1 Topic 6.6.1 Factoring Quadratics. 2 What it means for you: You’ll learn how to factor simple...

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Topic 6.6.1Topic 6.6.1

Factoring QuadraticsFactoring Quadratics

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What it means for you:You’ll learn how to factor simple quadratic expressions.

Factoring QuadraticsFactoring QuadraticsTopic

6.6.1

Key words:• quadratic• polynomial• factor• binomial

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.

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In Section 6.5 you worked out common factors of polynomials.


6.6.1

Factoring quadratics follows the same rules, but you have to watch out for the squared terms.

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Polynomials as Products of Two or More Factors


6.6.1

A quadratic polynomial has degree two, such as 2x2 – x + 7 or x2 + 12.

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).

So, x2 + bx + c = x2 + (m + n)x + mn

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

Therefore b = m + n and c = mn

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6.6.1

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.

b = m + nb = m + nc = mnc = mn

So, to factor x2 + bx + c, you need to find two numbers, m and n, that multiply together to give c…

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Example 1

Topic

6.6.1

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.

Solution follows…

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…

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Example 1

Topic

6.6.1

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)

= x2 + 3x + 2x + 6

= x2 + 5x + 6

This is the same as the original expression, so the factors are correct.

Using the distributive property

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Example 2

Topic

6.6.1

Factor x2 – x – 6.

Solution

Find two numbers that multiply to give –6 and add to give –1, the coefficient of x.

Solution follows…

Because c is negative (–6), one number must be positive and the other negative.

x2 – x – 6 = (x – 3)(x + 2)


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Example 2

Topic

6.6.1

Factor x2 – x – 6.


Check whether the binomial factors are correct:

(x – 3)(x + 2) = x(x + 2) – 3(x + 2)

= x2 + 2x – 3x – 6

= x2 – x – 6



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Example 3

Topic

6.6.1

Factor x2 – 5x + 6.

Solution

Find two numbers that multiply to give +6 and add to give –5, the coefficient of x.

Solution follows…

Because c is positive (6) but b is negative, the numbers must both be negative.

x2 – 5x + 6 = (x – 2)(x – 3)


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Factor x2 – 5x + 6.


Example 3

Topic

6.6.1


Check whether the binomial factors are correct:

(x – 2)(x – 3) = x(x – 3) – 2(x – 3)

= x2 – 3x – 2x + 6

= x2 – 5x + 6



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6.6.1

Factor x2 + 2x – 8.

SolutionFind two numbers that multiply to give –8 and add to give +2, the coefficient of x.

Solution follows…

x2 + 2x – 8 = (x + 4)(x – 2)

Check whether the binomial factors are correct: (x + 4)(x – 2) = x(x – 2) + 4(x – 2)

= x2 – 2x + 4x – 8

= x2 + 2x – 8



Example 4

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Factor each expression below.

Guided Practice


6.6.1

Solution follows…

1. a2 + 7a + 10 2. x2 + 7x + 12

3. x2 – 17x + 72 4. x2 + x – 42

5. b2 + 2b – 24

2 × 5 = 10; 2 + 5 = 7; so (a + 2)(a + 5)

–8 × –9 = 72; –8 + –9 = –17; so (x – 8)(x – 9)

–4 × 6 = –24; –4 + 6 = 2; so (b – 4)(b + 6)

3 × 4 = 12; 3 + 4 = 7; so (x + 3)(x + 4)

–6 × 7 = –42; –6 + 7 = 1; so (x – 6)(x + 7)

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Guided Practice


6.6.1

Solution follows…

6. a2 – a – 42 7. x2 – 15x + 54

8. m2 + 2m – 63 9. t2 + 16t + 55

10. p2 + 9p – 10

–7 × 6 = –42; –7 + 6 = –1; so (a – 7)(a + 6)

9 × –7 = –63; 9 + –7 = 2; so (m + 9)(m – 7)

–1 × 10 = –10; –1 + 10 = 9; so (p – 1)(p + 10)

–6 × –9 = 54; –6 + –9 = –15; so (x – 6)(x – 9)

5 × 11 = 55; 5 + 11 = 16; so (t + 5)(t + 11)

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Guided Practice


6.6.1

Solution follows…

11. x2 – 3x – 18 12. p2 + p – 56

13. x2 – 2x – 15 14. n2 – 5n + 4

15. x2 – 4

–6 × 3 = 18; –6 + 3 = –3; so (x – 6)(x + 3)

3 × –5 = –15; 3 + –5 = –2; so (x + 3)(x – 5)

–2 × 2 = –4; –2 + 2 = 0; so (x – 2)(x + 2)

–7 × 8 = –56; –7 + 8 = 1; so (p – 7)(p + 8)

–1 × –4 = 4; –1 + –4 = –5; so (x – 1)(x – 4)

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Guided Practice


6.6.1

Solution follows…

16. x2 – 25 17. 4x2 – 64

18. 9a2 – 36 19. x2 – 49a2

20. 4a2 – 100b2

–5 × 5 = –25; –5 + 5 = 0; so (x – 5)(x + 5)

First factor the 9: 9(a2 – 4)Now 2 × –2 = –4; 2 + –2 = 0; so 9(a – 2)(a + 2)

First factor the 4: 4(a2 – 25b)Now –5b × 5b = –25b2; –5b + 5b = 0; so 4(a – 5b)(a + 5b)

First factor the 4: 4(x2 – 16)Now 4 × –4 = –16; 4 + –4 = 0; so 4(x – 4)(x + 4)

7a × –7a = –49a2; 7a + –7a = 0; so (x – 7a)(x + 7a)

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Find the value of ? in the problems below.


Independent Practice

Solution follows…

Topic

6.6.1

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 + ?) 10

–4

–16

–1

17

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Solution follows…

Topic

6.6.1

6. x2 – 121 7. 4c2 – 64

8. 16a2 – 225 9. x2 + 2x + 1

10. x2 + 8x + 16 11. b2 – 10b + 25

12. d2 + 21d + 38 13. x2 – 13x + 42

14. a2 – 18a + 45 15. 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 – 4)(a – 12)(a – 3)(a – 15)

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Solution follows…

Topic

6.6.1

16. x2 + 18x + 17 17. x2 – 24x + 80

18. a2 + 5a – 24 19. b2 – 19b – 120

20. x2 + 14x – 72

(x + 1)(x + 17)

(b – 24)(b + 5)

(x – 4)(x – 20)

(a – 3)(a + 8)

(x – 4)(x + 18)

21. Determine whether (x + 3) is a factor of x2 – 2x – 15.

22. If 2n3 – 5 is a factor of 12n5 + 2n4 – 30n2 – 5n, find the other factors.

(x2 – 2x – 15) = (x + 3)(x – 5), so (x + 3) is a factor.

n, (6n + 1)

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24. If (2x + 5) is a factor of 2x3 + 15x2 + 13x – 30, find the other factors.



Solution follows…

Topic

6.6.1

25. If (a – 1) is a factor of a3 – 6a2 + 9a – 4, find the other factors.

26. If (x – 2) is factor of x3 + 5x2 – 32x + 36, find the other factors.

23. If (8n – 3) is a factor of 8n3 – 3n2 – 8n + 3, find the other factors.(n – 1), (n + 1)

(x – 1), (x + 6)

(a – 1), (a – 4)

(x – 2), (x + 9)

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Solution follows…

Topic

6.6.1

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.

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Topic

6.6.1

Round UpRound 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.

1 Topic 6.6.1 Factoring Quadratics. 2 What it means for you: You’ll learn how to factor simple...

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Transcript of 1 Topic 6.6.1 Factoring Quadratics. 2 What it means for you: You’ll learn how to factor simple...