :-)

Answer 46

I think I'm gonna move on to english now...that at least is something I'm darn good at

Answer 47

but thank you for the help!
I became a fan because you know...your patience exceeds my imbecility

Answer 48

or at least my sincere lack of competence in this field

Answer 49

:-)

it's funny, my sweetie says that in real life I have no patience whatsoever :-)

Answer 50

you're biased with her, it's a whole different thing when you're dealing with a stranger

Answer 51

there's another method of doing this called synthetic division. It only works for dividing by things without exponents, like \(x+5\), not like \(x^2+3x+5\)

You write down all of the coefficients, including the "missing ones" (where you write down a 0 to represent a term that wasn't present, such as in \(x^2 + 16 = x^2 + 0x + 16\))

1 8 13
-5
------------
1

you write down just the number from the divisor, and you write it with the opposite sign. so here I wrote -5 because our divisor is \(x+5\)

finally, you drop the coefficient from the first column down below the line, as I did.

Here's the drill. You take that right-most number from below the line, multiply it by the number "on the shelf" (-5, here), and write that underneath the number in the next column:

1 8 13
-5 -5
------------
1

now you add straight down:

1 8 13
-5 -5
------------
1 3

and repeat!

1 8 13
-5 -5 -15
------------
1 3 -2

we ran out of columns and have \(-2\) as our remainder. the "1 3" across the bottom is the coefficients of the quotient, so \(1x+3 = x+3\) just like we got up above. This goes very quickly, as you can see!

Answer 52

if the the last number ends up being 0, there is no remainder.

Answer 53

I remember doing this...I've taken Algebra 1, algebra 2, geometry, precalc and now I'm taking this math for college readiness

Answer 54

For example:

\[\frac{x^2+8x+15}{x+5}\]

1 8 15
-5 -5 -15
-----------
1 3 0

so our answer is \[\frac{x^2+8x+15}{x+5} = x+3\]we can check by multiplying:
\[(x+3)(x+5) = x(x+5) + 3(x+5) = x*x + 5*x + 3*x + 3*5 = x^2 + 5x + 3x + 15\]\[\qquad=x^2+8x+15\]just as it should be!

Answer 55

it covers everrrrrryyyything...thank fully I only need one semester and not two since I have all my other math credits :D

Answer 56

oops, ran off the edge of the "page" there, but no matter

Answer 57

Well, I'll look forward to seeing you now and then ;-)

Answer 58

oh yeah, you will...bugger this stupid credit...

Answer 59

If you're good in English, you should try answering questions here in the English section. There are plenty of people who need help...

Answer 60

Someone just tagged me with an English question, for example :-)

Answer 61

I have actually :) it's real fun! and tagged you?

Answer 62

I just tagged you on the question...