1. Factoring Quadratics
Topic 6.6.1

2. Factoring Quadratics 6.6.1 California Standard: What it means for you:
Topic
6.6.1
Factoring Quadratics
California Standard:
11.0 Students apply basic factoring techniques to second and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials.
What it means for you:
You’ll learn how to factor simple quadratic expressions.
Key words:
quadratic
polynomial
factor
binomial

3. Topic
6.6.1
Factoring Quadratics
In Section 6.5 you worked out common factors of polynomials.
Factoring quadratics follows the same rules, but you have to watch out for the squared terms.

4. So, x2 + bx + c = x2 + (m + n)x + mn
Topic
6.6.1
Factoring Quadratics
Polynomials as Products of Two or More Factors
A quadratic polynomial has degree two, such as 2x2 – x + 7 or x
Some quadratics can be factored — in other words they can be expressed as a product of two linear factors.
Suppose x2 + bx + c can be written in the form (x + m)(x + n).
Then:
x2 + bx + c = (x + m)(x + n)
b, c, m, and n are numbers
= x(x + n) + m(x + n)
Expand out the parentheses using the distributive property
= x2 + nx + mx + mn
So, x2 + bx + c = x2 + (m + n)x + mn
Therefore b = m + n and c = mn

5. Factoring Quadratics 6.6.1 c = mn b = m + n
Topic
6.6.1
Factoring Quadratics
So, to factor x2 + bx + c, you need to find two numbers, m and n, that multiply together to give c…
So, to factor x2 + bx + c, you need to find two numbers, m and n, that multiply together to give c, and that also add together to give b.
c = mn
b = m + n

6. Factoring Quadratics 6.6.1 Factor x2 + 5x + 6. Solution
Topic
6.6.1
Factoring Quadratics
Example 1
Factor x2 + 5x + 6.
Solution
The expression is x2 + 5x + 6, so find two numbers that add up to 5 and that also multiply to give 6.
The numbers 2 and 3 multiply together to give 6 and add together to give 5.
You can now factor the quadratic, using these two numbers:
x2 + 5x + 6 = (x + 2)(x + 3)
Solution continues…
Solution follows…

7. Factoring Quadratics 6.6.1 Factor x2 + 5x + 6. Solution (continued)
Topic
6.6.1
Factoring Quadratics
Example 1
Factor x2 + 5x + 6.
Solution (continued)
To check whether the binomial factors are correct, multiply out the parentheses and then simplify the product:
(x + 2)(x + 3) = x(x + 3) + 2(x + 3)
 Using the distributive property
= x2 + 3x + 2x + 6
= x2 + 5x + 6
This is the same as the original expression, so the factors are correct.

8. Factoring Quadratics 6.6.1 Factor x2 – x – 6. Solution
Topic
6.6.1
Factoring Quadratics
Example 2
Factor x2 – x – 6.
Solution
Find two numbers that multiply to give –6 and add to give –1, the coefficient of x.
Because c is negative (–6), one number must be positive and the other negative.
x2 – x – 6 = (x – 3)(x + 2)
Solution continues…
Solution follows…

9. Using the distributive property
Topic
6.6.1
Factoring Quadratics
Example 2
Factor x2 – x – 6.
Solution (continued)
Check whether the binomial factors are correct:
(x – 3)(x + 2) = x(x + 2) – 3(x + 2)
Using the distributive property
= x2 + 2x – 3x – 6
= x2 – x – 6
This is the same as the original expression, so the factors are correct.

10. Factoring Quadratics 6.6.1 Factor x2 – 5x + 6. Solution
Topic
6.6.1
Factoring Quadratics
Example 3
Factor x2 – 5x + 6.
Solution
Find two numbers that multiply to give +6 and add to give –5, the coefficient of x.
Because c is positive (6) but b is negative, the numbers must both be negative.
x2 – 5x + 6 = (x – 2)(x – 3)
Solution continues…
Solution follows…

11. Using the distributive property
Topic
6.6.1
Factoring Quadratics
Example 3
Factor x2 – 5x + 6.
Solution (continued)
Check whether the binomial factors are correct:
(x – 2)(x – 3) = x(x – 3) – 2(x – 3)
Using the distributive property
= x2 – 3x – 2x + 6
= x2 – 5x + 6
This is the same as the original expression, so the factors are correct.

12. Using the distributive property
Topic
6.6.1
Factoring Quadratics
Example 4
Factor x2 + 2x – 8.
Solution
Find two numbers that multiply to give –8 and add to give +2, the coefficient of x.
x2 + 2x – 8 = (x + 4)(x – 2)
Check whether the binomial factors are correct:
(x + 4)(x – 2) = x(x – 2) + 4(x – 2)
Using the distributive property
= x2 – 2x + 4x – 8
= x2 + 2x – 8
This is the same as the original expression, so the factors are correct.
Solution follows…

13. Factoring Quadratics 6.6.1 Guided Practice
Topic
6.6.1
Factoring Quadratics
Guided Practice
Factor each expression below.
1. a2 + 7a x2 + 7x + 12
3. x2 – 17x x2 + x – 42
5. b2 + 2b – 24
2 × 5 = 10; = 7; so (a + 2)(a + 5)
3 × 4 = 12; = 7; so (x + 3)(x + 4)
–8 × –9 = 72; –8 + –9 = –17; so (x – 8)(x – 9)
–6 × 7 = –42; –6 + 7 = 1; so (x – 6)(x + 7)
–4 × 6 = –24; –4 + 6 = 2; so (b – 4)(b + 6)
Solution follows…

14. Factoring Quadratics 6.6.1 Guided Practice
Topic
6.6.1
Factoring Quadratics
Guided Practice
Factor each expression below.
6. a2 – a – x2 – 15x + 54
8. m2 + 2m – t2 + 16t + 55
10. p2 + 9p – 10
–7 × 6 = –42; –7 + 6 = –1; so (a – 7)(a + 6)
–6 × –9 = 54; –6 + –9 = –15; so (x – 6)(x – 9)
9 × –7 = –63; 9 + –7 = 2; so (m + 9)(m – 7)
5 × 11 = 55; = 16; so (t + 5)(t + 11)
–1 × 10 = –10; – = 9; so (p – 1)(p + 10)
Solution follows…

15. Factoring Quadratics 6.6.1 Guided Practice
Topic
6.6.1
Factoring Quadratics
Guided Practice
Factor each expression below.
11. x2 – 3x – p2 + p – 56
13. x2 – 2x – n2 – 5n + 4
15. x2 – 4
–6 × 3 = 18; –6 + 3 = –3; so (x – 6)(x + 3)
–7 × 8 = –56; –7 + 8 = 1; so (p – 7)(p + 8)
3 × –5 = –15; 3 + –5 = –2; so (x + 3)(x – 5)
–1 × –4 = 4; –1 + –4 = –5; so (x – 1)(x – 4)
–2 × 2 = –4; –2 + 2 = 0; so (x – 2)(x + 2)
Solution follows…

16. Factoring Quadratics 6.6.1 Guided Practice
Topic
6.6.1
Factoring Quadratics
Guided Practice
Factor each expression below.
16. x2 – x2 – 64
18. 9a2 – x2 – 49a2
20. 4a2 – 100b2
–5 × 5 = –25; –5 + 5 = 0; so (x – 5)(x + 5)
First factor the 4: 4(x2 – 16)
Now 4 × –4 = –16; 4 + –4 = 0; so 4(x – 4)(x + 4)
First factor the 9: 9(a2 – 4)
Now 2 × –2 = –4; 2 + –2 = 0; so 9(a – 2)(a + 2)
7a × –7a = –49a2; 7a + –7a = 0; so (x – 7a)(x + 7a)
First factor the 4: 4(a2 – 25b)
Now –5b × 5b = –25b2; –5b + 5b = 0; so 4(a – 5b)(a + 5b)
Solution follows…

17. Factoring Quadratics 6.6.1 Independent Practice
Topic
6.6.1
Factoring Quadratics
Independent Practice
Find the value of ? in the problems below.
1. x2 + 3x – 4 = (x + 4)(x + ?)
2. a2 – 2a – 8 = (a + ?)(a + 2)
3. x2 + 16x – 17 = (x + ?)(x – 1)
4. x2 – 14x – 32 = (x + 2)(x + ?)
5. a2 + 6a – 40 = (a – 4)(a + ?) –1 –4 17
–16 10
Solution follows…

18. Factoring Quadratics 6.6.1 Independent Practice
Topic
6.6.1
Factoring Quadratics
Independent Practice
Factor each expression below.
6. x2 – c2 – 64
8. 16a2 – x2 + 2x + 1
10. x2 + 8x b2 – 10b d2 + 21d x2 – 13x + 42
14. a2 – 18a a2 – 16a + 48
(x + 11)(x – 11)
4(c – 4)(c + 4)
(4a – 15)(4a + 15)
(x + 1)2
(x + 4)2
(b – 5)2
(d + 2)(d + 19)
(x – 6)(x – 7)
(a – 3)(a – 15)
(a – 4)(a – 12)
Solution follows…

19. (x2 – 2x – 15) = (x + 3)(x – 5), so (x + 3) is a factor.
Topic
6.6.1
Factoring Quadratics
Independent Practice
Factor each expression below.
16. x2 + 18x x2 – 24x a2 + 5a – b2 – 19b – 120
20. x2 + 14x – 72
(x + 1)(x + 17)
(x – 4)(x – 20)
(a – 3)(a + 8)
(b – 24)(b + 5)
(x – 4)(x + 18)
21. Determine whether (x + 3) is a factor of x2 – 2x – 15.
(x2 – 2x – 15) = (x + 3)(x – 5), so (x + 3) is a factor.
22. If 2n3 – 5 is a factor of 12n5 + 2n4 – 30n2 – 5n, find the other factors.
n, (6n + 1)
Solution follows…

20. Factoring Quadratics 6.6.1 Independent Practice
Topic
6.6.1
Factoring Quadratics
Independent Practice
23. If (8n – 3) is a factor of 8n3 – 3n2 – 8n + 3, find the other factors.
(n – 1), (n + 1)
24. If (2x + 5) is a factor of 2x3 + 15x2 + 13x – 30, find the other factors.
(x – 1), (x + 6)
25. If (a – 1) is a factor of a3 – 6a2 + 9a – 4, find the other factors.
(a – 1), (a – 4)
26. If (x – 2) is factor of x3 + 5x2 – 32x + 36, find the other factors.
(x – 2), (x + 9)
Solution follows…

21. Factoring Quadratics 6.6.1 Independent Practice
Topic
6.6.1
Factoring Quadratics
Independent Practice
What can you say about the signs of a and b when:
27. x2 + 9x + 16 = (x + a)(x + b),
28. x2 – 4x + 6 = (x + a)(x + b),
29. x2 + 10x – 75 = (x + a)(x + b),
30. x2 – 4x – 32 = (x + a)(x + b)?
Both are positive.
Both are negative.
The larger of a and b is positive, the other negative.
The larger of a and b is negative, the other positive.
Solution follows…

22. Factoring Quadratics 6.6.1 Round Up
Topic
6.6.1
Factoring Quadratics
Round Up
The method in this Topic only works for quadratic expressions that have an x2 term with a coefficient of 1 (so it’s usually written just as x2 rather than 1x2).
In the next Topic you’ll see how to deal with other types of quadratics.