Question

find the product or quotient for each equation.

1. $-5x^2(2x^2 + 3x - 1) = $
2. $\frac{1}{x^3}(x^5 - x + x^3) = $
3. $(3x^2 - 4x) \div x^3 = $
4. $-x(2x^2 + 5x + 1) = $
5. $\frac{3x^3 - x^2 - x - 1}{-x^3} = $
6. $-3x^4(x^4 - \frac{1}{x^2} + 2x) = $
7. $(2x^2 - 2x - 2) \div 2x = $
8. $-9x(\frac{1}{x} + 5 - 9x) = $
9. $(x^3 - 5x^2 + 4) \div x^2 = $
10. $(2x^5 - 3) \div a^2 = $
11. $-2x^3(-2x^3 + 3x^2 - 6x + 6) = $
12. $(x^2 - x - 1) \div 3x^2 = $
13. $3x^3(2x^4 + 3x - \frac{1}{x^2} + 4) = $
14. $\frac{5x^2 + 3x^4 - 5x^3 + 6}{x^2} = $
15. $x^2(9x^{2a + 2} - 4x^2) = $
16. $\frac{16x^4 + 8x^3 + 24x^2}{4x^2} = $
17. $a^2b^2c^2(-2ab - 5ac + 3bc) = $
18. $\frac{2}{3}x(9x - 6) = $

Explanation:

Step1: Distribute $-5x^2$ to each term

$-5x^2 \cdot 2x^2 + (-5x^2) \cdot 3x + (-5x^2) \cdot (-1)$
$=-10x^4 -15x^3 +5x^2$

Step2: Distribute $\frac{1}{x^3}$ to each term

$\frac{x^5}{x^3} - \frac{x}{x^3} + \frac{x^3}{x^3}$
$=x^2 - x^{-2} + 1$

Step3: Split into separate fractions

$\frac{3x^2}{x^3} - \frac{4x}{x^3}$
$=\frac{3}{x} - \frac{4}{x^2}$

Step4: Distribute $-x$ to each term

$-x \cdot 2x^2 + (-x) \cdot 5x + (-x) \cdot 1$
$=-2x^3 -5x^2 -x$

Step5: Split into separate fractions

$\frac{3x^3}{-x^3} - \frac{x^2}{-x^3} - \frac{x}{-x^3} - \frac{1}{-x^3}$
$=-3 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}$

Step6: Distribute $-3x^4$ to each term

$-3x^4 \cdot x^4 + (-3x^4) \cdot (-\frac{1}{x^2}) + (-3x^4) \cdot 2x$
$=-3x^8 + 3x^2 -6x^5$

Step7: Split into separate fractions

$\frac{2x^2}{2x} - \frac{2x}{2x} - \frac{2}{2x}$
$=x - 1 - \frac{1}{x}$

Step8: Distribute $-9x$ to each term

$-9x \cdot \frac{1}{x} + (-9x) \cdot 5 + (-9x) \cdot (-9x)$
$=-9 -45x +81x^2$

Step9: Split into separate fractions

$\frac{x^3}{x^2} - \frac{5x^2}{x^2} + \frac{4}{x^2}$
$=x -5 + \frac{4}{x^2}$

Step10: Split into separate fractions

$\frac{2x^5}{a^2} - \frac{3}{a^2}$

Step11: Distribute $-2x^3$ to each term

$-2x^3 \cdot (-2x^3) + (-2x^3) \cdot 3x^2 + (-2x^3) \cdot (-6x) + (-2x^3) \cdot 6$
$=4x^6 -6x^5 +12x^4 -12x^3$

Step12: Split into separate fractions

$\frac{x^2}{3x^2} - \frac{x}{3x^2} - \frac{1}{3x^2}$
$=\frac{1}{3} - \frac{1}{3x} - \frac{1}{3x^2}$

Step13: Distribute $3x^3$ to each term

$3x^3 \cdot 2x^4 + 3x^3 \cdot 3x + 3x^3 \cdot (-\frac{1}{x^2}) + 3x^3 \cdot 4$
$=6x^7 +9x^4 -3x +12x^3$

Step14: Split into separate fractions

$\frac{5x^2}{x^2} + \frac{3x^4}{x^2} - \frac{5x^3}{x^2} + \frac{6}{x^2}$
$=5 +3x^2 -5x + \frac{6}{x^2}$

Step15: Distribute $x^2$ to each term

$x^2 \cdot 9x^{2a+2} - x^2 \cdot 4x^2$
$=9x^{2a+4} -4x^4$

Step16: Split into separate fractions

$\frac{16x^4}{4x^2} + \frac{8x^3}{4x^2} + \frac{24x^2}{4x^2}$
$=4x^2 +2x +6$

Step17: Distribute $a^2b^2c^2$ to each term

$a^2b^2c^2 \cdot (-2ab) + a^2b^2c^2 \cdot (-5ac) + a^2b^2c^2 \cdot 3bc$
$=-2a^3b^3c^2 -5a^3b^2c^3 +3a^2b^3c^3$

Step18: Distribute $\frac{2}{3}x$ to each term

$\frac{2}{3}x \cdot 9x - \frac{2}{3}x \cdot 6$
$=6x^2 -4x$

Answer:

1. $\boldsymbol{-10x^4 -15x^3 +5x^2}$
2. $\boldsymbol{x^2 - x^{-2} + 1}$
3. $\boldsymbol{\frac{3}{x} - \frac{4}{x^2}}$
4. $\boldsymbol{-2x^3 -5x^2 -x}$
5. $\boldsymbol{-3 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}}$
6. $\boldsymbol{-3x^8 + 3x^2 -6x^5}$
7. $\boldsymbol{x - 1 - \frac{1}{x}}$
8. $\boldsymbol{81x^2 -45x -9}$
9. $\boldsymbol{x -5 + \frac{4}{x^2}}$
10. $\boldsymbol{\frac{2x^5}{a^2} - \frac{3}{a^2}}$
11. $\boldsymbol{4x^6 -6x^5 +12x^4 -12x^3}$
12. $\boldsymbol{\frac{1}{3} - \frac{1}{3x} - \frac{1}{3x^2}}$
13. $\boldsymbol{6x^7 +12x^3 +9x^4 -3x}$
14. $\boldsymbol{3x^2 -5x +5 + \frac{6}{x^2}}$
15. $\boldsymbol{9x^{2a+4} -4x^4}$
16. $\boldsymbol{4x^2 +2x +6}$
17. $\boldsymbol{-2a^3b^3c^2 -5a^3b^2c^3 +3a^2b^3c^3}$
18. $\boldsymbol{6x^2 -4x}$