1) The document discusses various methods for manipulating and solving algebraic expressions, including adding, subtracting, and factoring polynomials. 2) Factoring techniques include grouping like terms, using the difference of squares formula, and recognizing perfect square trinomials. 3) The quadratic formula is introduced as a way to solve quadratic equations of the form ax2 + bx + c = 0.

1. Math Science Option 406
Math-Sci
Algebraic Expressions

2. Adding Like Terms
 Simplify (2x + 5y) + (3x – 2y)
 I'll clear the parentheses, group like terms,
and then simplify:
 (2x + 5y) + (3x – 2y)
= 2x + 5y + 3x – 2y
= 2x + 3x + 5y – 2y
= 5x + 3y

3. Adding Vertically
 Simplify (2x + 5y) + (3x – 2y) Vertically
 I'll put each variable in its own column; in
this case, the first column will be the x-
column, and the second column will be the
y-column:
 2x + 5y
 + 3x - 2y
 5x + 3y same as before

4. Adding Polynomials
 Simplify (3x3 + 3x2 – 4x + 5) +
(x3 – 2x2 + x – 4)
 I can add horizontally:
 (3x3 + 3x2 – 4x + 5) + (x3 – 2x2 + x – 4)
= 3x3 + 3x2 – 4x + 5 + x3 – 2x2 + x – 4
= 3x3 + x3 + 3x2 – 2x2 – 4x + x + 5 – 4
= 4x3 + 1x2 – 3x + 1

5. Subtracting Polynomials
 (x3 + 3x2 + 5x – 4) – (3x3 – 8x2 – 5x + 6)
 =(x3 + 3x2 + 5x – 4) – 1(3x3 – 8x2 – 5x + 6)
 =(x3 + 3x2 + 5x – 4) – 1(3x3) – 1(-8x2) – 1(-
5x) – 1(6)
 = x3 + 3x2 + 5x – 4 – 3x3 + 8x2 + 5x – 6
 = x3 – 3x3 + 3x2 + 8x2 + 5x + 5x – 4 – 6
 = -2x3 + 11x2 + 10x –10

6. Manipulating Algebraic Expressions
 BEDMAS – Brackets, Exponents, Division,
Multiplication then Addition, Subtraction – Left to
Right in the order they occur.
 (4x)2x – (x-3)(3x-1)
 Multiply and FOIL First Outside Inside Last
 = 8x2 – (3x2-1x-9x+3)
 = 8x2 – (3x2-10x+3)
 Change signs of bracket contents
 = 8x2 – 3x2 +10x-3
 = 5x2 +10x - 3

7. Polynomial Division
 Polynomials can be divided by single terms
 (3x2 – 6x) / x
 = 3x2 – 6x
x
 = 3x2 – 6x
x x
 3x – 6
 Polynomials can be factored. Factoring means dividing
out, BUT KEEPING, a common factor.
 3x2 – 6x x is a common factor
 = x(3x – 6) If you multiply, you will get the original.

8. Numerical Division
 Polynomials can be divided by binomials (2 term
expressions)
 The rule to follow is DMSB which stands for
Divide, Multiply, Subtract, Bring down
 21 Divide 16 by 8  gives 2
 8√168 Multiply the 2 by 8  gives 16
 -16 Subtract 16-16  gives 0
 08 Bring down the 8 and repeat DMSB
– 8
0

9. Polynomial Division
 Polynomials can be divided by binomials (2 terms)
 The rule to follow is DMSB which stands for Divide,
Multiply, Subtract, Bring down
 x + 3 Divide x2 by x  gives x
 x+8√x2+11x + 24 Multiply the x by (x+8)  gives x2+8x
 - x2+ 8x Subtract  gives 3x
 3x + 24 Bring down the 24 and repeat DMSB
– 3x + 24 Divide 3x by x  gives 3
0 Multiply 3 by (x+8)  gives 3x +24
Subtract gives 0

10. Activities
 Do Worksheet 10.1: Q 1, 3
 Do Worksheet 10.2: Q 1,2,3
 Do Worksheet 10.3: Q
 Do Worksheet 10.4: Q

11. Factoring 1: Grouping
 Factoring by Salt ‘n Pepper is the easiest
and first method you should try.
 If that does not work, try factoring by
grouping.
 E.g. 6xy + 3x – 8y – 4
 Factor each pair: 3x (2y + 1) – 4 (2y + 1)
 Group common brackets: (2y + 1)(3x –4)

12. Activity
 Factor by grouping:
 20x2 + 5x – 12x – 3
 The order does not matter…most of the time

13. Factoring 2a: Salt ‘n Pepper
 If we FOIL multiply two binomials
(x+2)(x+5), we get x2 + 7x + 10.
 Notice that 2 + 5 = 7 and 2 x 5 = 10.
 In a 2nd order trinomial, x2 + bx + c, the
middle term, b, is the result of the sum of the
two last terms of the binomials and the last
term c is the result of the product of the last
two terms of the binomials.
 Salt ‘n Pepper refers to Sum & Product

14. Activity
 Using Sum & Product, factor x2 + 11x + 24
 Determine two numbers that add to 11 and
multiply to 24
 (x + ? ) (x+ ?)
 Note that when we use S & P, the number in
front of the x2 term is 1.

15. Factoring 2b: a ≠ 1
 Sometimes we can factor out a common
factor.
 E.g. 2x2 + 14x + 20 = 2 (x2 + 7x + 10)
 So it factors to 2(x+2)(x+5)
 E.g. x3 + 7x2 + 10x = x(x2 + 7x + 10)
= x(x+2)(x+5)
 Factor 3x2 + 30x + 63

16. Factoring 2c: No common factor
 This is a modified Salt ‘n Pepper
 Consider 6x 2 + 11x + 4
 Note that there are no common factors
 Step 1: Write 6x 2 + ___ + ___ + 4
 Step 2: Determine 2 numbers that multiply to
6x4=24 and that add to 11x
 Step 3: Insert them 6x 2 + ___ + ___ + 4
 Step 4: Factor by grouping.
 Step 5: Answer is ( )( )

17. Factoring 4: Special Cases
 Perfect Square: (x + a) (x +a)
 Product: x2 + 2ax + a2
 So to know if it is a perfect square we look
for the final term to be a square and the
middle term to be ½ the square root of the
final term.
 E.g. x2 + 10x + 25: Note √25 is 5 and 10 = 2x5
 = (x + 5)(x+ 5)
 = (x+5)2

18. Factoring 4b: Special Cases
 Difference of Squares: (x-a)(x+a)
 Product: x2 – a2
 To determine if a binomial is a difference of
squares, we look for the first and last terms
to be squares and a subtraction!.
 E.g. 100x 2 – 64y2
 (10x – 8y)(10x + 8y) = 2(5x -4y)2(5x+4y)
 Fully factored = 4 (5x-4y)(5x + 4y)

19. Activities
 Go to page 170-1
 Do Q. 4, 6, 7

20. Solving Equations
 Equations can be solved using factoring.
 E.g. x2 + 7x + 10 = 0
 Factored (x+5)(x+2) = 0
 If two numbers multiply to zero, one or both
equal zero. Therefore the terms equal zero.
 So x+5 =0 and x +2 = 0
 Solving these gives us x = -5 and x = -2.
 Check: Sub x = -5, -2 into equation.

21. Activity
 Solve x2 -13x – 30 = 0
 Solve 2x2 - 20x = 32

22. Problem Solving
 The length of a rectangle is 2x + 3;
 The width is x - 2
 If the area is 9 cm2, what are the dimensions
(length, width) of the rectangle?
 A = l w

23. Solution
 A = length x width
 9 = (2x+3)(x-2)
 Or (2x +3)(x-2) =9
 FOIL: 2x2 + 3x -4x -6 = 9
 Collect: 2x2 -1x -6 -9 = 0
 Grouping: 2x2 -1x -15 = 0
 : 2x2 - 6x + 5x -15 =0
 : 2x (x-3) + 5 (x- 3)

24. Continued
 (2x + 5)(x-3) =0
 So 2x + 5 = 0 and x – 3 =0
 2x = -5 x = 3
 x = -2.5
 We dismiss x = -2.5, because we cannot have a
negative length.
 We sub x = 3 into 2x+3 and x-2
 To get length 2(3) +3 and (3) – 2
 So the length is 9 cm and the width is 1 cm.

25. Activity
 Page 171
 Question 11 and 15

26. Quadratic Root Formula
 Theorem. If ax² + bx + c = 0,
 Theorem. Then
 x = -b ±√ (b2 – 4ac)
2a

27. Quadratic Root Formula
 Theorem. If ax² + bx + c = 0,
 Theorem. Then
 x = -b ±√ (b2 – 4ac)
2a
This can give either 0 or 1 or 2 roots
If b2-4ac > 0 2 roots
If b2-4ac = 0 1 root (the square root disappears)
If b2-4ac < 0 0 roots (can’t have a negative
square root)

28. Example
 Use the quadratic formula to solve this quadratic
equation:
 3x² + 5x − 8 = 0
 Solution. We have: a = 3, b = 5, c = −8.
 With x = -b ±√ (b2 – 4ac)
2a
 x=−5 + 11 or x = - 5 - 11
6 6
 x = 6 or x = -16
6 6
 x = 1 or x = −8
3

29. Rational Roots
 These are the two roots. And they are
rational. When the roots are rational, we
could have solved the equation by factoring,
which is always the simplest method.
 3x² + 5x − 8
 = (3x + 8)(x − 1)
 x = 1 or x = −8
3

30. Activity
 Try page 172, Q 12 d, g