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Solving Equations by Factoring

Page history last edited by Math in a Box - Susan Johnsey gm 5 years, 6 months ago

Solving Equations by Factoring

I have created a free course for learning to factor.  There are at least 5 major types to learn.

You will see factoring in many courses after Algebra 1 so be sure you can factor polynomials.

A link to the course is on the WIKI tab with the listing of my other courses.  CHECK it out.

 

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.

 

 

THESE are the EASY ONES!  

 

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

 

  t2  - 4t  -12 = 0

   2(3m -2)(3m+2) = 0

   (t  -6)(t +2) = 0

  3m -2 =0  or 3m+ 2 =0

   t  - 6 =0  or t+2 = 0

3m = 2   or 3m = -2

   t= 6  or t= -2

two solutions

m = 2/3  or m = - 2/3

two solutions

 

 I have created two videos for  factoring.  

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

You will see the two videos listed  for FACTORING QUADRATICS .

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