Ms. Rigano's Math Classes - Home



NAME: ______________ Factoring Quadratics Packet PERIOD:____

Finding the GCF (Greatest Common Factor)

So what does GCF mean? It’s the largest number or variable that goes into both terms.

Ask yourself what numbers or variables go into both terms. Then pull in out and see what you have left (the opposite of distribution).

Steps: - Write down all of the factors (separate all of the x’s and y’s)

- Find the biggest number under both terms, circle it, and cross off all of the other pairs of numerical factors.

- Circle any variables that are under both terms.

- Anything that’s circled is the GCF and anything that’s not circled or crossed out is what is left over and we can rewrite in the ( ). Remember to separate the leftovers from the two different terms with either a + or a - depending on whether or not the second term is positive or negative. If there is no +/-, then just use ‘and.’

EX: 15x2y4 – 105x3y2 15x2y4 105x3y2 GCF = 15x2y2

1 15 1 105 5 21

3 5 3 35 7 15

x x y y y y x x x y y 15x2y2(y2 – 7x)

Note: We only need to look at the factors of 105 that are also factors of 15

In Class Practice:

|16x2 and 8 |24x2 and 18x |10x3 + 35x |22 and 4 |

| | | | |

|8(2x2 and 1) |6x(4x and 3) |5x(2x2 + 7) |2(11 and 2) |

|15x3 + 8x2 |21xy + 14x2y2 |9x2 - 11x |30xy3 - 20x2y2 |

| | | | |

|x2(15x + 8) |7xy(3 + 2xy) |x(9x – 11) |10xy2(3y – 2x) |

Review: FOIL (First Outer Inner Last)

1) (x + 3)(x + 5) 2) (x – 3)(x + 3)

x2 + 8x + 15 x2 - 9

3) (2x + 1)(x – 5) 4) (2x – 1)2 + 5

2x2 – 9x – 5 4x2 – 4x + 6

Using the concepts of GCF with quadratics to help factor them.

The problem … x2 + 3x + 2 … does any number or variable go into all three of these terms?

So, we need to look at the expression differently. We have to figure out how to un-FOIL, or factor the problem.

Questions to ask yourself to help figure out what to do…

[pic]

When you get to “DONE” double check your answer by FOILing out your ( ).

First…Complete this Sum & Product Puzzle: Set 1

In each diagram below, write the two numbers on the sides of the “X” that are multiplied together to get the top number of the “X,” but added together to get the bottom number of the “X.”

EX:

Factors of -36: 1 -36 -1 36

2 -18 -2 18

9 -4 3 -12 -3 12

4 -9 -4 9

6 -6

We pick -4 and 9 because they add up to 5

1. 2. 3.

-3 -3 2 2 2 -15

4 -13

4. 5. 6.

12 -7 -8 3 -3 -2

7. 8. 9.

1 -15 -15 5 3 4

x2 + 3x + 2 : a x2 + b x + c (remember the standard equation for a quadratic?)

BIG X: Steps to factor…

Step 1: Complete the X with ac at the top and b at the bottom.

Step 2: Replace b with the two factors.

Step 3: Group the first two terms and the last two terms, then factor out a GCF for each side.

Step 4: Factor the expression into two binomials.

Ex. x2 + 3x + 2 a = 1 b= 3 c = 2

ac = 2

*Big X Short cut…when a = 1 Big X: when a ≠ 1 Big X Alternative: when a ≠ 1

| | | |

|( x + 1 )( x + 2 ) |x2 + 3x + 2 |1 ( from big X ( 2 |

| |x2 + 1 x + 2 x + 2 |1 ( a ( 1 |

| |x2 + 1 x + 2 x + 2 |simplify the fractions |

| |_x_( x + 1 ) + _2_( x + 1 ) ( x + 1 )( x|[pic] |

| |+ 2 ) | |

x2 + 3x + 2 : a x2 + b x + c (remember the standard equation for a quadratic?)

Little x: Steps to factor…

Step 1: Write down all of the factors of a, and all of the factors of c, separately.

Step 2: Set up the factors using an x so that they multiply diagonally and add to b.

Step 3: Factor the expression so that the top numbers are one set of parentheses and the bottom numbers are the other.

Ex. x2 + 3x + 2 a = 1 b= 3 c = 2

Factors of a = 1 Factors of c = 2

a c 1 and 1 1 and 2

X X X X

Note…you may have to play

1 + 2 = 3 = b with where the factors go

+ + +

So

= ( 1x + 2 )( 1x + 1 ) = ( x + 2 )( x + 1 ) NOTE: This method never

changes, no matter what a equals.

[pic]

Factoring Practice HW:

1) 7x2 - 31x – 20 2) 2x2 + 17x + 21 3) x2 – 7x – 18

(7x + 4)(x – 5) (2x + 3)(x + 7) (x – 9)(x + 2)

4) x2 – 16x + 63 5) x2 – 5x – 14 6) 5x2 – x – 18

(x – 7)(x – 9) (x – 7)(x + 2) (5x + 9)(x – 2)

7) [pic] 8) [pic] 9) 9x2 + 6x + 1

(x + 9)(x + 5) (y – 3)(y – 3) (3x + 1)(3x + 1)

More Factoring Practice: Factor the following quadratics as much as possible.

1. 2x2 - 72 2. 2x2 + 3x + 1 3. x3 + 6x2 + 9x

2(x + 6)(x – 6) (2x + 1)(x + 1) x(x + 3)(x + 3)

4. 18x + 12x2 + 2x3 5. 6x2 + 13 x + 6 6. 20x2 + 34x + 6

2x(x + 3)(x + 3) (3x + 2)(2x + 3) 2(2x + 3)(5x + 1)

7. [pic] 8. 9x2 – 10x + 1 9. x2 + 10x +25

(m2 – p)(m2 + p) (9x – 1)(x – 1) (x + 5)(x + 5)

Algebra 2: Factoring Quadratics WS (Factor Frenzy HW)

Factor the following quadratics as much as possible. If it is unfactorable, write prime.

|1. x2 + 5x + 6 |5. [pic] |9. [pic] |

| | | |

|(x + 3)(x + 2) |3(2m – 1)(m + 2) |5b(2b + 5)(2b + 5) |

| | | |

|2. 9x2 - 16 | | |

| | | |

|(3x + 4)(3x – 4) |6. [pic] |10. [pic] |

| | | |

| |3(k – 4)(k + 4) |2k(2k – 1) |

| | | |

|3. x2 - 14x + 49 | | |

| | | |

|(x – 7)(x – 7) |7. [pic] |11. [pic] |

| | | |

| |2(x – 10)(x – 10) |4(x + 1)(x +1) |

| | | |

|4. x2 - 4 | | |

| | |12. [pic] |

|(x – 2)(x + 2) |8. [pic] | |

| | |5(c + 6)(c + 6) |

| |3(2z + 3)(2z + 3) | |

Mod 3.2 Notes – Complex Numbers

|Review |ADD |SUBTRACT |MULTIPLY |

| |(3 + x) + (5 + x) |(4 + x) – (6 + x) |(3 + x)(5 + x) |

| |3 + x + 5 + x |4 + x – 6 – x |3(5) + 3(x) + 5(x) + x(x) |

| |3 + 5 + x + x |4 – 6 + x – x |15 + 3x + 5x + x2 |

| |8 + 2x |-2 + 0x |15 + 8x + x2 |

| | |-2 |Or x2 + 8x + 15 |

|Now |Replace ‘x’ with ‘i’ ( what is i? i is an imaginary number ( i = |

|New |(3 + i) + (5 + i) |(4 + i) – (6 + i) |(3 + i)(5 + i) |

| |3 + i + 5 + i |4 + i – 6 – i |3(5) + 3(i) + 5(i) + i(i) |

| |3 + 5 + i + i |4 – 6 + i – i |15 + 3i + 5i + i 2 |

| |8 + 2i |-2 + 0i |15 + 8i + i2 |

| | |-2 |Or i 2 + 8i + 15 |

| | | | |

| | | |* you can’t leave i2 ( so… |

| | | |i = |

| | | |i2 = -1 |

| | | | |

| | | |15 + 8i + (-1) |

| | | |15 – 1 + 8i |

| | | |14 + 8i |

TRY: (4 – i)(3 + i)

13 + i

Homework (pg 96 #3 – 18, 23 in the textbook)

Identify the real and imaginary parts of the given number. Then tell which of the following sets the number belongs to: real numbers, imaginary numbers, and complex numbers.

3. 5 + i 4. 7 - 6i 5. 25 6. i

complex complex real imaginary

(has real and imaginary part)

Add

7. (3 + 4i) + (7 + 11i) 8. (2 + 3i) + (6 – 5i)

10 + 5i 8 - 2i

9. (-1 – i) + (-10 + 3i) 10. (-9 – 7i) + (6 + 5i)

-11 + 2i -3 – 2i

Subtract

11. (2 + 8i) – (7 + 6i) 12. (4 + 5i) – (14 - i)

-5 – 3i -10 + 6i

13. (-8 – 3i) – (-9 – 5i) 14. (5 + 2i) – (5 – 2i)

1 + 2i 4i

Multiply

15. (2 + 3i)(3 + 5i) 16. (7 + i)(6 – 9i)

-9 + 19i 51 - 57i

17. (-4 + 11i)(-5 – 8i) 18. (4 - i)(4 + i)

108 - 23i 17

23. Match each product on the right with the corresponding expression on the left.

A. (3 – 5i)(3 + 5i) a. __B__ -6 + 24i

B. (3 + 3i)(3 + 5i) b. __D__ -34

C. (-3 – 5i)(3 + 5i) c. __A__ 34

D. (3 – 5i)(-3 – 5i) d. __C__ 16 - 30i

[pic]

SOLVING for x: Review x2 – 9 = 0

+9 +9

x2 = 9

[pic] *solving for x2 means I get

x = ± 3 a ±n my answer*

Now solving with complex numbers (i) – Mod 3.1 Notes

|x2 + 9 = 0 |2x2 + 1 = 0 |x2 + 25 = 0 |

|[pic] |[pic] |[pic] |

[pic]

Homework (pg 88 #3 – 6, 17 – 20 in the textbook)

Solve the quadratic equation by taking square roots.

3. 4x2 - 24 = 0 4. [pic] + 15 = 0

x = ±[pic] x = ±[pic]

5. 2(5 – 5x2) = 5 6. 3x2 – 8 = 12

x = [pic] x = [pic] = [pic]

Solve the quadratic equation by taking square roots. Allow for imaginary solutions.

17. x2 = -81 18. x2 + 64 = 0

x = ± 9i x = ± 8i

19. 5x2 – 4 = -8 20. 7x2 + 10 = 0

x = ±[pic] x = ±[pic]

SOLVING for x: ax2 + bx + c = 0 … Follow the same methods you have already learned.

|Quadratic Equation |Factored Form |Solutions |

|x2 + 3x – 10 = 0 |(x + __5__ )(x + __-2__ ) = 0 |__-5___, __2___ |

|x2 – 2x – 8 = 0 |(x + __-4__ )(x + __2__ ) = 0 |__4___, __-2___ |

|3x2 – 7x – 20 = 0 |(3x + __5__ )(x + __-4__ ) = 0 |__-5/3___, ___4__ |

|6x2 + x – 2 = 0 |(2x + __-1__ )(3x + __2__ ) = 0 |__1/2___, __-2/3___ |

HW: Solving quadratic equations by graphing, square roots, and factoring.

Find the x-intercepts of the following functions by factoring. Be sure to write your answers as points since they are intercepts. (There will be one that you CAN’T DO)

1) f(x) = x2 + 6x + 8 = 0 2) f(x) = x2 – 9x + 18

(-2, 0) and (-4, 0) (6, 0) and (3, 0)

3) f(x) = 3x2 – 16x – 12 4) f(x) = x2 + 8x

(-2/3, 0) and (6, 0) (0, 0) and (-8, 0)

5) 6m2 + 4m = 0 6) 2m2 + 8m = 5m + 20

(0, 0) and (-2/3, 0) (5/2, 0) and (-4, 0)

7) 3m2 + 7m = 9 8) m2 – 8m + 7 = 0

(.92, 0) and (-3.25, 0) (7, 0) and (1, 0)

9) The perimeter of a rectangle is 64 cm. The area is 240cm2. What are its dimensions?

L = 20 cm W = 12 cm

10) A ball is thrown from a height of 25 feet. The path of the ball is modeled by the function

f(x) = -x2 + 25 where x represents seconds and f(x) represents the height of the ball. How long does it take the ball to hit the ground after it is thrown?

0 = -x2 + 25 = (x – 5)(x + 5) ( x = 5 sec

Completing the Square:

|The original problem |[pic] |

|Move the “c” term over to the other side—only terms with variables should remain. | |

| |[pic] |

|Find [pic]. |[pic] |

|Find [pic]. |[pic] |

|Add [pic]to BOTH sides of the equation. | |

| |[pic] |

|Convert the side with 3 terms to squared binomial form – that is[pic]. | |

| |[pic] |

|Take the square root of both sides – do not forget the [pic]!!! |[pic] |

| |[pic] |

|Get x alone. Remember that [pic]means you have two answers. Simplify as necessary. | |

|The original problem (where [pic]) |2x2 + 16x – 4 = 0 |

|Divide all terms by the “a” value. |x2 + 8x – 2 = 0 |

|Move the “c” term over to the other side—only terms with variables should remain.| |

| |x2 + 8x = 2 |

|Find [pic]. |[pic] |

|Find [pic]. |[pic]= [pic] = [pic] |

|Add [pic]to BOTH sides of the equation. | |

| |x2 + 8x + 16 = 2 + 16 |

|Convert the side with 3 terms to squared binomial form – that is[pic]. | |

| |(x + 4)2 = 18 |

|Take the square root of both sides – do not forget the [pic]!!! | |

| |[pic] = [pic] |

| |x + 4 = [pic] |

| | – 4 – 4 |

|Get x alone. Remember that [pic]means you have two answers. Simplify as | |

|necessary. |x = -4 [pic] |

| | |

| |which means x = -4 + [pic] or x = -4 – [pic] |

| |(since we can’t simplify [pic], we can leave our answer with the [pic]. |

Practice:

A) x2 – 4x – 12 = 0 B) x2 – 14x – 44 = 0

x = 6 and x = -2 x = 7 + [pic] and x = 7 - [pic]

C) x2 + 2x + 93 = 0 D) x2 – 8x + 36 = 0

x = -1 + 2i[pic] and x = -1 - 2i[pic] x = 4 + 2i[pic] and x = 4 - 2i[pic]

Completing the Square HW

1) x2 + 3x - 4 = 0 2) x2 - 2x - 8 = 0

x = -4 and x = 1 x = 4 and x = -2

3) x2 + 8x + 12 = 0 4) x2 – 4x – 5 = 0

x = -6 and x = -2 x = 5 and x = -1

5) 2x2 + 16x + 1 = 0 6) 3x2 – 12x - 2 = 0

x = -4 +/-[pic] x = 2 +/- [pic]

|The Quadratic Formula | |

|Math Form: |Song Form: |

|[pic] |To the tune of “Pop Goes the Weasel” |

| | |

| |X equals negative b |

| |Plus or minus the square root |

| |of b squared minus 4ac |

| |ALL over 2a |

A negative boy couldn’t decide whether to go to a radical party. He decided to be square and missed out on 4 awesome chicks, and the whole night was over at 2am.

Ex: 0 = 4x2 + 3x - 12

a = 4 b = 3 c = -12 x = [pic]

Practice:

A) 6x2 + 7x – 1 = 0 B) 5x2 + 10x – 1 = 0 C) 3x2 – 10x + 5 = 0

x = [pic] x = [pic] x = [pic]

Then simplify

Quadratic Formula HW

1) 0 = 3x2 – 10x + 10 2) 2x2 – 20x + 3 = 0

x = [pic] x = [pic]

3) x2 – 10x + 2 = 0 4) 0 = x2 + 3x – 1 5) x2 + 5x – 2 = 0

x = [pic] x = [pic] x = [pic]

The discriminant is a small part of the quadratic formula: b2 – 4ac

[pic]so no x-intercepts

[pic] so 1 x-intercepts

[pic] so 2 x-intercepts

[pic]

Algebra 2 Worksheet: Practice Solving Quadratics

Evaluate the discriminant of each of the following equations. Tell how many solutions and what type each equation has.

1. [pic]

56 ( 2 real roots

2. [pic]

0 ( 1 real root (repeating)

Solve each of the following equations using any of the four methods we’ve learned in class. (**Know how to use ALL four methods for solving quadratics!!)

|3. [pic] |8. [pic] |

| | |

|x = 3 or x = -3 |x = -3 and x = -1 |

| | |

| | |

| | |

|4. [pic] |9. [pic] |

| | |

|x = 25 and x = 4 |x = [pic] |

| | |

| | |

| | |

|5. [pic] | |

| | |

|x = 0 and x = -7 | |

| | |

| |10. x2 – 14x – 95 = 0 |

| | |

|6. [pic] | |

| |x = 19 and x = -5 |

|x = -1 and x = 22 | |

| | |

| |11. [pic] |

| | |

|7. [pic] |x = 7 and x = -7 |

| | |

|x = 10i[pic] and x = -10i[pic] | |

| |12. [pic] |

| | |

|13. [pic] |x = -5 and x = -2 |

| | |

|x = -1 and x = -1/4 | |

| |16. [pic] |

| | |

|14. [pic] | |

| | |

| | |

| | |

| |17. [pic] |

| | |

|15. [pic] |x = 3 and x = -2 |

| | |

| | |

|x = 1 and x = 1/2 |18. [pic] |

| | |

| | |

| |x = -2 and x = 1 |

Extra Practice with solving for x: Use either factoring, completing the square, or the quadratic formula.

1. x2 – 7x + 10 = 0 2. x2 – 14x + 45 = 0

x = 2 and x = 5 x = 9 and x = 5

3. 4x2 + 8x – 77 = 0 4. 16x2 – 8x – 3 = 0

x = -11/2 and x = 7/2 ` x = 3/4 and x = -1/4

5. x2 = 15 + 2x 6. k2 – 10k + 35 = 7k – 35

x = 5 and x = -3 x = 7 and x = 10

7. n2 + 7n + n2 – 10 = 5 8. 3 = 16x – 20x2

x = 3/2 and x = -5 x = 1/2 and x = 3/10

9. 2x2 + 5 + 20x = 20 – 8x2 + x 10. x2 – 3x = 0

x = 3/5 and x = -5/2 x = 0 and x = 3

11. x2 + x + 10 = 0 12. 4x2 – 2x – 12 = 0

x = (-1 ± i√39)/2 x = -3/2 and x = 2

[pic]

-----------------------

-30

4

9

-6

4

-24

-5

-5

6

5

-84

-10

-75

7

12

-14

-15

[pic]

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download