VCE Specialist 3/4 Question Thread!

[Zues]

Thanks Conic!

[cosine]

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how can I solve this, using synthetic division or something similar, because I havent been taught long division or polynomial division ?

[keltingmeith]

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how can I solve this, using synthetic division or something similar, because I havent been taught long division or polynomial division ?

You have to learn an extension upon synthetic division, that I both don't know and am not willing to teach.

My advice is that you have a lot of time on your hands - learn polynomial division.

[brightsky]

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how can I solve this, using synthetic division or something similar, because I havent been taught long division or polynomial division ?

By inspection: -75*z3 = 150 so z3 = - 2.

[Zues]

how do i do 7,8?

thanks

[keltingmeith]

how do i do 7,8?

thanks

7) Divide P(z) by (z-1). Then, what comes out will be Q(z). All you have to do after that is evaluate Q(i)
Because all the coefficients are real, we know the polynomial will have form:



Expanding gives:

, which is E.

[Zues]

7) Divide P(z) by (z-1). Then, what comes out will be Q(z). All you have to do after that is evaluate Q(i)
Because all the coefficients are real, we know the polynomial will have form:



Expanding gives:

, which is E.

with 7, i dont know how to long divide when theres those brackets involved, could you please show me?

[DSubShell]

with 7, i dont know how to long divide when theres those brackets involved, could you please show me?

Didn't really go indepth into trying to the division, but it did look tricky.

An alternative idea that I got random inspiration for:









(Substitute P(i) and continue...)

Should be easier... maybe!

[cosine]

Hey guys

Factoring this expression, I didnt complete the square, and tried another method idk what its called but this is







Is the reason why we cant do this method, because we cant introduce 'i' into the factors, because the first bracket isnt a difference of two squares? Hence why we use complete the square to factor it? Also, if the above method does yield a difference of two squares, and introducing 'i' was possible, could we use it instead of completing the square? Thanks guys!

[brightsky]

The objective of factorisation is to break a polynomial expression down into linear factors. The expression (2z + 1) is already a linear factor so there is no need to break it down any further. And yes, (2z - i^2) is not a difference of perfect squares.

[cosine]

Why is the reason for factorising these two seperate expressions, the first one, although we have a complex factor/root/zero, we still introduce 'i' in the quadratic expression and then complete the square. But for the second example, we also have a complex linear factor, but the answer does not introduce 'i' in the quadratic expression.


Expression 1:

Answer:


Expression 2:

The answer for this one does not include 'i' in the two factors that are yielded by the quadratic. Thanks

[brightsky]

In the first polynomial expression, the quadratic factor has discriminant less than 0, and so is broken down into two complex linear factors. In the second polynomial expression, the quadratic factor has discriminant greater than 0, and so is broken down into two real linear factors.

[cosine]

In the first polynomial expression, the quadratic factor has discriminant less than 0, and so is broken down into two complex linear factors. In the second polynomial expression, the quadratic factor has discriminant greater than 0, and so is broken down into two real linear factors.

Ah i get it now, would you recommend to use the discriminant from now on, to determine whether i need to introduce 'i' into the equation? I mean, if i did introduce i in the second expression, the answer would totally be wrong, right?

Thanks for the help brightsky, you truly 'brighten' my day

[keltingmeith]

Ah i get it now, would you recommend to use the discriminant from now on, to determine whether i need to introduce 'i' into the equation? I mean, if i did introduce i in the second expression, the answer would totally be wrong, right?

Thanks for the help brightsky, you truly 'brighten' my day

You can do that, but personally I think it's easier to just jump to the quadratic equation if you can't factorise it by sight. IMHO, there's no point introducing i into something unless it's z^2+a=0 (and even then it's not necessary at all).

[cosine]

You can do that, but personally I think it's easier to just jump to the quadratic equation if you can't factorise it by sight. IMHO, there's no point introducing i into something unless it's z^2+a=0 (and even then it's not necessary at all).

What do you mean factorise if by sight? Just remember what you see is different to what i see in an expression haha