Question

exponent rules
name\trule\texamples
adding & subtracting monomials\tcombine like terms!!! (do not change common variables and exponents!)\t1. $9x^2y - 10x^2y = $

1. subtract $6vw$ from $8vw$.

product rule\t$x^a \cdot x^b = x^{a + b}$\t1. $h^2 \cdot h^6 = $

1. $(-2a^2b) \cdot (7a^3b) = $

power rule\t$(x^a)^b = x^{a(b)}$\t1. $(x^2)^3 = x^6$

1. $(-2m^5)^2 \cdot m^3 = $

quotient rule\t$\frac{x^a}{x^b} = x^{a - b}$\t1. $\frac{27x^5}{42x} = $

1. $\frac{(y^2)^2}{y^4} = $

negative exponent rule\t$x^{-a} = \frac{1}{x^a}$\t1. $-5x^{-2} = -5$

1. $\frac{4k^2}{8k^5} = $

zero exponent rule\t$x^0 = 1$\t1. $7x^0 = $

1. $\frac{(w^4)^2}{w^8} = $

© gina wilson (all things algebra®, llc), 2012 - 2017

Explanation:

Step1: Combine like monomials

$9x^2y - 10x^2y = (9-10)x^2y$

Step2: Subtract coefficients

$(9-10)x^2y = -1x^2y = -x^2y$

Step3: Subtract like monomials

$8w - 6w = (8-6)w$

Step4: Subtract coefficients

$(8-6)w = 2w$

Step5: Apply Product Rule

$h^2 \cdot h^6 = h^{2+6}$

Step6: Add exponents

$h^{2+6} = h^8$

Step7: Multiply coefficients, apply Product Rule

$(-2a^2b) \cdot (7a^3b) = (-2 \times 7)a^{2+3}b^{1+1}$

Step8: Simplify terms

$(-2 \times 7)a^{2+3}b^{1+1} = -14a^5b^2$

Step9: Apply Power Rule

$(x^2)^3 = x^{2 \times 3}$

Step10: Multiply exponents

$x^{2 \times 3} = x^6$

Step11: Apply Power Rule, then Product Rule

$(-2m^5)^2 \cdot m^3 = ((-2)^2m^{5 \times 2}) \cdot m^3 = 4m^{10} \cdot m^3$

Step12: Add exponents

$4m^{10} \cdot m^3 = 4m^{10+3} = 4m^{13}$

Step13: Apply Quotient Rule

$\frac{27x^5}{42x} = \frac{27}{42}x^{5-1}$

Step14: Simplify fraction and exponent

$\frac{27}{42}x^{5-1} = \frac{9}{14}x^4$

Step15: Apply Power Rule, then Quotient Rule

$\frac{(y^2)^2}{y^4} = \frac{y^{2 \times 2}}{y^4} = \frac{y^4}{y^4}$

Step16: Subtract exponents

$\frac{y^4}{y^4} = y^{4-4} = y^0 = 1$

Step17: Apply Negative Exponent Rule

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

Step18: Rewrite simplified form

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

Step19: Apply Quotient Rule

$\frac{4k^2}{8k^5} = \frac{4}{8}k^{2-5}$

Step20: Simplify fraction and exponent

$\frac{4}{8}k^{2-5} = \frac{1}{2}k^{-3} = \frac{1}{2k^3}$

Step21: Apply Zero Exponent Rule

$7x^0 = 7 \times 1$

Step22: Multiply to simplify

$7 \times 1 = 7$

Step23: Apply Power Rule, then Quotient Rule

$\frac{(w^4)^2}{w^8} = \frac{w^{4 \times 2}}{w^8} = \frac{w^8}{w^8}$

Step24: Subtract exponents

$\frac{w^8}{w^8} = w^{8-8} = w^0 = 1$

Answer:

Adding & Subtracting Monomials

1. $-x^2y$
2. $2w$

Product Rule

1. $h^8$
2. $-14a^5b^2$

Power Rule

1. $x^6$
2. $4m^{13}$

Quotient Rule

1. $\frac{9}{14}x^4$
2. $1$

Negative Exponent Rule

1. $-\frac{5}{x^2}$
2. $\frac{1}{2k^3}$

Zero Exponent Rule

1. $7$
2. $1$