SOLUTION: I have been struggling with this math problem for awhile now and I was wondering if anyone could help me? I would deeply appreciate it!! Please and Thank you!! Multiplying Pol

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Question 131981This question is from textbook Algebra Structure and Method
: I have been struggling with this math problem for awhile now and I was wondering if anyone could help me? I would deeply appreciate it!! Please and Thank you!!
Multiplying Polynomials
Solve.
(2n-3)(n^2+3n-2)=(n-1)(2n^2+5n-4) This question is from textbook Algebra Structure and Method

Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
%282n-3%29%28n%5E2%2B3n-2%29=%28n-1%29%282n%5E2%2B5n-4%29

On each side of the equation, multiply each term of the binomial by each term of the associated trinomial. Be careful about signs.

2n%5E3%2B6n%5E2-4n-3n%5E2-9n%2B6=2n%5E3%2B5n%5E2-4n-2n%5E2-5n%2B4

Now collect like terms on each side:
2n%5E3%2B3n%5E2-13n%2B6=2n%5E3%2B3n%5E2-9n%2B4

Add -2n%5E3 to both sides and add -3n%5E2 to both sides:
-13n%2B6=-9n%2B4

Add 9n to both sides and add -6 to both sides:
-4n=-2

Divide by -4:
n=%28-2%29%2F%28-4%29=1%2F2

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(2n-3)(n^2+3n-2)=(n-1)(2n^2+5n-4)
2n^3+6n^2-2n-3n^2-9n+6 = 2n^3+5n^2-4n-2n^2-5n+4
2n^3 + 3n^2 - 11n + 6 = 2n^3 + 3n^2 -9n + 4
2n = 2
n = 1
==========
Cheers,
Stan H.