5 x 3Coefficient

Exponent/ Power/ Index

Variable/ Literal/ Base

ALGEBRAThe science of equations/ expression.

The letters used in algebra is known as literal or variable.

Note: 1x1 can be written as x

–1xy can be written as –xy

Exercise1.Form Algebraic expressions

1. 7 times a number x2. 4 times a number x minus 5

3. The sum of two numbers x and y divided by another number p

4. Half of a number y5. Double the product of the numbers x,y and z.

6. Double the product of a and b and divide by 5

7. 3 times a number minus 4 times a number y.

8. Ali bought p pencils for Rf. 3 each and q pens for Rf. 7 each. What is the total cost?

9. 7ab times twice xy

SIMPLIFICATION OF ALGEBRAIC EXPRESSION. Simplifying like terms To multiply and divide algebraic expression follow the indices rule. You can add and subtract only like terms. That is, the variables and their exponents should be

same.

Eg: a) 3ab + 2ab = 5ab but 3ab +2b cannot be added togetherb) 3a2 – 2a2 = a2

3a – 2a2 cannot be added together

EXERCISE 1Simplify

1. 9ab – 5ab2. 6xy – 12xy + 2xy3. 4gh + 5jk – 2gh +74. 2x2y – x2y + 3xy2 –2 y2x5. 7cd – 8dc + 3cd6. 4ab + 10bc – 2ab – 5cd7. 3ba – ab + 3ab – 5ab8. 2p2 – 5p2 + 2p – 4p9. 3a2b – 2ab +4 a2b – ba

10.x5 – 5x3 + 2 – 2x3

11. h3+5h – 3 – 4h2 – 2h +7 +5h2 12. 23z2 + 17k2 – 3z2 +813. x5+ x5+ x5+x5

14. x2+ 5x +4 – x2 + 6x –315. 2 + 3x2+ x4 – 3x2 +116. 7x × 2y2 ×(2y)2

17. (3x)2 × 3x2 × 5y18. 2xy2 × 3x2y + 4x3y3

Expand and simplify [Single bracket]

Expand the bracket by multiplying the term out side the bracket with each term inside the bracket.

8 x−2 (3 x+5 )8 x−6 x−10

And then simplify by collecting all like terms.8 x−6 x−10

2 x−10

EXERCISE 2Expand and Simplify

1. 4( 2x – 5) 2. 5x (3x +5y)3. 3x + 2( x+1)4. 7+ 3(2x – 3)5. 3x – 4(2x – 5)6. 3ab – 2a(b – 2) 7. x(x – 2) + 3x(x – 3)8. y(3y – 1) – (3y – 1)9. 7b(a + 2) – a(3b +3)10. 3(x – 2) – (x – 2)11. 7(2x +2) – (x – 2)

Expand and simplify [Multiplying two expressions in brackets]

Multiply each term in the first bracket with each term in the second bracket.[Remember to take the terms with the sign]

(3+2 y )( 4 y−6 )

3 (4 y−6 )+2 y (4 y−6)Remove both the brackets

3 (4 y−6 )+2 y (4 y−6 )

12 y−18+8 y2−12 y

Simplify the like terms

12 y−18+8 y2−12 y

8 y2−18

EXERCISE 3Expand and Simplify

1. (x +1) (x +3)2. (y +4) (y+5)3. (x – 2) (x +2)4. (x+5) (x – 2)5. (x – 3) (x – 2)6. (3x +y) (x + 2y)

7. (5x – y) (3y – x)8. 2(x– 1) (x+2)9. 4(2y – 3) (3y +2)10. 2x(2x – 1) (2x+1) 11. (x + 4)2

12. (2x +y)2

13. (x+ 2)2 + (2x + 1)2 14. (x – 3)2 – (x + 1)2

Expanding brackets using IDENTITIES(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

EXERCISE 4 Expand using identities

1. (3 + x)2

2. (5 + y)2

3. (2p + 5)2

4. (3x + 2y)2

5. (3 – x)2

6. (7 – y)2

7. (x – 6)2

8. (2x – 2y)2

9. (5m – 4n)2

10. (x + 2)2 + (x + 4)2

11. (2y – 1)2 + (y – 3)2

12. (5– x)2 + (6 + x)2

13. (4 + 2x)2 – (x + 1)2

14. (2g – 6)2 – (g + 1)2

15. (x + 7)2 – (2x + 3)2

16. (y + 3)2 – (y – 3)2

Find the unknown p and/or q of the following using identities.

1. (x + p)2 = x2 + 8x + 16

2. (x + q)2 = x2 +12x + 36

3. (x – p)2 = x2 – 10x + 25

4. (px + q)2 = 4x2 +12x +9

5. (px – 1)2 = 16x2 – qx + 1

6. (px – q)2 = 4x2 – 20x + 25

SUBSTITUTION

EXERCISE 1. Evaluate the following given that x = –3 and y = 2.

a) x2 b) x2 – 2y c) xy2 d) (xy)2 e) x2

yf) 2y2 g) x + 3y

h) (2x)2 – (3y)2 i) 3x2 + 3 j) x2 – y2 k) 2x + 3y4

2. Evaluate the following given that a = 4, b = –2 and c = –3.

a) a(b + c) b) a2 (b – c) c) 2c(a – c) d) b2(2a +3c)

e) b2 + 2b + af) √(a2+c2) g) b2

a+ 2 c

b h) √ab+c2

3. Evaluate the following given that k = –3, m = 1 and n = –4.

a) 2 k+mk−n

b) 5m√k2+n2 c) kn−k

2md)

k+m+n

k2+m2+n2 e) m√k−n

f) m2(2k2 – 3n2) g) k2m2( m – n)

CHANGING THE SUBJECT OF THE FORMULA

EXERCISE 1

1. Make x the subject of the following.

a) 2x = 5

b) Ax = B

c) 9x = T + N

d) x + B = T

e) N = x + D

f) N2 + x = P + Q

g) x – A = E + K

h) F = B – x

2. Make y the subject of the following

a) Ay + C = N

b) Ny – F = H

c) Z – Py = T

d) j = my + c

e) A(y + B) = C

f) b(y – d) = q

g) n = r(y + t)

h) g = m(y + n)

EXERCISE 2

1. Make a the subject of the following.

a) aD

= B

b) a−A

B = T

c) g = a−R

D

d) Aa+B

D = B

e) n = ea−f

h

f) M (a+B)

N = T

g) ra

= B

h) pa

+ q = M

i) C – da

= B

j) Y

a−C – T = 0

2. Make x the subject of the following

a) x2 = B

b) x2 + A = B

c) b = a+ x2

d) C– x2 = m

e) mx2 = n

f) ax2 – t = m

g) m

x2 = a + b

h) x2

a + b = c

i) √ x = 2

j) √ ( x+1 ) = 5

k) √ ( x−D ) = A2

l) g = √ (c−x )

m) √ ( M−N x ) = P

n) a√ ( x2−k ) = b

o) √ Mx

= N

p) √ ( x2−4 ) = 6

L.C.M AND H.C.F OF ALGEBRAIC EXPRESSIONS

L.C.M. of algebraic expressions should include all the terms with their highest power.

H.C.F. of algebraic expressions should include the common terms with their lowest power

EXERCISE

Find the LCM and HCF of the following algebraic expressions.

1. a2 , a3 , a3 ,a5

2. 10a3b2c3, 8a2b2c2

3. a2 , b , c3

4. p3qr2, pq2r3 , 5p2q3r5. p3q2 , p2 q3 , p2q6. a2b3c3, a3b3 ab2c2

7. 3mn2, 6mnp, 12m2np8. 2ab, 5b, 7ab2

9. 3x2yz, 6xy2z3, 3xyz2

10. x2, x3, x5

11. a3b, a2b2, ab3

12. 6x2, 9x3, 15x

13. 2y4, 8y, 10y2

14. 5p2, 3q3, 7pq15. 5xy, 10x2y, 15xy2

16. 10a2b, 15ab2, 20a2b17. 3x, 618. (x+7)2, (x+7)2, (x+7)19. x, (x−1)20. (m−n3), (m−n)21. (x+1)2, (x−1), (x+1) (x−1)2

22. (2x+3)2, (x+4), (2x+3)(x+4)23. (x+1)(x−3), (x−3)(x+4), (x+4)(x−2)24. (y + 2)2, (y + 2)(y – 3), 3(y + 2)

FACTORIZATION

SIMPLE FACTORIZATION

6 x2+14 x

Step 1:

Take out the highest common factors [HCF] of the variables and the coefficients

2 ×3 x× x2+2 ×7 x

HCF =2 x

Step 2:

Write the remaining terms inside the bracket

2 ×3 x× x2+2 ×7 x

2 x (3 x2+7)

EXERCISE

Factorize the following

1) 4x – 6

2) 8m +12n +16r

3) 3pqr – 9pqs

4) 8x2y – 4xy2

5) 8pq + 6pr – 4ps

6) 5pq – 10qr + 15qs

7) 4ut – 16t + 20rt

8) m3 – m2n + mn2

9) 56x2y – 28xy2 10) 72m2n + 36mn2 – 18m2n2

FACTORIZATION BY GROUPING

x2−xy−2 x+2 y

Step 1:

Group the expression into two parts

x2−xy−2 x+2 y

Step 2:

Apply simple factorization to the groups separately

x2−xy−2 x+2 y

x (x− y )−2(x− y)

Step 3:

Apply simple factorization to the answer

x (x− y )−2(x− y)

( x− y )(x−2)

EXERCISE

1) 3m + 3n + mx + nx

2) 6x + xy + 6z + yz

3) rs – 2ts + rt – 2t2

4) ab – 4bc + ac – 4c2

5) mn – 2mr – 3rn – 6r2

6) sx – tx + sy – ty

7) 2mh – 2mk + nh – nk

8) 2ax – 2ay – bx + by

9) ms +2mt2 – ns – 2nt2

10) pr – 4p – 4qr + 16q

FACTORIZATION OF TRINOMIALS

3 x2+7 x−6

Step 1:

Multiply the x2 term and constant

3 x2+7 x−6

3 x2×−6=−18 x2

Step 2:

Find factors of the product such that they add up to the x term.

Step 3:

Split the middle term using the factors

3 x2+7 x−6

= – 18x2

9x – 2x7

x9

3 x2+9 x−2x−6

Step 4: Factorize by Grouping

3 x2+9 x−2x−6

3x(x + 3) – 2 (x + 3)

(x + 3)(3x – 2)

EXERCISE

1) x2 + 8x + 12

2) x2 + 6x + 5

3) x2 + 13x + 12

4) x2 + 6x + 9

5) x2 – 7x + 12

6) x2 – 8x + 12

7) x2 – 12x + 36

8) x2 – 15x + 36

9) x2 + 2x – 15

10) x2 + x – 12

11) x2 + 3x – 54

12) x2 + 2x – 8

13) x2 – 2x – 8

14) x2 – x – 20

15) x2 – 2x – 15

16) x2 – x – 12

17) 2x2 + 4x + 2

18) 2x2 + 7x + 6

19) 2x2 + x – 6

20) 2x2 – 7x + 6

21) 9x2 – 6x + 1

22) 6x2 – x – 1

23) 3x2 + 8x + 4

24) 2x2 – 3x – 5

FACTORIZATIONS BY DIFFERENCE IN TWO SQUARE

25 y2−144

Using the identity: a2 – b2 = (a + b)(a – b)

(5y)2 – (12)2

(5y + 12)(5y – 12)

EXERCISE

1) x2 – 16

2) x2 – y2

3) 4x2 – 1

4) 16x2 – y2

5) 25p2 – 36q2

6) 9m2 – 4n2

7) m2n2 – 9 p2

8) 9

16 a2 – 25

9) 144x2 – 1

10) 4a4 – 16b2

11) x2 – x

12) 9x3 – 36x

13) 3y3 – 12y

14) a3 – ab2

15) 18m3 – 8mn2