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Algebra 2 Solving Quadratic Equations by Factoring

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Section 5: Quadratic Functions and Inequalities: Lecture 3 | 19:53 min

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	2 answers Last reply by: DJ Sai Mon Sep 3, 2018 12:12 PM Post by Jerry Xu on July 31, 2018 All examples say example 1... also 3rd example (3x^2 = 27) would be easier i you just divide both sides by 3 to get x^2 = 9
	0 answers Post by Mia Moore on November 2, 2016 omg I'm ur biggest fan... like totes fangirling right now #beliebe
	0 answers Post by Mia Moore on November 2, 2016 hello everyone
	1 answer Last reply by: Dr Carleen Eaton Thu Jan 7, 2016 6:28 PM Post by Manuel Ramirez on November 20, 2015 5xsquare-12=11x what happen if u got the x on the other side
	2 answers Last reply by: Mia Moore Wed Nov 2, 2016 6:21 PM Post by Manuel Ramirez on November 20, 2015 3xsquare-9x=0... having trouble

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Solving Quadratic Equations by Factoring

Use factoring techniques to factor a quadratic expression. For trinomials, use the guess and check technique. Factor a greatest common factor first. Look for a perfect square trinomial and the difference of two squares.
Use the zero product property to set each factor equal to 0 and find the solutions.

Solving Quadratic Equations by Factoring

Factor x² − 9

Notice that this problem can be done using Difference of Squares
x² − y² = (x + y)(x − y)

x² − 9 = x² − 3² = (x + 3)(x − 3)

Factor k² + 8k + 7

Remember that when our quadratic expression is in standard form such that a = 1, we are
always looking for two numbers n and m such that:
n*m = 7
n + m = 8
The only numbers that work are 1 and 7 because
1*7 = 7 and 1 + 7 = 8

Final solution is then (k + 1)(k + 7)

Factor 4x³ − 144x

Since this problem is not a trinomial, the best way to solve it is to find the GCF.
The GCF is 4x. Which leaves you with:
4x(x² − 36)
Notice how the factored term can be factored further by differences of squares.

4x(x² − 36) = 4x(x² − 6²) = 4x(x + 6)(x − 6)

Solve 3n⁴ + 12n³ − 36n² = 0 by factoring

Find the GCF
The GCF is 3n²
That leaves 3n²(n² + 4n − 12).
Now, given that trinomial is in standard form and a = 1, find the two numbers n and m such that
n*m = − 12
n + m = 4
the only two numbers that work in this case are 6 and − 2 because
6* − 2 = − 12
6 - 2 = 4
That leaves you with 3n²(n + 6)(n − 2) = 0
Using the Zero Product Property, that leaves you with three equations.
3n² = 0, n + 6 = 0, n − 2 = 0
Solve

n = 0, n = − 6, n = 2

Solve 2r³ − 4r² = 0 by factoring

Find the GCF
The GCF is 2r²
That leaves you with 2r²(r − 2) = 0
Use the zero product property to solve for r.
2r² = 0, r − 2 = 0
Solve

r = 0, r = 2

Solve 9x³ + 3x² − 30x = 0 by factoring.

Find the GCF
The GCFis 3x
3x(3x² + x − 10) = 0
Notice that the resulting trinomial, although in standard form, a is no longer equal to 1.
That requires an additional step we'll call Dividing by a, a in this case equals to 3.
Look for a number n and m such that
n*m = a*c = 3* − 10 = − 30
n + m = b = 1
The only numbers that work are − 5 and 6 because
− 5*6 = − 30
− 5 + 6 = 1 which is exactly what we need.
Now that we've found n and m, we have
3x(x − 5)(x + 6) = 0 but because a is not 1, we must divide the numbers we found by a
3x(x − 5)(x + 6) = 3x(x − [5/3])(x + [6/3])
Simplify
3x(x − [5/3])(x + [6/3]) = 3x(3x − 5)(x + 2) = 0. Notice how the 3 in the denominator went infront of the x since [5/3] cannot be reduced.
Solve using Zero Product Property.
3x = 0, 3x − 5 = 0 x + 2 = 0
Solve

x = 0, x = [5/3], x = − 2

Solve 5x³ − 18x² + 9x = 0 by factoring.

Find the GCF
The GCFis x
x(5x² − 18x + 9) = 0
Notice that the resulting trinomial, although in standard form, a is no longer equal to 1.
That requires an additional step we'll call Dividing by a, a in this case equals to 5.
Look for a number n and m such that
n*m = a*c = 5*9 = 45
n + m = b = − 18
The only numbers that work are − 3 and − 15 because
− 3* − 15 = 45
− 3 + ( − 15) = − 18 which is exactly what we need.
Now that we've found n and m, we have
x(x − 3)(x − 15) = 0 but because a is not 1, we must divide the numbers we found by a
x(x − 3)(x − 15) = x(x − [3/5])(x − [15/5]) = 0
Simplify
x(x − [3/5])(x − [15/5]) = x(5x − 3)(x − 3) = 0 Notice how the 5 in the denominator went infront of the x since [3/5] cannot be reduced.
Solve using Zero Product Property.
x = 0, 5x − 3 = 0, x − 3 = 0
Solve

x = 0, x = [3/5], x = 3

Solve 3x³ − 11x² − 20x = 0 by factoring.

Find the GCF
The GCFis x
x(3x² − 11x − 20) = 0
Notice that the resulting trinomial, although in standard form, a is no longer equal to 1.
That requires an additional step we'll call Dividing by a, a in this case equals to 3.
Look for a number n and m such that
n*m = a*c = 3* − 20 = − 60
n + m = b = − 11
The only numbers that work are 4 and − 15 because
43* − 15 = − 60
4 + ( − 15) = − 11 which is exactly what we need.
Now that we've found n and m, we have
x(x + 4)(x − 15) = 0 but because a is not 1, we must divide the numbers we found by a
x(x + 4)(x − 15) = x(x + [4/3])(x − [15/3]) = 0
Simplify
x(x + [4/3])(x − [15/3]) = x(3x + 4)(x − 5) = 0 Notice how the 3 in the denominator went infront of the x since [4/3] cannot be reduced.
Solve using Zero Product Property.
x = 0, 3x + 4 = 0, and x − 5 = 0
Solve

x = 0, x = − [4/3], and x = 5

Solve 4x² + 10x + 6 = 0 by factoring.

Find the GCF
The GCFis 2
2(2x² + 5x + 3) = 0
Notice that the resulting trinomial, although in standard form, a is no longer equal to 1.
That requires an additional step we'll call Dividing by a, a in this case equals to 2.
Look for a number n and m such that
n*m = a*c = 2*3 = 6
n + m = b = 5
The only numbers that work are 3 and 2 because
3*2 = 6
3 + 2 = 5 which is exactly what we need.
Now that we've found n and m, we have
2(x + 3)(x + 2) = 0 but because a is not 1, we must divide the numbers we found by a
2(x + 3)(x + 2) = 2(x + [3/2])(x + [2/2]) = 0
Simplify
2(x + [3/2])(x + [2/2]) = 2(2x + 3)(x + 1) = 0 Notice how the 2 in the denominator went infront of the x since [3/2] cannot be reduced.
Solve using Zero Product Property.
2 = 0, 2x + 3 = 0, and x + 1 = 0
Solve

x = − [3/2] and x = − 1

Solve 9x² − 24x + 12 = 0 by factoring.

Find the GCF
The GCFis 3
3(3x² − 8x + 4) = 0
Notice that the resulting trinomial, although in standard form, a is no longer equal to 1.
That requires an additional step we'll call Dividing by a, a in this case equals to 3.
Look for a number n and m such that
n*m = a*c = 3*4 = 12
n + m = b = − 8
The only numbers that work are − 2 and − 6 because
− 2* − 6 = 12
− 2 + ( − 6) = − 8 which is exactly what we need.
Now that we've found n and m, we have
3(x − 2)(x − 6) = 0 but because a is not 1, we must divide the numbers we found by a
3(x − 2)(x − 6) = 3(x − [2/3])(x − [6/3]) = 0
Simplify
3(x − [2/3])(x − [6/3]) = 3(3x − 2)(x − 2) = 0 Notice how the 3 in the denominator went infront of the x since [2/3] cannot be reduced.
Solve using Zero Product Property.
3 = 0, 3x − 2 = 0, and x − 2 = 0
Solve

x = [2/3] and x = 2

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Answer

Solving Quadratic Equations by Factoring

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Factoring Techniques

Zero Product Rule

Example: Zero Product

Example 1: Solve by Factoring

Example 1: Solve by Factoring

Example 1: Solve by Factoring

Example 1: Solve by Factoring

Intro 0:00
Factoring Techniques 0:15

Greatest Common Factor (GCF)
Difference of Two Squares
Perfect Square Trinomials
General Trinomials

Zero Product Rule 5:22

Example: Zero Product

Example 1: Solve by Factoring 7:46
Example 1: Solve by Factoring 9:48
Example 1: Solve by Factoring 12:34
Example 1: Solve by Factoring 15:28

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Solving Quadratic Equations by Factoring

Welcome to Educator.com.0000

In a previous lesson, we talked about solving quadratic equations by graphing.0002

And we are going to go on to discuss another technique, an algebraic technique, which is solving quadratic equations by factoring.0007

So, first I am starting out with just a brief review of the different factoring techniques.0015

And each of these is covered in-depth in the Algebra I series.0022

So, if you are unsure on any of these, make sure you review that before you go on,0025

because we are just using these techniques as a tool, now, to actually solve quadratic equations.0029

OK, first, remember the greatest common factor: the greatest common factor of two or more numbers is the product of their common prime factors.0036

So, the GCF of two or more numbers is the product of their common prime factors.0048

And, in factoring, often, when you are factoring an equation, if there is a GCF, you want to factor that out first,0067

because then you are working with something less complicated.0074

For example, if you had 4x² + 16x + 36, you will recognize that there is a GCF of 4.0077

So, I am going to factor that out and get 4 times x squared, plus 4x, plus 9.0087

Factoring out the 4 gives me x² + 4x + 9; and this is much easier to work with.0097

So, remember to factor out the GCF first, if there is one.0103

Recall the difference of two squares: you will recognize this, because it is going to be in the form a² - b².0109

So, an example would be x² - 4; and that is equal to (x + 2) (x - 2).0121

So, this is the difference of two squares; here is the plus and the minus.0134

And so, when you are factoring, if you recognize this, you can quickly factor it as the difference of two squares,0139

knowing that it fits into this formula, a² - b².0144

Perfect square trinomials: perfect square trinomials would be the result of squaring a binomial.0150

For example, if you have x² + 6x + 9, this is actually equal to (x + 3)².0164

So, if you take a binomial and square it, you will get a perfect square trinomial.0175

Take a binomial; multiply it times itself; that is what a perfect square trinomial is.0180

And so, recognizing that makes for easy factoring.0185

General trinomials are a little more complex to factor; again, you can review all of this in Algebra I--techniques for factoring.0189

And an example of a general trinomial would be x² + x - 12;0198

it doesn't fit into any of these special cases, like the difference of two squares, or perfect square trinomials.0203

So, I have to do a little more work in factoring that out.0210

Recall from earlier lessons that you need to look at the signs.0213

I have a positive sign here; I have a negative here.0216

And if I have a negative, that must be the result of multiplying a positive and a negative.0218

I then look at what the factors of 12 are; and the factors of 12 are 1 and 12, 2 and 6, and 3 and 4.0226

And one will be positive, and one will be negative; and I need to have factors that sum to this middle term, which is actually 1.0239

And I can try different combinations; and I know that -1 and 12 is not going to work; -12 and 1 certainly will not work.0249

And I go on down, and from that, you can see that, if I want them to sum to 1, I am going to need factors that are close to each other.0259

And the closest two I have here are 3 and 4.0268

So, if I make 4 positive and 3 negative, then when I add the outer term and the inner terms, I will get my middle term x.0271

And you can always check this using the FOIL method: First terms (that is x²),0286

Outer terms (-3x), Inner terms (4x), and then the Last is -12.0292

Simplifying this, I get my original back.0303

General trinomials take a little more work, and you can always check those by multiplying it back out to make sure you did it correctly.0306

So, we are making sure you have all these and know how to use them well.0313

And then, we are going to be using them today to actually solve some quadratic equations.0317

Now, once you have factored, you need to use the zero product rule to actually find the solutions.0322

And what the zero product rule says is that, for any number a and b, if ab is zero, then either a is zero or b is zero--0331

because if a equals 0, then you would get 0 times b; that would work as a solution;0339

if b is 0, then 0 times a would give you 0, and that is also a solution.0346

For example, if I was given x² - 16 = 0 and asked to solve that, I would first recognize that it is0353

in the form a² - b², and that it is therefore the difference of two squares.0361

That allows me to factor it pretty quickly into (x + 4) (x - 4).0372

So, I am factoring this, and it still equals 0.0380

Now, use the zero product rule: the zero product rule tells me that, if this is 0 or this is 0, then this entire thing will equal 0, which is what I want.0383

So, if x + 4 equals 0, this will be solved; if x - 4 equals 0, this will be solved.0397

So, I am going to set this factor equal to 0 and solve for x; I am going to set this factor equal to 0 and solve for x.0407

And that is going to give me...let's see...x = 4.0418

And if you wanted to check that, you could go back up here and say, "OK, let x equal -4."0428

So, that is -4² - 16 = 0; that is 16 - 16 = 0, and that checks out; that is a valid solution.0434

I could do the same thing for 4; x = -4, or I could say x = 4; and that is going to give me 4² - 16 = 0.0445

16 - 16 does equal 0, so there are two solutions here.0454

And I was able to find those using factoring and the zero product rule.0458

So, trying this out: first I am just asked to factor; and recall that the first thing you want to do is factor out any greatest common factor,0467

because that is going to make whatever is left much easier to work with.0478

And I see that I have a greatest common factor of 4.0482

This factors into 4 times x² - x - 6.0490

Now, all I have here is a general trinomial, so I want to think about what I have.0496

And I want to make sure that I bring my 4 along with me, because that is part of the solution.0502

I have a negative sign here; and the only way you are going to end up with that is if one of these is positive and one is negative.0510

Now, I am going to think about what my factors of 6 are; I have 1 and 6, and 2 and 3.0521

I need factors that sum to a middle term of -1; and that is not a very large number, so I am going to look for factors that are close together.0529

I am going to try these first: now one is negative, and the other is positive.0536

Since this is negative, I am going to look for making the larger number negative; so let me try if I have -3 plus 2.0542

That equals -1, and that gives me the coefficient of that middle term; so this is what I have.0549

And I can always check that by using FOIL to go back and multiply these out;0566

and then I would have to multiply the 4 back into it.0570

But this is factored; so I first factored out the GCF, and then I saw that I can take it farther, because this is not factored out all the way.0574

It is a general trinomial that factors into (x + 2) (x - 3), times that greatest common factor, 4.0581

OK, now we are asked to actually solve; and this is 3x² = 27.0589

Before you can solve a quadratic equation, you need to put it in standard form.0597

Recall that standard form is ax² + bx + c = 0.0605

So, I have 3x² = 27; to put this in standard form, I am going to subtract 27 from both sides.0613

And I have an ax² term; b must be 0, because there is no x term; and then I have a c of -27.0625

So, I am going to solve by factoring; I have this in standard form--now I am going to factor out the GCF.0634

And the GCF is 3; I am going to pull that out.0643

Now, I am looking at this, and I see that what I have is something in the form a² - b², which is the difference of two squares.0653

Always make sure you bring this GCF down--don't leave it behind.0665

(x - 3) (x + 3) = 0; so now, I am going to use the zero product property (or zero product rule) to find my solutions.0669

And the zero product rule says that if a times b equals 0, then a equals 0 or b equals 0, or they both have to be 0.0687

So, first I am going to have 3 times (x - 3) equals 0; and if I divide 0/3, I am just going to get x - 3 = 0, or x = 3.0699

So, that is one solution; the other solution is going to be x + 3 = 0.0717

Using the zero product rule, that tells me that x equals -3; so my two solutions are that x equals 3 and x equals -3.0722

These are my two solutions for this quadratic equation.0733

And I found that just by factoring: x = 3 and x = -3; and I made my factoring a lot easier by first pulling out the GCF.0743

Again, solve by factoring: and we have another situation where it is not in standard form.0756

So, I am going to put it in standard form, which is ax² + bx + c = 0.0760

All right, that is going to give me 4x² - 24x + 36 = 0, just subtracting 24x from both sides.0772

This is another situation where I have a greatest common factor; so I have a GCF equal to 4--factor that out.0784

Pull that out: that is 4, times x² - 6x + 9, equals 0.0792

Again, I have something much easier to work with since I have pulled that out.0800

So, figuring out how to work with this, I am going to go ahead and factor this out, because it is not all the way factored.0805

And here, I have (x - 3) times (x -3); and it is actually a perfect square trinomial; this is really just (x - 3)².0817

OK, so this is a perfect square trinomial, because you have x², and then (just check this using FOIL)0838

I would have x² - 3x - 3x + 9; so I know I did that correctly.0850

Now, I am going to use the zero product rule, which tells me that, if one of these is equal to 0, then the entire thing is equal to 0.0862

So, I can use that to find the solution.0879

Now, you can just go ahead and divide both sides by 4, in which case this will move over to here, and that just stays 0.0881

Really, I just need to work with this and this: x - 3 = 0, and this is the same thing: x - 3 = 0.0889

So, actually, when I figure this out, I just get the same thing for both; and it is x = 3.0899

So, I only have one solution, or one real root, in this case; so, x equals 3.0906

And again, put it in standard form; factor out the greatest common factor; complete your factoring;0912

and then use the zero product rule to determine that there is one real solution, and that is that x equals 3.0919

OK, again, solve by factoring; this is already in standard form; however, I have to get rid of this greatest common factor, which is 2.0929

So, I am pulling that out to get 2x² + 5x - 12.0938

When the leading coefficient is not 1, the factoring is a bit more difficult.0945

So, let's move over here and work on this.0950

When the leading coefficient is 2, I am going to have something in this form.0953

And I have a negative here, so I also know that one of these is going to be a positive, and one is going to be a negative.0959

But I don't actually know which one yet; but I know I am going to have +/-, or I am going to have this.0965

OK, let's look at some factors of 12: factors of 12 would be 1 and 12, 2 and 6, and 3 and 4.0974

Now, 5 is not that large of a number, especially when we think about the fact that we are going to have to be also multiplying by 2.0985

So, if I go and use something like 6, and it ends up getting multiplied by 2, it is going to be very large; the difference is going to be great.0992

I want factors that are smaller, since the difference between those, even with this 2x thrown in, is only going to be 5x.0999

So, I am going to start with these, because they don't have a large difference between them.1008

So, I am going to start out just trying (2x + 3) (x - 4) and seeing what I get.1012

I am not worried about the first term; I am worried about the outer terms added to the inner terms.1022

And see if I get the correct middle term.1029

I am looking for the middle term equal to 5x.1031

This is going to give me 2x times -4; that is -8x, plus 3x; that is 5x.1036

I have the right idea, but I actually have the wrong signs here.1049

So, I am going to try reversing the signs, because (this is -5x) I want this to be 5x, because here I have -8x + 3x is -5x.1053

So, the same idea: let's try different signs, though.1063

This time, I am making this negative and this positive; this is going to give me 2x(4); that is going to give me +8x; -3x is 5x.1070

So, this is correct; I got the correct middle term, so this is the correct factoring.1083

And this can be a lot of work to factor these; so it is important to go logically--1088

for example, seeing that I don't have a very large term here, especially when I am dealing with the 2x also1093

(it is going to amplify things), to look for factors that are not very far apart.1099

OK, so now, I am back here, and I am solving by factoring.1103

This is going to give me (2x - 3) (x + 4) = 0.1108

Dividing both sides by 2, this 2 is just going to drop out.1117

So, when I use the zero product property, I am going to get 2x - 3 = 0, and I am also going to get x + 4 = 0.1123

So, I just need to go ahead and solve those to get 2x = 3, or x = 3/2.1133

Here, I just have x + 4 = 0, and that is simple: it is x = -4.1151

I have two solutions: x = 3/2 and x = -4.1159

I solved this by pulling out the greatest common factor, factoring that out, then factoring this trinomial into this,1170

and using the zero product rule to give me 3/2 for a solution from here, and x = -4 for a solution from here.1177

Thanks for visiting Educator.com, and I will see you next lesson!1189

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Dr. Carleen Eaton

Solving Quadratic Equations by Factoring

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro