Explanation of this solution

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stevecowall

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(3tx - 3)(x² - x - 1) = 3tx³ - 3tx² - 3tx - 3x² + 3x + 3 = 3tx³ - (3t + 3)x² + (3 - 3t)x + 3

3 - 3t = 0, t = 1

2u = -(3t + 3) = -(3(1) + 3) = -6, u = -3

10t + u = 10(1) + (-3) = 7.

The part that I don't understand is how 3 - 3t = 0.
 
stevecowall said:
(3tx - 3)(x² - x - 1) = 3tx³ - 3tx² - 3tx - 3x² + 3x + 3 = 3tx³ - (3t + 3)x² + (3 - 3t)x + 3

3 - 3t = 0, t = 1

2u = -(3t + 3) = -(3(1) + 3) = -6, u = -3

10t + u = 10(1) + (-3) = 7.

The part that I don't understand is how 3 - 3t = 0.
Click to expand...

I don't think you have presented the complete problem. Where did '2u' come from?

Please check your post and confirm.
 
The question was:
One factor of 3tx^3 + 2ux^2 + 3 is x^2 - x -1. Find the value of 10t + u.
 
I get the same answer from a different process:

since (x2 - x -1) is a factor of (3tx3 + 2ux2 + 3), we can write:


3tx3 + 2ux2 + 3

= 3tx(x2 - x -1) + 3tx2 + 3tx + 2ux2 + 3

= 3tx(x2 - x -1) + (3t + 2u)(x2 -x -1) + x(3t + 2u) + (3t +2u) + 3tx + 3

Then

x(6t + 2u) + (3t + 2u + 3) = 0

then

6t + 2u = 0

and

3t + 2u + 3 = 0

then we get u = -3 and t = 1 and

10t + u = 7
 
Subhotosh Khan said:
since (x2 - x -1) is a factor of (3tx3 + 2ux2 + 3), we can write:


3tx3 + 2ux2 + 3

= 3tx(x2 - x -1) + 3tx2 + 3tx + 2ux2 + 3

= 3tx(x2 - x -1) + (3t + 2u)(x2 -x -1) + x(3t + 2u) + (3t +2u) + 3tx + 3
Click to expand...
Thanks a lot Subhotosh Khan. But I don't understand how you got:
3tx(x2 - x -1) + 3tx2 + 3tx + 2ux2 + 3
 
Subhotosh Khan said:
I get the same answer from a different process:

since (x2 - x -1) is a factor of (3tx3 + 2ux2 + 3), we can write:


3tx3 + 2ux2 + 3

= 3tx(x2 - x -1) + 3tx2 + 3tx + 2ux2 + 3

= 3tx(x2 - x -1) + (3t + 2u)(x2 -x -1) + x(3t + 2u) + (3t +2u) + 3tx + 3

Then

x(6t + 2u) + (3t + 2u + 3) = 0

then

6t + 2u = 0

and

3t + 2u + 3 = 0

then we get u = -3 and t = 1 and

10t + u = 7
Click to expand...


Why does x(6t+2u)+(3t+2u+3)=0?
 
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