Question

POLYNOMIALS... help with quotient, remainders, finding coefficients :)

Answer 1

POLYNOMIALS! A battery produces a voltage that is dependent on the temperature t in ºC. The voltage formula is modelled by the formula \[V(t)=2t^3+bt^2+c \]
for \[-1 \le t \le 2.5\] where b and c are real numbers.

(a) find the quotient and remainder when V(t) is divided by (t-2)

(b) given that the graph of V(t) has a repeated root at t=2, find the values of b and c.

Answer 2

i used synthetic division to find the remainder and quotient, but i dont think its correct.

I got R= c + 16 + 4b
\[Q(x)=2t^2+ (4+b)t+(8+2b)\]

Answer 3

Looks correct !

Answer 4

since t-2 is a double root, i then synthetically divided the quotient by 2 and i got
\[\frac{ Q(x) }{ 2 }=2+8+b+24+4b\]

Answer 5

\(
\begin{array}{}

2~|&2&b&0&c\\
&&4&8+2b&16+4b\\
\hline
&&&&\\
&2&4+b&8+2b&|c+16+4b

\end{array}
\)


Quotient = \(2t^2 + (4+b)t + (8+2b)\)
Remainder = \(c+16 + 4b\)

Answer 6

for part b,
quotients doesnt matter... just find out the remainder and set it equal to 0

Answer 7

Oh, I see, let me have a go...

Answer 8

hang on...

Answer 9

After second synthetic division, u wil get second remainder.

set the both remainders to 0
two equaitons, two unknowns. solve.

Answer 10

okk

Answer 11

yeh, i was trying to do that - my second remainder I got to be 24+4b=0 ... b=-6

Answer 12

but when i found the values and graphed it, there was no double root, so i must of done something wrong...

Answer 13

http://www.wolframalpha.com/input/?i=c%2B16%2B4b%3D0%2C2%282%29%5E2+%2B+%284%2Bb%29*2+%2B+8%2B2b%3D0

Answer 14

b = -6 is correct!

Answer 15

Did u graph the Voltage equation ?

Answer 16

really? how come my graph is not right... hmm (then i found c=-18)

Answer 17

yep, V(t)

Answer 18

your c is wrong

Answer 19

you should get :

b = -6
c = 8

Answer 20

NOOOooooooOooo stupid me

Answer 21

hahah mistake in second sythetic division i guess... :)

Answer 22

why hello beautiful graph - thanks @ganeshie8 - yes that was it..

Answer 23

good :)

Answer 24

no it wasnt the division - i just looked at my calculator and i typed in a wrong digit -.-

Answer 25

problem solved.

Answer 26

lol, okay... and did u finish ur earlier problem about proving (x+1) is a factor of P(x) or somehting ?

Answer 27

...no! do you think you might have an idea? unlike @idkihavenoidea ? lol

Answer 28

lol i havent tried it yet... could u attach the quesiton here agian

Answer 29

sure hang on a sec

Answer 30

When P(x) is divided by
(x^2+x−2), the Quotient is Q(x)
and the remainder is (−2x+4)
When Q(x) is divided by (x+1) the remainder is 3.
prove that (x+1) is a factor of P(x)

Answer 31

Okay, basically you're given :

\(P(x) = (x^2+x-2)*Q(x) + (-2x+4)\)

\(Q(-1) = 3\)

Answer 32

You need to prove : \(P(-1) = 0\)

Answer 33

right ?

Answer 34

YES! :)

Answer 35

Evaluate \(P(-1)\)

Answer 36

\[P(-1) = (1+-1-2)*Q(x) + (2+4)\]
\[P(-1) = -2*Q(x) + 6\]

Answer 37

you need to replace x with -1 everywhere...

Answer 38

\(P(-1) = (1+-1-2)*Q(-1) + (2+4) \)

Answer 39

Yes!! I knew that..... dang it

Answer 40

\[P(-1) = -2*Q(-1) + 6\]

Answer 41

why u saving Q(-1) lol.... replace it with 3 !

Answer 42

and you're done :)

Answer 43

oooooooooooooooohhhhhhhhhhhhhhhhhhh ...there it is!

Answer 44

light bulb moment

Answer 45

When Q(x) is divided by (x+1) the remainder is 3.


=>

Q(-1) = 3 by remainder/factor theorem

Answer 46

when dividing : f(x)/g(x),


f(x) = g(x)*Q(x) + R(x)

Q(x) = quotient
R(x) = remainder

Answer 47

i feel remembering the division that way makes it easy to think the word problems.. .

Answer 48

that really helps, thanks a billion

Answer 49

np :)

Answer 50

shoot... it better be an easy one -.-

Answer 51

hehe - Given that P(x) is a cubic polynomial, with a leading coefficient of 1, find P(x) and all its zeros.

Answer 52

should be easy for u... after so much drilling of synthetic division lol

Answer 53

from part a, you already know that (x+1) is a factor of P(x),
what does that tell u about zeroes of P(x) ?

Answer 54

well, x=-1 is a zero and that there are two other zeros.... \[P(x)=(x+1)(x^2+bx+c)\]

Answer 55

we may not take that long route

Answer 56

i agree!

Answer 57

does the short route involve synthetic division?

Answer 58

one sec, let me think a bit...

Answer 59

Lets start here :

\(P(x) = (x^2+x-2)*Q(x) + (-2x+4)\)

Answer 60

Since \(P(x)\) has leading coefficient of 1,
say, \(Q(x) = x +a\)

then, P(x) becomes :
\(P(x) = (x^2+x-2)*(x+a) + (-2x+4) \)

Answer 61

find out "a" value, and plugin

Answer 62

use P(-1) = 0, to find the "a" value

Answer 63

ok ill do that now

Answer 64

i got a=4

Answer 65

Correct !

Answer 66

now expand and simplify P(x)?

Answer 67

thats it !

Answer 68

\(P(x) = (x^2+x-2)*(x+4) + (-2x+4) \)

simplify

Answer 69

I wish I had a mini you in my brain ....

Answer 70

aww you're very kind :) xD

Answer 71

haha .... I got \[P(x) = x^3+5x^2-4\]

Answer 72

looks good !!

Answer 73

the zeros are very awkward numbers - 0.884, -1, -4.828 oh well

Answer 74

hahah get use to them lol

Answer 75

wait a sec, wolfram gives different numbers for zeroes :

http://www.wolframalpha.com/input/?i=%28x%5E2%2Bx-2%29*%28x%2B4%29+%2B+%28-2x%2B4%29

Answer 76

nvm, i see u made a typo... thats all i guess :)

Answer 77

mine are correct, they have it in a nicer form (i clicked approximation on wolfram and the zeros i found checked).

Answer 78

0.884

is not same as : 0.8284

:P

Answer 79

(oops!) Well, thanks for your time @ganeshie8 I have to go now - thanks for all your tips aswell! Have a great day :) ....U DA BOMB

Answer 80

its a typo from u, feel it ;)

Answer 81

felt... haha

Answer 82

have a nice day :) cya !