# A couple Algebra 1 problems (polynomial division, mixture problem)

• Apr 29th 2012, 04:18 PM
Tsuzuku
A couple Algebra 1 problems (polynomial division, mixture problem)
Divide. (x^3 - 3x^2 + 3x + 1) divided by (x - 2)

A solution containing 20% acid is mixed with a solution containing 10% acid to make 200 liters of a solitution containing 12% acid. How much of the 10% solution was used?

Also... would √(t-2)^2 be just t-2 or the absolute value of t-2?
Thanks!
• Apr 29th 2012, 04:23 PM
skeeter
Re: A couple Algebra 1 problems (polynomial division, mixture problem)
Quote:

Originally Posted by Tsuzuku
Divide. (x^3 - 3x^2 + 3x + 1) divided by (x - 2)

use synthetic division

A solution containing 20% acid is mixed with a solution containing 10% acid to make 200 liters of a solitution containing 12% acid. How much of the 10% solution was used?

(200-x)(.20) + x(.10) = 200(.12)

solve for x

Also... would √(t-2)^2 be just t-2 or the absolute value of t-2?

√(t-2)^2 = |t-2|

...
• Apr 29th 2012, 05:00 PM
Tsuzuku
Re: A couple Algebra 1 problems (polynomial division, mixture problem)
Quote:

Originally Posted by skeeter
...

what is synthetic division
• Apr 29th 2012, 05:23 PM
skeeter
Re: A couple Algebra 1 problems (polynomial division, mixture problem)
Quote:

Originally Posted by Tsuzuku
what is synthetic division

Synthetic Division
• Apr 29th 2012, 07:07 PM
HallsofIvy
Re: A couple Algebra 1 problems (polynomial division, mixture problem)
If you don't know "synthetic division", just use regular division (synthetic division is really just a short cut). How many times does x divide into $x^3$? Now, just as you do in division of numbers multiply that quotient by the divisor, x- 2, and subtract from $x^3- 3x^3- 3x+ 1$ to get the next dividend and continue.

"A solution containing 20% acid is mixed with a solution containing 10% acid to make 200 liters of a solution containing 12% acid. How much of the 10% solution was used?"
Let x be the amount of 10% solution. Then 200- x is the amount of 20% solution. The amount of acid in the 10% solution is .1x and the amount of acid in the 20% solution is .2(200- x)= 40- .2x so the total amount of acid is .1x+ 40- .2x= 40- .1x. The amount of acid in 200 liters of 12% solution is .12(200)= 24 so you must have 40- .1x= 24. Solve for x.