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Mathematics 20 Online
OpenStudy (anonymous):

Medal for right answer :))))))))))) Find the remainder of (h^4 + h^2 – 2) ÷ (h + 3)

OpenStudy (megatronprime):

Here's what I found: H^4+h^2−2=h+3 Subtract h+3 from both sides. h^4+h^2−2−(h+3)=h+3−(h+3) h^4+h^2−h−5=0 Use quartic formula. h=−1.227079,1.444794

OpenStudy (anonymous):

I HAVE 4 MORE QUESTIONS AFTER THIS ONE :) lmao

Nnesha (nnesha):

there are two ways to find remainder 1st) divide by using long or synthetic division method 2nd) solve the divisor `h+3` for the variable h and then substitute h for its value into the polynomial

OpenStudy (anonymous):

how should i write that though like whats the final answer it says type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

OpenStudy (michele_laino):

hint: it is suffice to replace \(h=-3\) into this polynomial: \(h^4+h^2-2\)

OpenStudy (anonymous):

so what’s the final and complete answer :)??????????

OpenStudy (michele_laino):

please try to do this computation: \((-3)^4+(-3)^2-2=...?\)

OpenStudy (anonymous):

88?

OpenStudy (michele_laino):

that's right!

OpenStudy (anonymous):

so thats the final answer?

OpenStudy (michele_laino):

yes! We have applied the so called \(Theorem \;of\; Ruffini\)

OpenStudy (anonymous):

I have more questions so more medals to give

Nnesha (nnesha):

these kids should learn how to substitute numbers.

OpenStudy (anonymous):

Using synthetic division, find (2x^4 + 4x^3 + 2x^2 + 8x + 8) ÷ (x + 2).

OpenStudy (anonymous):

anyone??

OpenStudy (anonymous):

thx btw

OpenStudy (michele_laino):

I know how to do such division, nevertheless I have a slightly different method, since I'm italian

OpenStudy (anonymous):

cool cool so????

OpenStudy (michele_laino):

if you want I can apply my method

OpenStudy (anonymous):

yeah i just need the answer :)

OpenStudy (michele_laino):

I'm sorry, I can't give the direct answer, since it is against the Code of Conduct

OpenStudy (anonymous):

well just make it more simple

OpenStudy (michele_laino):

ok! first step we have to make such division: \(2x^4:x=...?\)

OpenStudy (anonymous):

2^3

OpenStudy (anonymous):

so is it 2x^3+2x^2 or just 2x?

OpenStudy (michele_laino):

we have \(2x^3\) am I right?

OpenStudy (anonymous):

yeah?

OpenStudy (anonymous):

but the ned number? is it 2x?

OpenStudy (anonymous):

next*

OpenStudy (michele_laino):

so we can write this: |dw:1450290685256:dw|

OpenStudy (anonymous):

wait so whats next?

OpenStudy (michele_laino):

next we have to do this computation: \[ - 2{x^3}\left( {x + 2} \right) = ...?\]

OpenStudy (anonymous):

is the answer 2x^3+2x^2+4

OpenStudy (michele_laino):

yes! It is

OpenStudy (anonymous):

yaws

OpenStudy (anonymous):

wait a sec

OpenStudy (anonymous):

k so theres a triangle so 3 sides

OpenStudy (anonymous):

one side is A=2c^3+16

OpenStudy (anonymous):

b=c+2

OpenStudy (anonymous):

so what is the height or h

OpenStudy (anonymous):

?/////????

OpenStudy (michele_laino):

we have to compute the area of such triangle, first

OpenStudy (anonymous):

Hint: The formula for area of a triangle is A = bh, where b is the length of the base and h is the height of the triangle.)

OpenStudy (michele_laino):

yes! I know, nevertheless, we have to use the Eron's formula first

OpenStudy (anonymous):

oaky

OpenStudy (anonymous):

sry I’m pretty bad at this

OpenStudy (michele_laino):

firs step, we have to compute the half perimeter: \(p=(a+b+c)/2=(2c^3+16+c+2+c)/2=...?\)

OpenStudy (michele_laino):

first*

OpenStudy (anonymous):

okay

OpenStudy (michele_laino):

please try to complete such computation

OpenStudy (anonymous):

is it 2c^2-4c+8? theres also another one which ends with some more numbers

OpenStudy (anonymous):

like after 8 theres ((32/c+2))

OpenStudy (anonymous):

so which is it

OpenStudy (michele_laino):

hint: the perimeter is: \[\begin{gathered} \left( {2{c^3} + 16} \right) + \left( {c + 2} \right) + c = 2{c^3} + 2c + 18 = \hfill \\ \hfill \\ = 2\left( {{c^3} + c + 9} \right) \hfill \\ \end{gathered} \] so, what is the half-perimeter?

OpenStudy (anonymous):

I’m not sure? does the answer not start with 2c^2 is it maybe 4c^2

OpenStudy (michele_laino):

it is not the answer it is the perimeter

OpenStudy (anonymous):

so what’s the height? I’m confused

OpenStudy (michele_laino):

as you well know, a triangle has three heights

OpenStudy (michele_laino):

is the triangle a right triangle?

OpenStudy (anonymous):

what but i know a and b so whats c

OpenStudy (anonymous):

right?

OpenStudy (michele_laino):

like this one: |dw:1450291781898:dw|

OpenStudy (anonymous):

the h that I’m trying to find is a 90 degree

OpenStudy (anonymous):

the one to the left

OpenStudy (michele_laino):

please can you make a drawing of the triangle?

OpenStudy (anonymous):

yeah wait

OpenStudy (anonymous):

OpenStudy (anonymous):

there it is

OpenStudy (michele_laino):

is \(A\) an area?

OpenStudy (anonymous):

what

OpenStudy (anonymous):

I’m trying to find h

OpenStudy (anonymous):

Find the height, h, of the triangle shown. h = . (Hint: The formula for area of a triangle is A = bh, where b is the length of the base and h is the height of the triangle.)

OpenStudy (michele_laino):

then we have to solve such equation: \((c+2) \ h=2c^3+16\)

OpenStudy (anonymous):

if i give you 4 options can u tell me which on is correct

OpenStudy (michele_laino):

namely we have to find the value of \(h\): \[h = \frac{{2{c^3} + 16}}{{c + 2}} = ...?\]

OpenStudy (anonymous):

here are the answer options

OpenStudy (anonymous):

is it a b c or d??????????

OpenStudy (anonymous):

plzzzzzz

OpenStudy (michele_laino):

hint: we can use this identity: \[\begin{gathered} 2{c^3} + 16 = 2\left( {{c^3} + 8} \right) = \hfill \\ \hfill \\ = 2\left( {c + 2} \right)\left( {{c^2} - 2c + 4} \right) \hfill \\ \end{gathered} \] so, after a substitution, we get: \[h = \frac{{2{c^3} + 16}}{{c + 2}} = \frac{{2\left( {c + 2} \right)\left( {{c^2} - 2c + 4} \right)}}{{c + 2}} = ...?\]

OpenStudy (anonymous):

ok so the answer is 2c^2-4c+8?

OpenStudy (michele_laino):

that's right!

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