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3:6 Solving EquationsbyFactoring

Page history last edited by Math in a Box - Susan Johnsey gm 3 years, 11 months ago

LESSON 3.6  Solving Equations by Factoring

 

Do you remember multiplying with ZERO: 5(0)=0 or 0(-9)=0 or 0(0)=0. 

 

That's easy!!     

It  is simply arithmetic,  but it is important in our next use of factoring.  I hope you recall factoring.   

We have had several lessons on that topic.  

 

Factoring is writing an algebra expression as MULTIPLICATION.

 

When we multiply with zero our answer is always zero. 

Do you recall 0 times 0=0   or that 5 times 0 = 0 ?

No other number has that quality.

 

If (x-5) times (x+9)=0   then x-5=0  or x+9=0   or both equal 0!

  Right?

 

Thus (x-5)(x+9)=ZERO means that    x- 5=0 or that the x+9=0.

 

NOW  Solve the above two algebra equations:

 

x-5=0 

so x= +5

x+9=0

so x= -9.

 

x=5 and x=-9 are the solutions for (x-5)(x+9)=0 . 

 

Check them by letting x=5:  

(x-5)(x+9)=

(5-5)(5+9)=

    (0)(14)=0 .

  TRUE!

 

Now you check the -9.  

It too will make our equation true.

Solve (x-5)(x+9)=0

 means x= 5 or x=-9 . 

These are called solutions.

 

 

 

CONSIDER  x2+4x - 45 =0.  

Do you know where I got this quadratic trinomial?   

 

I hope so.  If not then multiply (x-5)(x+9).   YOU must know how; some call it FOIL  others call it the distributive property.

(x-5) times (x+9) =  x2+9x-5x-45 simplifies to x2+4x -45.     

 

If I take the trinomial x2+4x-45 and substitute -9 for each x then

           I will have (-9)2+4(-9)-45

                             =81-36-45 which is 0.

 

If I take the trinomial x2+4x - 45 and substitute a 5 for each x then                          

          I will have  (5)2+4(5)-45

                              =25+20-45 which is 0.

So you see if we have (x-5)(x+9) or x2+4x-45 equals ZERO

 

then x=-9           or         x=5.       

 

These are the only numbers that will make (x-5)(x+9) or x2+4x-45 equal ZERO.

 

Solve x2+4x-45=0.  The values of x that make this true were 5 and -9.  

We call those numbers the solution to x2+4x-45. 

 

 Let us solve some polynomial equations.   Here are the steps that we used above.
We will  :

    1.  rewrite equation = 0     (must equal 0),

 

   2.  factor the expression   (written as multiplies),

 

 

   3.  solve each of the  simple equations for x.

            ( WE will have several linear equations to solve.)

 

 

EXAMPLE:   x2 -17x +72 = 0 

becomes (x- 9)(x -8) =0. 

WE Factored.                          

we have x-9 =0 and x-8 =0. Now solve for x.

Solutions are x=9 and x=8.

 

You try these.   Write these in your notebook so that you have good examples to follow.

t2  - 12 = 4t  

We must begin by rewriting.

18m2 - 8 = 0 Note the common factor of 2.       

     2(9m2 - 4=0

 

t4- 13t2 + 36 = 0                  

this is not a quadratic; 

it is a quartic.

  t2  - 4t  -12 = 0

   2(3m -2)(3m+2) = 0

 

 The 2 cannot=0, but the 3m-2 and the 3m+2 can =0.

Use Algebra 1 to solve for m.

  (t2-9) (t2-4)  =0

   (t  -6)(t +2) = 0

  3m -2 =0  or 3m+ 2 =0

  t2-9   = 0  or t2-4  = 0  

   t  - 6 =0  or t+2 = 0

3m = 2   or 3m = -2

   (t - 3)(t+3)=0    or    (t - 2)(t+2)=0

   t= 6  or t= -2

m = 2/3  or m = - 2/3

t=3  or    t= -3    or    t=2  or t= -2

4 solutions! 

The highest exponent will tell you!!

 

 I have created two videos for working "word problems" with factoring.  

Please visit my web site www.mathinabox.com and click the tab "Lessons / Videos".  

You will see the two videos listed  at the bottom of the middle list.  

Click here to go to http://www.mathinabox.com/LessonsVideos.html 

 

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EXAMPLE with WORD problem:

The length of a rectangle is 8 more than its width

Its area is 240 ft2.  

Find the equation to express its area in terms of W,  then solve to find its width and length. 

 

W= width, W+8=length          See above that it state the length is 8 more than the width.


Area = length times width

   240  = (W+8) times W      

       

This becomes: 240 = W2 + 8w

 

This is a quadratic and we can find the solutions by

rewriting=0,     

factoring and then solving.

 

0 = W2 + 8W -240  Be sure you know why the 240 is negative now.

0=(W +20)(W - 12)

0=W+20 or W-12=0

W=-20 or W=12    

Since the width of a rectangle can not be negative the width must be 12 feet.  

Length is 8 more than width=20 feet.


 

Solve this.   Please email the work and answers to sojohnsey@gmail.com .

1.  Five times an integer plus three times the square of the integer is twenty-two. Find the integer.   

 

There are many problems that can be set  up and solved using factoring.  

The world of science, economics, engineering, architecture, electricity are only a few to consider.

 

Remember multiplying with 0 yields 0 . 

If 2x (x+4) = 0 then one of the factors has to be zero.  

To multiply and get zero for an answer always means one or both factors we multiplied with was equal to zero.   

Therefore either 2x =0 or x+4 = 0 or both are equal to zero. 

Solve these equations:

2x =0        or      x+4 = 0

 x = 0        or      x = -4.


 

MORE EXAMPLES

 

Find the solutions.

Example :  Easy one, since it is already equal to zero and factored .      

2x(x+4)=0    so      2x=0  or x+4 =0           

x=0  or x=- 4 are the solutions for 2x(x+4)=0 .  

Check them by substituting -4 for the x in the original problem. 

2x(x+4)       

2(-4)[-4+4] =-8[0]=0  

Yes that is what we wanted, zero.                               

You check the x=0.


 

Example: 2x3- 9x2=0 

or I could write it as 2x3=9x2

Either way I MUST follow these steps: 

1.  rewrite it so that it =zero, 

2.  then factor,

3.  then solve each EQUATION.

 


 

EXAMPLE:     x4- 9x2=0  There is a common factor  of x2             

(I hope you recall  always do COMMON FACTOING first IF THERE IS ANY)

So this can be:

x2 (x2-9) =0    NOW FACTOR THE x2 - 9        

x2 (x-3)(x+3) =0      

 

x2= 0  or   x-3=0  or        x+3 =0      

There are 3 equations to solve.       

        

The x2 = 0 means x = 0.                

The x - 3 = 0 means x= 3 .

The  x+3 = 0 means x = - 3.

 

Our solutions are x=0 or x=3 or x= -3.  

Can you check these by substitution into the original problem?  Let x be a 0 in the equation.   Does it make the equation true?  

Let x be a 3  or -3 in the equation.   Does it make the equation true?   It should.


 

Solve by factoring. 

 Please email these to me at sojohnsey@gmail.com   Write Factoring class 3.6 in the subject line of the email.

Send your factored form and your solutions.  

If you can not factor them then send what you have of each problem. 

You should rewrite it first(if needed) then factor and then solve for the x's.

Although I have this in 3 columns please send your work in only one long column.

1. 

Completely FACTOR:

x4 - 27x

 

2.  SOLVE:   t2 - 16 = 0 

3.  SOLVE:  x2 + 8x + 16 = 0

4.  SOLVE:  p2 + 3p = 4 

5.  SOLVE:  v2 + 12 = 8v      

6. SOLVE:   3x2 - 7x - 6 = 0

7.  SOLVE:  3x2 -10x -8= 0   

8. SOLVE:   x - 9=x( x - 5)        

HINT:   Multiply these out and then rewrite =0 so that they look like #3 or #7 above.

9.  SOLVE:  t + 3 = t( t + 3)  

HINT:   Multiply these out and then rewrite =0  so that they look like #3 or #7 above.

 

10.  SOLVE:   u-3u+9=( u - 1)(2 u + 3)   

HINT:   Multiply these out and then rewrite so that they look like #3 or #7 above.

 

Do Assignment  3.6 in SECTION 1.

 

I hope you feel you have benefited from my lessons.   Let me know your questions about other topics too.

 

You can find me with Google.  

Susan Johnsey      Math in a Box        www.mathinabox.com 

 

Please recommend my lessons to other students or teachers.

 

 

 

 

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