Sec3 AMaths

Only-Way-4-Destiny! SG

Factorise - 2 - 7x - 2x² + 3x³ completely. Hence, solve 3x - 4x² = 28x³ + 16x^4 .

I don't know how to do this. Can someone help me? Thanks

wee_ws SG

Hi,

Notice that x + 1 is a factor of - 2 - 7x - 2x² + 3x³.

The second expression does not seem to resemble the first in any way. Usually it should, because of 'Hence'. Do check it out.

Thanks!

Cheers,
Wen Shih

-Wanderer- SG

Hi,

Notice that x + 1 is a factor of - 2 - 7x - 2x² + 3x³.

The second expression does not seem to resemble the first in any way. Usually it should, because of 'Hence'. Do check it out.

Thanks!

Cheers,
Wen Shih

My roots for the 1st equation are (x + 1), (x - 2) & (3x + 1). I believe that is the easiest part so not much explanation is needed. And I agree with the mod on the 2nd expression not resembling the 1st in any way. They don't even have the same roots! Maybe you copied the question wrongly???

Mikethm SG

Factorise - 2 - 7x - 2x² + 3x³ completely. Hence, solve 3x - 4x² = 28x³ + 16x^4 .

I don't know how to do this. Can someone help me? Thanks

- 2 - 7x - 2x² + 3x³

=(x-2)(x+1)(1+3x)

3x - 4x² = 28x³ + 16x^4

3x - 4x² -28x³ -16x^4=0

x^4(3x^-3-4x^-2 - 28x^-1 -16)=0

x^4 = 0 or 3x^-3-4x^-2 - 28x^-1 -16=0

let x^-1 = 2y

3(8y^3)-4(4y^2)-28(2y)-16=0 .... divide by 8

3y^3-2y^2-7y-2=0

(y-2)(y+1)(3y+1)=0

y = 2, -1 or -1/3

2y = 4, -2 or -2/3

x^-1 = 4, -2 or -2/3

x = 1/4, -1/2 or -3/2

Therefore x = 0, 1/4, -1/2 or -3/2

dkcx SG

Wow, interesting. Didn't know nowadays sec 3 A maths requires so much work...

wee_ws SG

Hi,

An eye-opener, although it is too much to ask for from a sec 3 student.

Thanks!

Cheers,
Wen Shih

Only-Way-4-Destiny! SG

My teacher said the question was correct, the question is correct and it was a killer question. So idk how to do. Perhaps need substituion?

Mikethm SG

My teacher said the question was correct, the question is correct and it was a killer question. So idk how to do. Perhaps need substituion?

Hi I have provided the solution already. Is it unclear in any way? Perhaps you can tell me which part of the solution you do not understand and I can explain it.

Forbiddensinner SG

My teacher said the question was correct, the question is correct and it was a killer question. So idk how to do. Perhaps need substituion?

Scroll up. Mikethm already posted the correct answer.

Dejomel SG

I'm more curious on how many marks this question is worth.

Only-Way-4-Destiny! SG

Not stated on my paper, the marks

Anyway, I have manipulated the hence part.

3x - 4x² = 28x³ + 16x^4

3x = 16x^4 + 28x³ + 4x²

3x = 4x(4x³ + 7x² + x)

3x/4x = 4x³ + 7x² + x

3/4 = 4x³ + 7x² + x

16x³ + 28x² + 4x - 3 = 0

I got the 2nd cubic equation but I am supposed to use the first part to solve. I solved the equation without the first part. I got

x = 0.25, -1.5 or -0.5

dkcx SG

its x^4 so you need 4 values of x. U missed out x = 0

The correct method is posted already anyway

Only-Way-4-Destiny! SG

This is the alternative method of solving but without using the first part. There's "HENCE" , making the question damn hard

-Wanderer- SG

- 2 - 7x - 2x² + 3x³

=(x-2)(x+1)(1+3x)

3x - 4x² = 28x³ + 16x^4

3x - 4x² -28x³ -16x^4=0

x^4(3x^-3-4x^-2 - 28x^-1 -16)=0

x^4 = 0 or 3x^-3-4x^-2 - 28x^-1 -16=0

let x^-1 = 2y

3(8y^3)-4(4y^2)-28(2y)-16=0 .... divide by 8

3y^3-2y^2-7y-2=0

(y-2)(y+1)(3y+1)=0

y = 2, -1 or -1/3

2y = 4, -2 or -2/3

x^-1 = 4, -2 or -2/3

x = 1/4, -1/2 or -3/2

Therefore x = 0, 1/4, -1/2 or -3/2

Wow! That's a really good job in spotting the pattern. =D

Forbiddensinner SG

This is the alternative method of solving but without using the first part. There's "HENCE" , making the question damn hard

x^4(3x^-3-4x^-2 - 28x^-1 -16)=0

x^4 = 0 or 3x^-3-4x^-2 - 28x^-1 -16=0

let x^-1 = 2y

3(8y^3)-4(4y^2)-28(2y)-16=0 .... divide by 8

3y^3-2y^2-7y-2=0

This is the part where you use the hence. Instead of slowly working out the factors, you just throw (y-2)(y+1)(3y+1)=0 back to them directly. This is what is used from the 1st question.

Only-Way-4-Destiny! SG

- 2 - 7x - 2x² + 3x³

=(x-2)(x+1)(1+3x)

3x - 4x² = 28x³ + 16x^4

3x - 4x² -28x³ -16x^4=0

x^4(3x^-3-4x^-2 - 28x^-1 -16)=0

x^4 = 0 or 3x^-3-4x^-2 - 28x^-1 -16=0

let x^-1 = 2y

3(8y^3)-4(4y^2)-28(2y)-16=0 .... divide by 8

3y^3-2y^2-7y-2=0

(y-2)(y+1)(3y+1)=0

y = 2, -1 or -1/3

2y = 4, -2 or -2/3

x^-1 = 4, -2 or -2/3

x = 1/4, -1/2 or -3/2

Therefore x = 0, 1/4, -1/2 or -3/2

Wow! The answer was here! Very impressive. Thanks a lot, Mikethm!

Impressive

Mikethm SG

Glad to be of service.

SBS261P SG

your alternative method miss out x=0 because you take 3x / 4x = 3/4. This is BOOBOO!

Mikethm, how did you spot the pattern? perhaps can share

HiAy3Captain SG

maybe he set the question?

Mikethm SG

@SBS261P

Ignoring the x 1st and concentrating on how to get the coefficients to match. Once that is done, you will see that the "x"s are in the opposite order of what they should be. Then it is a matter of flexibility.

@HiAy3Captain

I am offended that you think I was the one who set this stupid question which require a unnecessarily complicated solution to a simple question. :)