Question

Show all work in dividing 9 x squared minus 1 all over 8 x minus 4 divided by the fraction 3 x squared plus 5 x minus 2 all over 2 x squared plus x minus 1 (list restrictions).
[Part 2 (2 points)] Use complete sentences to explain your process.

Answer 1

so you are given: \[\left(\frac{ 9x^2 -1 }{ 8x - 4 } \right) \div \left( \frac{ 3x^{2}+5x -2 }{ 2x^{2} +x -1 } \right)\]

Answer 2

Dividing by a fraction is the same thing as multiplying by the reciprocal so than you have \[\left( \frac{ 9x^2 -1 }{ 8x -4 } \right) * \left( \frac{ 2x^2 + x - 1 }{ 3x^2 +5 } \right)\]
simply multiply straight across to get:
\[\left( \frac{ 18x^4 +9x^3 -9x^2 -2x^2 -x +1 }{ 24x^3 +40x -12x^2 -20 } \right)\]

Answer 3

How about just finding the roots of:
The denominator of the first divisor
The numerator of the dividend
The denominator of the dividend

Answer 4

Next you collect the similar terms \[\left( \frac{ 18x^4 +9x^3 -11x^2 -x + 1 }{ 24x^3 -12 x^2 +40x -20 } \right)\]

Answer 5

@wio ah yeah can just factor the expressions to make division simpiler

Answer 6

The denominators are clear restrictions. The numerator is a restriction because you can't divided by 0.

Answer 7

I'm afraid, though I may be wrong, that doing the actual algebra may get rid of discontinuities that exist, in the same way that \(x/x = 1\) get's rid of the \(x\neq 0\) restrictions.

Answer 8

The whole expression factors to \[\left( \frac{ (x+1)(2x-1)(3x-1)(3x+1) }{ 4(2x-2)(3x^2+5) } \right)\]

Answer 9

now cancel the same terms on top and bottom to simplify to: \[\left( \frac{ (x+1)(3x-1)(3x+1) }{ 4(3x^2 +5) } \right)\]