Poly question (extension Cambridge q) (1 Viewer)

[hsc3hard5me]

How would I do this?

 

[Kurosaki]

Try using the division transformation .

 

[hsc3hard5me]

Try using the division transformation .

Sorry, but what is that? :S

Also how was your 3u hsc exam hehe

 

[Kurosaki]

Division transformation is that you can express a polynomial A (the dividend), with divisor D, quotient Q and remainder R as:
, and the degree of R is less than that of D. So try thinking about what D and R are . If you still can't figure it out I'll show you how to do it .

And the MX1 exam was pretty easy, not too hard.

 

[Joshmosh2]

How would I do this?

Cool question

 

[iStudent]

Division transformation is that you can express a polynomial A (the dividend), with divisor D, quotient Q and remainder R as:
, and the degree of R is less than that of D. So try thinking about what D and R are . If you still can't figure it out I'll show you how to do it .

And the MX1 exam was pretty easy, not too hard.

You mean you smashed it and should get 70/70 and that you would've state ranked if not for internals

 

[Drongoski]

Let P(x) = (x-p)(x-q)Q(x) + ax+b

.: P(p) = 0 + ap+b = p^3
& P(q) = 0 + aq+b = q^3

Solving the 2 simultaneous equations for a and b, you get:

a = p^2 + pq + q^2 and b = -pq(p+q)

.: remainder is (p^2+pq+q^2)x - pq(p+q)

 

[hsc3hard5me]

Thanks, I have another question I'm struggling to get, it looks really easy but I don't know how to get the answer:

 

[Carrotsticks]

You can find M easily by using the fact that there is a root at x=-2, in other words P(-2)=0.

Now that you have M, you can use the sum of roots.

Note that the roots are -2, A and A, so the sum of roots is -2+2A. Compare that to the sum of roots by observing the coefficients and you can solve for A.

 

[Axio]

Let P(x) = (x-p)(x-q)Q(x) + ax+b
.: P(p) = 0 + ap+b = p^3
& P(q) = 0 + aq+b = q^3

Solving the 2 simultaneous equations for a and b, you get:

a = p^2 + pq + q^2 and b = -pq(p+q)

.: remainder is (p^2+pq+q^2)x - pq(p+q)

With this, how do you know that the remainder is linear? (Or does it not matter)

 

[Trebla]

With this, how do you know that the remainder is linear? (Or does it not matter)

Division by a quadratic always yields a remainder of a lower degree than the quadratic (ie a linear or constant function). If the 'remainder' has a degree greater than or equal to the degree of the quadratic then it is not really a true 'remainder' because further division can be carried out

 

[hsc3hard5me]

You can find M easily by using the fact that there is a root at x=-2, in other words P(-2)=0.

Now that you have M, you can use the sum of roots.

Note that the roots are -2, A and A, so the sum of roots is -2+2A. Compare that to the sum of roots by observing the coefficients and you can solve for A.

I tried subbing in x=-2 but it m ends up cancelling out and the equation ends up as 0=0

 

[Carrotsticks]

Oh dear, my response was incorrect anyway. I did not see that the second term was an x term, not an x^2 term. Ignore that please =)

The sum of roots is zero since the x^2 term has a zero coefficient (which is why it is not there).

2A - 2 = 0

A=1.

Now that you have all three roots, you can use the product of roots to calculate M.

 

[Joshmosh2]

I tried subbing in x=-2 but it m ends up cancelling out and the equation ends up as 0=0

um.. do what carrot wrote. Basically, let alpha = -2, and beta, gamma = A

Find sum and product of roots and solve simultaneously

 

[hsc3hard5me]

Oh dear, my response was incorrect anyway. I did not see that the second term was an x term, not an x^2 term. Ignore that please =)

The sum of roots is zero since the x^2 term has a zero coefficient (which is why it is not there).

2A - 2 = 0

A=1.

Now that you have all three roots, you can use the product of roots to calculate M.

Thanks, that explained it quickly!