Quadratic Zeroes

Presentation

•

Mathematics

•

University

•

Hard

James Gonzalez

FREE Resource

9 Slides • 33 Questions

Quadratic equation: zero principle

The first part of this lesson covers factoring completely. The second part will be the graphic organizer for the quadratic equation. The last part of this lesson is the review for chapter 6. All parts must be complete to receive full credit.

Factoring completely

Over the last few lessons we factored using the GCF, then by grouping and finally we looked at special cases of factoring. When factoring, there are instances that you may have to use all the methods mentioned. This will be called factoring completely.
Factoring completely:
1. Identify if there are any GCF in the equation and factor them pulling the GCF to the front of the equation.
2. Identify how many terms you are factoring and apply the correct method.
3. Can those factors be factored again? If so factor them.

Ex:

24m^2\ -42my\ +\ 9y^2

1. Is there a GCF?
$3\ \left(8m^2\ -14my\ +\ 3y^2\right)$
2. Identify how many terms and factor.
Multiples of 8x3 that add to -14
-12 x -2 = 24 -12 + -2 = -14
$8m^2\ -\ 12my\ -\ 2my\ +\ 3y^2$
Group the pairs and factor the GCF.

$\left(\left(8m^2\ -\ 12my\right)\ +\ \left(-2my\ +\ 3y^2\right)\right)$

4m(2n -3y) -1y(2m - 3y)
(4m - y)( 2m - 3y)
Complete factor: 3(4m - y)(2n-3y)

Multiple Choice

Factor completely
2c² + 4c - 84

(c-6)(2c+14)

(c+7)(2x-12)

(c-4)(2c+21)

2(c+7)(c-6)

Multiple Choice

Factor Completely:
3x² + 18x +15

3(x² + 6x + 5)

(x + 5)(x + 1)

3(x + 5)(x + 1)

Prime

Multiple Choice

Factor completely. 3𝑥³y² – 2𝑥²y + 5xy

y(3x³y-2x²+5)

xy(3x²y-2x+5)

x(3²y²-2xy+5y)

2xy(3/2x²y-x+5/2)

Solving quadratic equations using the zero-factor property

When given an equation set it equal to zero.
You may have to move terms to the left of the equation to set it to zero.
Factor completely. Since they are set to zero, each factored term can be set to zero to solve for the variable.
Your graphic organizer has split this process into two branches. Those that are already factored and only need to be solve which is on the left of the organizer and those that need to be set to zero which is on the right. Fill in the blanks on the organizer by looking at your book starting on page 463.

Ex:

y^2=\ 9y\ -\ 8

Put this back into standard form and set it equal to zero
$y^2\ -\ 9y\ +\ 8\ =0$
Now factor the equation: ( y -8)(y-1)
To solve for y: y-8=0 and y - 1 = 0
Solutions for y = {8, 1}
This is how to use the zero-factor property.
Remember to always get the equation in standard form then factor.

You will need to use all of your factoring knowledge to solve these equations.
There are cases where there is a double solution when solving the equation.
Ex:
$x^2\ +\ 16x\ +\ 64\ =\ 0$
When factored you will have $\left(x+8\right)^2$
The solution for x will just be {-8} which is a double solution.

Multiple Choice

Solve:

$x^2+6x+5=-3$

$x=1$ or $5$

$x=-5$ or $-1$

$x=-4$ or $-2$

$x=2$ or $4$

Multiple Choice

Solve:

$x^2-5x-14=0$

$x=2$ or $7$

$x=-7$ or $-2$

$x=-2$ or $7$

$x=-7$ or $2$

Multiple Choice

Solve:

$z\left(z-15\right)=0$

$z=0$ or $15$

$z=-15$ or $0$

$z=-15$

$z=15$

Multiple Choice

Solve:

$\left(x-5\right)\left(x-2\right)=0$

$x=-5$ or $2$

$x=2$ or $5$

$x=-2$ or $5$

$x=-5$ or $-2$

Multiple Choice

Solve:

$\left(x+4\right)\left(x+7\right)=0$

$x=4$ or $7$

$x=-4$ or $7$

$x=-7$ or $4$

$x=-7$ or $-4$

Multiple Choice

What is the first step in solving
$\left(x+2\right)\left(x-3\right)=0$ ?

Plug in zero for x

Solve for x

FOIL

Set each binomial factor equal to zero

Multiple Choice

The solutions of the quadratic equation
x²-5x+6=0 are

x=1 or x=-6

x=3 or x=2

x=-3 or x= 2

x=-1 or x=6

We have come to the end of chapter 6. The next slides will be review for the test on Thursday. You may take notes on the slides. Remember the test will be a zoom class with open camera and sound.

Multiple Choice

Factor completely.
5x² - 13x + 6

(x - 3)(5x - 2)

(5x - 3)(x - 2)

5(x - 3)(x + 2)

Not factorable

Multiple Choice

Factor:
4h² - 17h + 4

(2h - 2)²

4(h + 4)(h + 1)

(2h - 2)(2h + 2)

(h - 4)(4h - 1)

Multiple Choice

Factor
r² -16r + 60

(r - 15)(r + 4)

(r - 6)(r + 10)

(r - 6)(r - 10)

(r +30)(r - 2)

Multiple Choice

16n² - 25

(16n - 25)(16n + 25)

(8n - 5)(8n + 5)

(4n + 5)(4n - 5)

Not Factorable

Multiple Choice

x²+ 81

not factorable

( x - 9 ) ( x + 9 )

( x + 9 ) ( x + 9 )

( x - 9 ) ( x - 9 )

Multiple Choice

Factor:
x² - 9

( x -9) (x - 9)

(x + 3) (x + 3)

(x + 3) (x - 3)

( x + 9) (x-9)

Multiple Choice

Factor by grouping:
5n³-10n²+3n-6

(5n²+3)(n-2)

(5n²-3)(n-2)

(5n³+3)(n-2)

10n(n²-6)

Multiple Choice

What is the GCF of the trinomial?

6x⁵y⁴+24x⁶y²-36x³y³

6x³y²

6xy

6x⁶y⁴

6x³y

Multiple Choice

Factor by GCF:
4b⁵+ 4b³+ 16b²

4b³(b⁴+b²+4b)

4b³(b⁴+4b+5)

4(b⁵+b³+4b²)

4b²(b³+b+4)

Multiple Choice

Factor:
80x⁵-70x²-60x⁷

2x²(80x³-70-60x⁶)

10x²(80x⁴-70x-60x⁶)

10x²(8x³-7-6x⁶)

3x²(80x³-40-60x⁵)

Multiple Choice

Factor:
56a³-8a

8a²(56a³-8a)

8a(7a²-1)

8a(7a³-a)

8a²(35a²-a)

Multiple Choice

Factor completely:
x³+7x²-2x-14

(x²-2)(x-7)

(x²+2)(x-7)

(x²+2)(x+7)

(x²-2)(x+7)

Multiple Choice

Factor Completely:

2n³ + 7n² - 2n - 7

cannot be factored

(n - 1)²(2n + 7)

(n² - 1)(2n + 7)

(n + 1)(n - 1)(2n + 7)

Multiple Choice

Factor completely:
16x² - 20xy

4 (4x² - 5xy)

4x (4x - 5y)

4x² (4 - 5y)

2(duck)

Multiple Choice

Factor Completely:

x⁴- 9x²

x²(x + 3)(x - 3)

(x²+ 3x)(x²- 3x)

(x²+ 9)(x - 3)(x + 3)

x²(x²- 9)

Multiple Choice

Factor Completely:

x² - 17x + 72

(x - 9)(x + 8)

(x - 9)(x - 8)

x(x - 3)

(x + 9)(x + 8)

Multiple Choice

x³- 5x²+ 5x - 25

(x²- 5)(x - 5)

(x²+ 5)(x - 5)

(x²+ 5)(x + 5)

(x²- 1)(x - 25)

Multiple Choice

Factor by grouping

x³ + 4x² + 4x + 16

(x + 4)(x + 2)

(x - 4)(x² - 4)

(x + 4)(x² + 4)

(x - 4)(x - 2)

Multiple Choice

Solve the equation by factoring.

x = {-5,-3}

x = {5,-3}

x = {-5,3}

x = {5,3}

Multiple Choice

Solve the equation by factoring.

x = {1}

x = {-1}

x = { }

no solution

Multiple Choice

Solve the equation by factoring.

x = {3,8}

x = {-3,8}

x = {3,-8}

x = {-3,-8}

Multiple Choice

Solve
3x² - 192 = 0

x = -16, 16

x has no real solutions

x = -8, 8

x = -64, 64

Multiple Choice

Which of the following equations matches standard form of a quadratic?

ax + by = c

y = a(x - h)² + k

y = ax² + bx + c

y = (x + a)(x + b) + c

This completes today's lesson.

See you on Thursday in the zoom room.

Quadratic equation: zero principle

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