\"Know

2x+5=9

1

?

Subtract

5

from both sides.

Why did we take this step?

Because we have

2x+5

on the left side, and we want only

x

. Using Reverse PEMDAS, we ask the questions below in order.

Any

addition / subtraction

outside parentheses?

Yes, addition.

Any

multiplication / division

outside parentheses? --

Any

exponents

? --

Any

parentheses

? --

Therefore, we

subtract

to undo the addition.

1

To remove

5

on the left side, subtract

5

from both sides.

2x+5-5=9-5

2

Cancel

5

.

2x=9-5

2x=9-5

2

Simplify

9-5

to

4

.

2x=4

3

?

Divide both sides by

2

.

Why did we take this step?

Because we have

2x

on the left side, and we want only

x

.

Therefore, we

divide

to undo the multiplication.

1

To remove

2

on the left side, divide both sides by

2

.

\frac{2x}{2}=\frac{4}{2}

2

Cancel

2

.

x=\frac{4}{2}

x=\frac{4}{2}

4

Simplify

\frac{4}{2}

to

2

.

x=2

Done

3(3-2x)=5(7-5x)

1

?

Expand.

Why did we take this step?

Because by expanding, we

distribute the terms and remove the parentheses

, which usually allows us to simplify the expression further.

1

Let's expand the left side of the equation.

3(3-2x)

2

?

Expand by distributing terms.

Why did we take this step?

Because by using the distributive property, we expand the expression to

terms that are usually easier to simplify

.

1

Let's expand this expression.

3(3-2x)

2

Use this property:

k(a+b)=ka+kb

.

3\times 3+3\times -2x

3\times 3+3\times -2x

3

Simplify.

9-6x

4

Let's expand the right side of the equation.

5(7-5x)

5

?

Expand by distributing terms.

Why did we take this step?

Because by using the distributive property, we expand the expression to

terms that are usually easier to simplify

.

1

Let's expand this expression.

5(7-5x)

2

Use this property:

k(a+b)=ka+kb

.

5\times 7+5\times -5x

5\times 7+5\times -5x

6

Simplify.

35-25x

7

Putting both sides together.

9-6x=35-25x

9-6x=35-25x

2

?

Subtract

9

from both sides.

Why did we take this step?

Because this helps us cancel

9

. Since our goal is to solve for

x

,

canceling any term that is not

x

is helpful

.

1

To remove

9

on the left side, subtract

9

from both sides.

9-6x-9=35-25x-9

2

Cancel

9

.

-6x=35-25x-9

-6x=35-25x-9

3

Simplify

35-25x-9

to

-25x+26

.

1

Collect like terms.

1

Like terms must have the same variables and powers. Some examples:

Terms	Like Terms?	Reason
$2x$ and $3y$	No	Variables are different.
$2x$ and ${3x}^{2}$	No	Variables match, but powers don't.
$2x$ and $3x$	Yes	Both variables and powers match.
$2$ and $3$	Yes	Both have no variables.

2

Recognize that

35 \text{ and } {-}9

are like terms.

Like Terms =

35,-9

(35-9)-25x

2

Simplify.

26-25x

-6x=-25x+26

4

?

Add

25x

to both sides.

Why did we take this step?

Because in the previous step,

x

is on both sides of the equation. Since our goal is to solve for

x

,

we need it on one side only

.

1

To remove

-25x

on the right side, add

25x

to both sides.

-6x+25x=-25x+26+25x

2

Cancel

25x

.

-6x+25x=26

-6x+25x=26

5

Simplify

-6x+25x

to

19x

.

19x=26

6

?

Divide both sides by

19

.

Why did we take this step?

Because we have

19x

on the left side, and we want only

x

.

Therefore, we

divide

to undo the multiplication.

1

To remove

19

on the left side, divide both sides by

19

.

\frac{19x}{19}=\frac{26}{19}

2

Cancel

19

.

x=\frac{26}{19}

x=\frac{26}{19}

Done

Decimal Form: 1.368421

6x=12

1

?

Divide both sides by

6

.

Why did we take this step?

Because we have

6x

on the left side, and we want only

x

.

Therefore, we

divide

to undo the multiplication.

1

To remove

6

on the left side, divide both sides by

6

.

\frac{6x}{6}=\frac{12}{6}

2

Cancel

6

.

x=\frac{12}{6}

x=\frac{12}{6}

2

Simplify

\frac{12}{6}

to

2

.

x=2

Done

\sqrt{x+4}=x+5

1

Square both sides.

1

Let's square the left side of the equation.

{\sqrt{x+4}}^{2}

2

Use this rule:

{\sqrt{x}}^{2}=x

.

1

The square cancels the square root. For example:

{\sqrt{x}}^{2}={({x}^{\frac{1}{2}})}^{2}={x}^{\frac{1}{2}\times 2}={x}^{1}=x

2

Therefore, this simplifies to:

x+4

x+4

3

Let's square the right side of the equation.

{(x+5)}^{2}

4

Use Square of Sum:

{(a+b)}^{2}={a}^{2}+2ab+{b}^{2}

.

{x}^{2}+2x\times 5+{5}^{2}

5

Simplify

{5}^{2}

to

25

.

1

Expand

{5}^{2}

using multiplication.

5\times 5

2

Simplify.

25

{x}^{2}+2x\times 5+25

6

Simplify

2x\times 5

to

10x

.

{x}^{2}+10x+25

7

Putting both sides together.

x+4={x}^{2}+10x+25

x+4={x}^{2}+10x+25

2

Move all terms to one side.

1

Add

-{x}^{2}-10x-25

to both sides.

x+4-{x}^{2}-10x-25={x}^{2}+10x+25-{x}^{2}-10x-25

2

Cancel terms on the right side.

x+4-{x}^{2}-10x-25=0

x+4-{x}^{2}-10x-25=0

3

Simplify

x+4-{x}^{2}-10x-25

to

-9x-21-{x}^{2}

.

1

Collect like terms.

1

Like terms must have the same variables and powers. Some examples:

Terms	Like Terms?	Reason
$2x$ and $3y$	No	Variables are different.
$2x$ and ${3x}^{2}$	No	Variables match, but powers don't.
$2x$ and $3x$	Yes	Both variables and powers match.
$2$ and $3$	Yes	Both have no variables.

2

Recognize that

x \text{ and } {-}10x

are like terms.

Like Terms =

x,-10x

3

Recognize that

4 \text{ and } {-}25

are like terms.

Like Terms =

4,-25

(x-10x)+(4-25)-{x}^{2}

2

Simplify.

-9x-21-{x}^{2}

-9x-21-{x}^{2}=0

4

Use the Quadratic Formula.

1

In general, given

a{x}^{2}+bx+c=0

, there exists two solutions where:

x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}

2

In this case,

a=-1

,

b=-9

and

c=-21

.

{x}^{}=\frac{9+\sqrt{{(-9)}^{2}-4\times 21}}{-2},\frac{9-\sqrt{{(-9)}^{2}-4\times 21}}{-2}

3

Simplify.

x=\frac{9+\sqrt{3}\imath }{-2},\frac{9-\sqrt{3}\imath }{-2}

x=\frac{9+\sqrt{3}\imath }{-2},\frac{9-\sqrt{3}\imath }{-2}

5

Simplify solutions.

x=-\frac{9+\sqrt{3}\imath }{2},-\frac{9-\sqrt{3}\imath }{2}

Done

{x}^{4}+9{x}^{3}+9{x}^{2}-85x-150

1

Factor

{x}^{4}+9{x}^{3}+9{x}^{2}-85x-150

using Polynomial Division.

1

Factor the following.

{x}^{4}+9{x}^{3}+9{x}^{2}-85x-150

2

First, find all factors of the constant term 150.

1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150

3

Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

{1}^{4}+9\times {1}^{3}+9\times {1}^{2}-85\times 1-150 = -216

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

{(-1)}^{4}+9{(-1)}^{3}+9{(-1)}^{2}-85\times -1-150 = -64

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

{2}^{4}+9\times {2}^{3}+9\times {2}^{2}-85\times 2-150 = -196

Substitute -2 into x. Since the result is 0, x+2 is a factor..

{(-2)}^{4}+9{(-2)}^{3}+9{(-2)}^{2}-85\times -2-150 = 0

x+2

4

Polynomial Division: Divide

{x}^{4}+9{x}^{3}+9{x}^{2}-85x-150

by

x+2

.

	$x^3$	$7x^2$	$-5x$	$-75$
$x+2$	$x^4$	$9x^3$	$9x^2$	$-85x$	$-150$
	$x^4$	$2x^3$
		$7x^3$	$9x^2$	$-85x$	$-150$
		$7x^3$	$14x^2$
			$-5x^2$	$-85x$	$-150$
			$-5x^2$	$-10x$
				$-75x$	$-150$
				$-75x$	$-150$

5

Rewrite the expression using the above.

{x}^{3}+7{x}^{2}-5x-75

({x}^{3}+7{x}^{2}-5x-75)(x+2)

2

Factor

{x}^{3}+7{x}^{2}-5x-75

using Polynomial Division.

1

Factor the following.

{x}^{3}+7{x}^{2}-5x-75

2

First, find all factors of the constant term 75.

1, 3, 5, 15, 25, 75

3

Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

{1}^{3}+7\times {1}^{2}-5\times 1-75 = -72

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

{(-1)}^{3}+7{(-1)}^{2}-5\times -1-75 = -64

Substitute 3 into x. Since the result is 0, x-3 is a factor..

{3}^{3}+7\times {3}^{2}-5\times 3-75 = 0

x-3

4

Polynomial Division: Divide

{x}^{3}+7{x}^{2}-5x-75

by

x-3

.

	$x^2$	$10x$	$25$
$x-3$	$x^3$	$7x^2$	$-5x$	$-75$
	$x^3$	$-3x^2$
		$10x^2$	$-5x$	$-75$
		$10x^2$	$-30x$
			$25x$	$-75$
			$25x$	$-75$

5

Rewrite the expression using the above.

{x}^{2}+10x+25

({x}^{2}+10x+25)(x-3)(x+2)

3

?

Rewrite

{x}^{2}+10x+25

in the form

{a}^{2}+2ab+{b}^{2}

, where

a=x

and

b=5

.

Why did we take this step?

Because

{a}^{2}+2ab+{b}^{2}

is a common expression with a known factored form. This allows us to factor the expression in the next step.

1

Recognize that

{x}^{2}

is a perfect square.

{x}^{2}

=

x \times x

= Perfect Square

2

Recognize that

25

is a perfect square.

25

=

5 \times 5

= Perfect Square

({x}^{2}+2(x)(5)+{5}^{2})(x-3)(x+2)

4

Use Square of Sum:

{(a+b)}^{2}={a}^{2}+2ab+{b}^{2}

.

{(x+5)}^{2}(x-3)(x+2)

Done

{w}^{2}+8w-65

1

Ask: Which two numbers add up to

8

and multiply to

-65

?

1

List the factors of

-65

and their sums.

Factors of -65	Sum (must equal 8)
$1\times -65$	$1-65=-64$
$5\times -13$	$5-13=-8$
$-1\times 65$	$-1+65=64$
$-5\times 13$	$-5+13=8$ Found it!

2

Therefore, the two numbers are:

-5

and

13

-5

and

13

2

Rewrite the expression using the above.

(w-5)(w+13)

Done