1. 1 Product of a Power: problems STANDARD 10 Power of a Product: problems POLYNOMIALS: PROPERTIES OF EXPONENTS PROBLEM 1 PROBLEM 2 PROBLEM 3 PROBLEM 4 PROBLEM 1 PROBLEM 2 PROBLEM 3 PROBLEM 4 Product to Zero: problems Power with Negative Exponents: problems Quotient of Powers: problems Power of a Quotient: problems ALL PROPERTIES AT ONCE END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. 2 STANDARD 10: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multi-step problems, including word problems, by using these techniques. ESTÁNDAR 10: Los estudiantes suman restan, multiplican, y dividen monomios y polinomios. Los estudiantes resuelven problemas de múltiples pasos, incluyendo problemas escritos, usando estas técnicas. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 3 a m a n = a m+n Product of Powers: For any real number a and integers m and n = x 2+5 = x 7 x 2 5 y y y 2 57 = y 2+5+7 = y 14 STANDARD 10 Write the expressions as a single power of the base: Completing equations: y y = y 3 7? x ? 5 = x 9 ? + 5 = 9 4 + 5 = 9 x 4 5 = x 9 3 + 7 = ? 3 + 7 = 10 y 3 7 = y 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 a m n = a mn Power of a Power: = x (6)(6) (3)(3) = x 18 x 6 3 y 5 2 = y (5)(5) (2)(2) = y 10 For any real numbers a and b, and integers m and n STANDARD 10 Completing equations: Write the expression as a single power of the base: y ? 4 = y 28 y 9 6 = y ? 3(?) =15 3(5) =15 = x 15 x 3 ? = x 15 x 3 5 (?)(4) =28 = y 28 y 7 4 (7)(4) =28 (9)(6) =? = y 54 y 9 6 (9)(6) =54 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 Power of a Product: (ab) n = a b nn = x y 77 (xy) 7 (-3pr) 5 = (-3) p r 5 5 5 = -243p r 5 5 For any real numbers a and b, and integers m and n STANDARD 10 Simplify the expressions: (4 6) 3 = 4 6 33 = 64 216 = 13824 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. 6 STANDARD 10 2p (-5pq) 2 3 = 2p (-5) p q 3 2 22 = 2(-5) p p q 32 2 2 = 2(25) p q 3+2 2 = 50p q 5 2 Complete the statement using (7 4) 7 4 ? 2 3 7 4 ? 2 2 3 49 16 7 64 ? 784 448 > Simplify the expression: 4p (-3pq) 3 4 = 4p (-3) p q 4 3 33 = 4(-3) p p q 43 3 3 = 4(-27) p q 4+3 3 = -108p q 7 3 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 7 Power to the zero: a 0 = 1 (4y) 0 (-3kp) 0 = 1 STANDARD 10 0 0 UNDEFINED! PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 8 a = -n n 1 a = a -n n 1 a and Power with Negative Exponents: For any real number a, and any integer n, where a = 0 3 1 a 5 1 x a = -3 x = -5 y = -9 9 1 y = z -8 = b -6 8 1 z 6 1 b STANDARD 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. 9 STANDARD 10 8 4 -5 = 8 4 + (-5) = 8 1 8 = = 0.125 Evaluate the following expressions: (-4) -3 = 1 (-4) 3 = 1 -64 1 64 = - 1 (-3) -4 = (-3) 4 = 81 (3 ) -5 -2 = 3 (-5)(-2) = 3 10 = 59049 (-5 2 3) -3 (-5 2 3) 3 1 = (-5) 2 3 3 3 3 1 = 27000 1 = - -.016 125 8 27 1 = - PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. 10 STANDARD 10 Simplify the following expressing them with positive exponents: p k -2 3 p 1 k 2 3 = p k 3 2 = r t 3 -4 = r t 34 (7h m ) 2 3 -2 (7h m ) 1 = 2 3 2 1 = 7 h m 2 (2)(2) (3)(2) 2 6 4 7 h m 1 = 6 4 49h m 1 = PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11. 11 a m a n = a m-n Quotient of powers: For any real number a, except a=0, and integers m and n = x 9-3 = x 6 =y 7-6 x x 9 3 y y 7 6 = y STANDARD 10 Simplify the quotient: Complete the equation: 4 – 2 = ? 4 – 2 = 2 x x 4 2 = x 2 x x 4 2 ? ?– 5 = 9 14 – 5 = 9 x x 14 5 = x 9 x x ? 5 9 7 – ? = 3 7 – 4 = 3 x x 7 4 = x 3 x x 7 ? 3 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12. 12 a m a n = a m-n Quotient of powers: For any real number a, except a=0, and integers m and n STANDARD 10 x x 8 3 x 4 x x 8+4 3 = x x 12 3 = = x 12-3 = x 9 Simplify the quotient: x x 9 2 x 5 x x 9+5 2 = x x 14 2 = = x 14-2 = x 12 x 7 x 4 1 x x 7 4 = = x 7-4 = x 3 x 8 x 5 1 x x 8 5 = 8-5 = x 3 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13. 13 Suppose m and n are integers and a and b are real numbers. Then the following is true: Power of a Quotient: a b n = a b n n a b -n = b a n n b a n = = y x (2)(3) (3)(3) = y x 6 9 y x 2 3 3 y x 3 2 -5 x y 2 3 5 = = x y (2)(5) (3)(5) = x y 10 15 STANDARD 10 = y x (7)(2) (2)(2) = y x 14 4 y x 7 2 2 y x 5 -6 x y 1 5 6 = = x y (1)(6) (5)(6) = x y 6 30 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14. 14 = p q 12 6 p q 4 2 ? STANDARD 10 Complete the equations: (4)(?) = 12 (2)(?) = 6 (4)(3) = 12 (2)(3) = 6 = p q 12 6 p q 4 2 3 (2)(7) = 14 (3)(7) = ? (2)(7) = 14 (3)(7) = 21 = p q 16 20 p q ? ? 4 (?)(4) = 16 (?)(4) = 20 (4)(4) = 16 (5)(4) = 20 = p q 16 20 p q 4 5 4 = t r 14 21 t r 2 3 7 = t r 14 ? t r 2 3 7 -7 3 49 ? 2 = (-7) = 49 2 (3) = ? 2 (-7) = 49 2 (3) = 9 2 -7 3 49 9 2 = 3 2 27 8 ? = (3) = 27 ? (2) = 8 ? (3) = 27 3 (2) = 8 3 3 2 27 8 3 = PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15. 15 (4x y )(-2x y z ) 5 3 4 2 2 2 3 4 = (4)(-2)x x y y z 5 2 2+5 2 = -8x y z 3+4 = -8x y z 7 7 2 (4x y )(-2x y z ) 5 3 4 2 2 Simplify the following monomial: STANDARD 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16. 16 Simplify the following monomial: STANDARD 10 (3k n )(-7k n r ) 5 2 2 6 7 7 2 2 = (3)(-7)k k n n r 5 6 = -21k n r 6+5 7 2+2 = -21k n r 11 4 7 (3k n )(-7k n r ) 5 2 2 6 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17. 17 Simplify the following monomials (only positive exponents): STANDARD 10 x y z 3 54 xy 3 x yz 2 2 4 2 3 54 x y 3 x y z 2 2 4 2 1 1 = 3 54 x y 3 x y z 2 2 4 2 1 1 = x x y y z 3 5 4 3 2 2 4 2 11 = x y z 3+2 5+4 4 x y z 3+1 2 1+2 = x y z 5 9 4 4 2 3 = = x y z 5-3 9-4 4-2 = x y z 2 5 2 Let’s multiply the two fractions Let’s group variables together in both numerator and denominator Lets apply Product of Powers Property Lets apply Quotient of Powers Property PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

18. 18 Simplify the following monomials (only positive exponents): STANDARD 10 x y z 4 65 xy 2 x yz 4 3 5 4 4 65 x y 2 x y z 4 3 5 4 1 1 = 4 65 x y 2 x y z 4 3 5 4 1 1 = x x y y z 4 6 5 2 4 3 5 4 11 = x y z 4+4 6+5 5 x y z 2+1 4 1+3 = x y z 8 115 x y z 3 4 4 = = x y z 8-4 11-3 5-4 = x y z 4 8 Let’s multiply the two fractions Let’s group variables together in both numerator and denominator Lets apply Product of Powers Property Lets apply Quotient of Powers Property PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19. 19 -18p r w 3 42 36p r w x 5 3 4 2 18 2 9 3 33 1 2 9 3 33 1 36 2 Finding the GCF between 18 and 36: 2 2 3 2 3 2 18 = 3 2 2 36 = 2 2 3 2 We take all the numbers that repeat with the least exponent: 2 3 2 GCF= = 18 p r w x 4-2 3-3 2-4 -5 = -18.. 18 36.. 18 = p r w x 2 -5 0 -22 1 p 2w x 5 2 2 - = -18p r w 3 42 36p r w x 5 3 4 2 Simplify the following monomial: STANDARD 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

20. 20 -27a b c 5 79 48a b c d 2 6 8 3 27 3 9 3 33 1 24 2 12 6 3 48 2 Finding the GCF between 27 and 48: 3 3 27 = 3 3 We take all the numbers that repeat with the least exponent: 3 GCF= a b c d 7-3 5-6 9-8 -2 = -27.. 3 48.. 3 = a b c d -9 16 -2 14 -27a b c 5 79 48a b c d 2 6 8 3 1 2 2 2 4 3 48 = 2 4 3 a 16bd 2 4 - = c 9 Simplify the following monomial: STANDARD 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

21. 21 a b c 2 33 -3 a b c d 2 6 9 3 -3 1 = (-3) b c d (-4)(-3) (-2)(-3) (-1)(-3) (-6)(-3) = b c d 18 12 6 -27 = b c d 18 12 6 3 (-3) Simplify the following monomial: = a b c d -2 -4 -6 0 (-3) -3 a b c d 3-3 2-6 3-9 -2 -3 (-3) = STANDARD 10 a b c 2 33 -3 a b c d 2 6 9 3 -3 Let’s apply Quotient of Powers property Let’s Power to the zero Let’s apply Power to a Power property. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

22. 22 a b c 3 24 2 a b c d 5 6 7 2 -2 1 = 2 b c d (-3)(-2) (-5)(-2) (-1)(-2) (-3)(-2) = b c d 66 10 4 = b c d 66 10 2 2 Simplify the following monomial: = a b c d -5 -3 0 2 -2 a b c d 2-2 3-6 4-7 -5 -2 2 = STANDARD 10 a b c 3 24 2 a b c d 5 6 7 2 -2 Let’s apply Quotient of Powers property Let’s Power to the zero Let’s apply Power to a Power property. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved