Maths Question Help(polynomials) (1 Viewer)

[sungkwo]

Hi, I got a question on polynomials for homework, and i dont know how to do it.

It goes: Use the Remainder Theorem to find the remainder when x^4+3x^2-4x-1 is divided by x^2-3.

Hmm, I only learnt how to use the remainder theorem with linear polynomials (x-a) so...

Edit: I got another question I dont get, it goes solve the equation 9x^4-25x^2+10gx-g^2=0, when g = -7 or 2
PS sorry if this isnt the right section to post these questions, and feel free to move it to the right section, if it isnt.

Cheers,​

 

[bored of sc]

1) x^4+3x^2-4x-1 is divided by x^2-3. I don't know how to do it by the remainder therom but using long division you get -4x+17.

2) 9x^4-25x^2+10gx-g^2=0, when g = -7 or 2

When g = -7
9x⁴ -25x² -70x -49 = 0
Don't know, tried all factors of -49, e.g. 1, -1, 7, -7, 49, -49 and none of them work.

When g = 2
9x⁴ -25x² +20x -4 = 0
P(1) = 0
Therefore x = 1.

Wow, I need to study.

 

[xXmuffin0manXx]

Hi, I got a question on polynomials for homework, and i dont know how to do it.

It goes: Use the Remainder Theorem to find the remainder when x^4+3x^2-4x-1 is divided by x^2-3.

Hmm, I only learnt how to use the remainder theorem with linear polynomials (x-a) so...

Edit: I got another question I dont get, it goes solve the equation 9x^4-25x^2+10gx-g^2=0, when g = -7 or 2
PS sorry if this isnt the right section to post these questions, and feel free to move it to the right section, if it isnt.

Cheers,
​

lol..its just root 3..sub that in

 

[bored of sc]

lol..its just root 3..sub that in

Positive or negative root 3.

x² - 3 = 0
x² = 3
x = + square root 3

 

[xXmuffin0manXx]

Positive or negative root 3.

x² - 3 = 0
x² = 3
x = + square root 3

i assumed that was a given..


you wouldnt happen to go to matrix would you sungkwo ?

 

[tommykins]

Hi, I got a question on polynomials for homework, and i dont know how to do it.

It goes: Use the Remainder Theorem to find the remainder when x^4+3x^2-4x-1 is divided by x^2-3.

Hmm, I only learnt how to use the remainder theorem with linear polynomials (x-a) so...

Cheers,​

Let P(x) = x^4+3x^2-4x-1
When you divide a polynomial, you get the following result
P(a) = (x-a)Q.(x) + R(x)

R(x) is always one degree less than what you divided the polynomial by.

Since we're dividing by x^2 - 3, R(x) becomes Ax+B where A and B are constants.

It then becomes simultaneous equations.

P(sqrt3) = 17 - 4sqrt3 = sqrt3A + B
P(-sqrt3) = 17 + 4sqrt3 = -sqrtA + B

(1) 17 - 4sqrt3 = sqrt3A + B
(2) 17 + 4sqrt3 = -sqrtA + B

(1) + (2)
2B = 34
Thus, B = 17

sqrt3A + 17 = 17- 4sqrt3
sqrt3A = -4sqrt3
A = -4

Thus the remainder is -4x + 17 (as A = -4 and B = 17 in Ax + B)

Weird though, this is 4unit polynomials (using the remainder theorem in this respect).

 

[kaz1]

Let P(x) = x^4+3x^2-4x-1
When you divide a polynomial, you get the following result
P(a) = (x-a)Q.(x) + R(x)

R(x) is always one degree less than what you divided the polynomial by.

Since we're dividing by x^2 - 3, R(x) becomes Ax+B where A and B are constants.

It then becomes simultaneous equations.

P(sqrt3) = 17 - 4sqrt3 = sqrt3A + B
P(-sqrt3) = 17 + 4sqrt3 = -sqrtA + B

(1) 17 - 4sqrt3 = sqrt3A + B
(2) 17 + 4sqrt3 = -sqrtA + B

(1) + (2)
2B = 34
Thus, B = 17

sqrt3A + 17 = 17- 4sqrt3
sqrt3A = -4sqrt3
A = -4

Thus the remainder is -4x + 17 (as A = -4 and B = 17 in Ax + B)

Weird though, this is 4unit polynomials (using the remainder theorem in this respect).

Our teacher gave us a question like this in our last 3 unit test except that the polynomial was unknown and two factors were given. I got 0 for that question.

 

[tommykins]

I wouldn't expect the average 3uniter student to be able to think of this.

 

[clintmyster]

was in mai test the other day, a sim question...couldnt do it then but now its a synch

 

[sungkwo]

R(x) is always one degree less than what you divided the polynomial by.

Since we're dividing by x^2 - 3, R(x) becomes Ax+B where A and B are constants.

quote]

Is this something that is expected to know in Year 10? Because really idk what ur talking about???

Also I think its -5sqrt3, and 5sqrt3, or am I wrong, again

Edit: Anyone know the answer for the second question

 

[clintmyster]

R(x) is always one degree less than what you divided the polynomial by.

Since we're dividing by x^2 - 3, R(x) becomes Ax+B where A and B are constants.

quote]

Is this something that is expected to know in Year 10? Because really idk what ur talking about???

Also I think its -5sqrt3, and 5sqrt3, or am I wrong, again

Edit: Anyone know the answer for the second question

This is definetly 3U work...if your getting this, your textbook is either insanely advanced or your teacher gave this randomly to you because your insanely talented! either way you dont have to worry about this for about a year!

im pretty sure the answer posted above was right as well.

 

[addikaye03]

if the question was structured more i would almost consider it 2U, u dont need any 3U info

 

[tacogym27101990]

if the question was structured more i would almost consider it 2U, u dont need any 3U info

two unit don't learn the remainder theorem or long division do they??
how else could they do it??

 

[tommykins]

Is this something that is expected to know in Year 10? Because really idk what ur talking about???

Also I think its -5sqrt3, and 5sqrt3, or am I wrong, again

Edit: Anyone know the answer for the second question

No, you'll need to know the remainder theorem in year 11 3unit but the actual division between x^2-a^2 or (x-a)(x-b) thought requirement is within the Year 12 Extension 2 course

This is definetly 3U work...if your getting this, your textbook is either insanely advanced or your teacher gave this randomly to you because your insanely talented! either way you dont have to worry about this for about a year!

im pretty sure the answer posted above was right as well.

It's weird, you use 3unit techniques to do it but you DEFINITELY need 4unit thinking to be able to utilise the technique correctly.

 

[sungkwo]

so no-one knows how to do my second question???

i guess Ill post the full question then- maybe i got somethin wrong.

9x^4-25^2+10gx-g^2 is divisible by both x-1 and x+2

i) find the value of g
I got -7, 2 so you can start the question from here

ii)- this is the part i cant do: Using the value for g found in i, solve the equation 9x^4-25^2+10gx-g^2=0
I know with the 2 it x=1, but I cant do it with -7

Maybe another 3-unit theorem i need to know?

 

[blackglitter]

Weird though, this is 4unit polynomials (using the remainder theorem in this respect).

I thought so too