sunmoonlight
New Member
- Joined
- Feb 4, 2022
- Messages
- 24
- Gender
- Male
- HSC
- N/A
How to do this question?
-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Polynomial (1 Viewer)
- Thread starter sunmoonlight
- Start date
How to do this question?
Just looking at the coefficients - they are symmetrical and so instantly you can see that it is reducible quadratics. Just divide by z2 and regroup
Like z2(5(z2+1/z2)-11(z+1/z)+16) and let y=z+1/z
Then z2(5y2-11y+6), etc.
Like z2(5(z2+1/z2)-11(z+1/z)+16) and let y=z+1/z
Then z2(5y2-11y+6), etc.
Last edited:
Since all coefficients are real and rational, any complex roots will come in conjugate pairs.
It must be factorisable as a product of two quadratics, and with a leading term of 5x4, you could start from
(5x2 + Ax + B) (x2 + Cx + D)
expand, and equate coefficients to get equations in A, B, C, and D.
The answer is (5x2 - 6x + 5)(x2 - x + 1)
It must be factorisable as a product of two quadratics, and with a leading term of 5x4, you could start from
(5x2 + Ax + B) (x2 + Cx + D)
expand, and equate coefficients to get equations in A, B, C, and D.
The answer is (5x2 - 6x + 5)(x2 - x + 1)
member 6003
Member
- Joined
- Apr 20, 2022
- Messages
- 83
- Gender
- Male
- HSC
- 2023
that would be so slow, just do tywebb's method first to find complex roots then use that to find two quadratics
scaryshark09
∞∆ who let 'em cook dis long ∆∞
- Joined
- Oct 20, 2022
- Messages
- 1,618
- Gender
- Undisclosed
- HSC
- 1999
wait can someone do the full working for this question please
How to do this question?
scaryshark09
∞∆ who let 'em cook dis long ∆∞
- Joined
- Oct 20, 2022
- Messages
- 1,618
- Gender
- Undisclosed
- HSC
- 1999
whats after that
member 6003
Member
- Joined
- Apr 20, 2022
- Messages
- 83
- Gender
- Male
- HSC
- 2023
ill do it one sec
Well z can’t be 0 so you can get rid of the z2 and 5y2-11y+6=0, etc
Last edited:
scaryshark09
∞∆ who let 'em cook dis long ∆∞
- Joined
- Oct 20, 2022
- Messages
- 1,618
- Gender
- Undisclosed
- HSC
- 1999
yeh so maybe im dumb, but i cant do the algebra after that part so....
scaryshark09
∞∆ who let 'em cook dis long ∆∞
- Joined
- Oct 20, 2022
- Messages
- 1,618
- Gender
- Undisclosed
- HSC
- 1999
i found
y = (11+/- root 199 i )/10
y = (11+/- root 199 i )/10
member 6003
Member
- Joined
- Apr 20, 2022
- Messages
- 83
- Gender
- Male
- HSC
- 2023
this is how I did it. Maybe there's a faster way to prove the roots have magnitude 1 before finding them. I don't think there is but if that's given you could say z+1/z =2costheta and not have to solve the second quadratic.
Attachments
(5x2 + Ax + B) (x2 + Cx + D)
(x^2 + Cx + D) \\ &= 5x^4 + (5C + A)x^3 + (5D + B + AC)x^2 + (BC + AD)x + BD \\ \\ \text{On equating coefficients:} \qquad A + 5C &= -11 \qquad \text{(1)} \\ 5D + B + AC &= 16 \qquad \text{(2)} \\ BC + AD &= -11 \qquad \text{(3)} \\ BD &= 5 \qquad \text{(4)} \end{align*})
.
The values A, B, C, and D are likely all integers.
From (4), B and D have the same sign, as (likely) do A and C... and equation (1) prevents A and C both being positive.
So, (4) suggests two options: (a) B = 1 and D = 5 or (b) B = 5 and D = 1
Only A = -6 and C = -1 fits with A + 5C = -11. Thus, A = -6, B = 5, C = -1, and D = 1
(x^2 - x + 1))
The four solutions of
are
The values A, B, C, and D are likely all integers.
From (4), B and D have the same sign, as (likely) do A and C... and equation (1) prevents A and C both being positive.
So, (4) suggests two options: (a) B = 1 and D = 5 or (b) B = 5 and D = 1
- (a) requires both A + 5C and 5A + C to be -11 and A = C = -11 / 6, making AC = 121 / 36 in (2) ===> unlikely.
- (b) makes (3) and (1) the same, and requires AC = 16 - 5D - B = 6
Only A = -6 and C = -1 fits with A + 5C = -11. Thus, A = -6, B = 5, C = -1, and D = 1
The four solutions of
Not *that* slow, but I do agree that tywebb's approach is preferable.