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Precalculus Review
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0.3
Multiplying and Factoring Algebraic Expressions
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3. Multiplying Algebraic Expressions

4. Multiplying Algebraic Expressions
Distributive Law
The distributive law for real numbers states that
a(b ± c) = ab ± ac
(a ± b)c = ac ± bc
for any real numbers a, b, and c.
Quick Example
2(x – 3) is not equal to 2x – 3 but is equal to x – 2(3) = 2x – 6.

5. Multiplying Algebraic Expressions
There is a quicker way of expanding expressions, called the “FOIL” method (First, Outer, Inner, Last). Consider, for instance, the expression (x + 1)(x – 2).
The FOIL method says: Take the product of the first terms: x  x = x2, the product of the outer terms: x  (–2) = –2x, the product of the inner terms: 1  x = x, and the product of the last terms: 1  (–2) = –2, and then add them all up, getting x2 – 2x + x – 2 = x2 – x – 2.

6. Example 1 – FOIL a. (x – 2)(2x + 5) = 2x2 + 5x – 4x – 10
b. (x2 + 1)(x – 4) = x3 – 4x2 + x – 4
c. (a – b)(a + b) = a2 + ab – ab – b2
= a2 – b2

7. Example 1 – FOIL d. (a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2
cont’d
d. (a + b)2 = (a + b)(a + b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
e. (a – b)2 = (a – b)(a – b)
= a2 – ab – ab + b2
= a2 – 2ab + b2

8. Multiplying Algebraic Expressions
Special Formulas
(a – b)(a + b) = a2 – b2
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
Quick Examples
1. (2 – x)(2 + x) = 4 – x2
2. (1 + a)(1 – a) = 1 – a2
3. (x + 3)2 = x2 + 6x + 9
4. (4 – x)2 = 16 – 8x + x2
Difference of two squares
Square of a sum
Square of a difference

9. Factoring Algebraic Expressions

10. Factoring Algebraic Expressions
Factoring Using a Common Factor
To use this technique, locate a common factor—a term that occurs as a factor in each of the expressions being added or subtracted (for example, x is a common factor in 2x2 + x, because it is a factor of both 2x2 and x). Once you have located a common factor, “factor it out” by applying the distributive law.
Quick Examples
2x3 – x2 + x has x as a common factor, so
2x3 – x2 + x = x(2x2 – x + 1)

11. Factoring Algebraic Expressions
We would also like to be able to reverse calculations such as (x + 2)(2x – 5) = 2x2 – x – 10. That is, starting with the expression 2x2 – x – 10, we would like to factor it to get the expression (x + 2)(2x – 5).
An expression of the form ax2 + bx + c, where a, b, and c are real numbers, is called a quadratic expression in x.
Thus, given a quadratic expression ax2 + bx + c, we would like to write it in the form (dx + e)(f x + g) for some real numbers d, e, f, and g.

12. Factoring Algebraic Expressions
There are some quadratics, such as x2 + x + 1, that cannot be factored in this form at all.
Here, we consider only quadratics that do factor, and in such a way that the numbers d, e, f, and g are integers.
The usual technique of factoring such quadratics is a “trial and error” approach.

13. Factoring Algebraic Expressions
Factoring Quadratics by Trial and Error
To factor the quadratic ax2 + bx + c, factor ax2 as (a1x)(a2x) (with a1 positive) and c as c1c2, and then check whether or not ax2 + bx + c = (a1x ± c1)(a2x ± c2). If not, try other factorizations of ax2 and c.
Quick Example
To factor x2 – 6x + 5, first factor x2 as (x)(x), and 5 as(5)(1):
(x + 5)(x + 1) = x2 + 6x + 5.
(x – 5)(x – 1) = x2 – 6x + 5.
No good
Desired factorization

14. Example 3 – Factoring Quadratics
Factor the following:
a. 4x2 – 5x – 6
b. x4 – 5x2 + 6
Solution:
a. Possible factorizations of 4x2 are (2x)(2x) or (x)(4x).
Possible factorizations of –6 are (1)(–6), (2)(–3).

15. Example 3 – Solution
cont’d
We now systematically try out all the possibilities until we come up with the correct one.
(2x)(2x) and (1)(–6): (2x + 1)(2x – 6) = 4x2 – 10x – 6
(2x)(2x) and (2)(–3): (2x + 2)(2x – 3) = 4x2 – 2x – 6
(x)(4x) and (1)(–6): (x + 1)(4x – 6) = 4x2 – 2x – 6
(x)(4x) and (2)(–3): (x + 2)(4x – 3) = 4x2 + 5x – 6
Change signs: (x – 2)(4x + 3) = 4x2 – 5x – 6
No good
No good
No good
Almost!
Correct

16. Example 3 – Solution
cont’d
b. The expression x4 – 5x2 + 6 is not a quadratic, you say?
Correct. It’s a quartic (a fourth degree expression).
However, it looks rather like a quadratic. In fact, it is quadratic in x2, meaning that it is
(x2)2 – 5(x2) + 6 = y2 – 5y + 6
where y = x2.

17. Example 3 – Solution The quadratic y2 – 5y + 6 factors as
cont’d
The quadratic y2 – 5y + 6 factors as
y2 – 5y + 6 = (y – 3)(y – 2) so
x4 – 5x2 + 6 = (x2 – 3)(x2 – 2)
This is a sometimes useful technique.